Heat Capacity, Specific Heat Capacity And Molar Heat Capacity

Heat capacity of a substance

The heat capacity is the quantity of heat required to raise the temperature ofa substance by 1°C (1 K).

The heat capacity is usually denoted by If of the amount of heat required to raise the temperature of a given amount of substance, then its heat capacity, \(C=\frac{δ q}{d T} \).

characteristics of the heat capacity of a substance

The higher the heat capacity of a substance, the smaller the increase in temperature of a given amount of the substance when a certain amount of heat is added to it. For example, the heat capacity of water is higher than that of copper. So more heat will be required to raise the temperature of 1 g of water than lg of copper by 1 K (or 1°C).

The heat capacity of a substance depends on its nature.

The heat capacity ofa substance depends on its amount. Thus, it is an extensive property.

The heat capacity is a path-dependent quantity. The heat required to raise the temperature of a substance by IK depends on the process by which the substance is heated. For example, the amount of heat required to raise the temperature of1 mol of N2 gas by IK depends on whether the heating is done at constant volume or constant pressure.

Unit: \(\text { cal } \cdot{ }^{\circ} \mathrm{C}^{-1}\left[\text { or, cal } \cdot \mathrm{K}^{-1}\right] \text { or, } \mathrm{J} \cdot{ }^{\circ} \mathrm{C}^{-1}\left[\text { or, } \mathrm{J} \cdot \mathrm{K}^{-1}\right] \text {. }\)

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Specific heat or specific heat capacity of a substance

The amount of heat required to raise the temperature of the unit mass ofa substance by 1°C [or IK] is called specific heat or specific heat capacity ofthe substance. It is represented by d. Specific heat capacity is an intensive property ofthe system.

⇒ \(Unit: cal \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} or, cal \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1} or, \mathrm{J} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} or, \mathrm{J} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}.\).

The specific heat capacity of water (4.]8J-g-1.°C-1) is considerably higher than that of other common substances. Thus, a large amount of heat as well as time is required to warm a given amount of water. For the same reason, hot water takes a long time to cool.

PSEB Class 11 Chemistry Chapter 6 Chemical Thermodynamics Notes

Molar heat capacity

The amount of heat required to raise the temperature of I mol of a substance by 1°C (or IK) is called the molar heat capacity of the substance. The molar heat capacity is denoted by’ Cm’ (the suffix’ m ’ refers to molar), it is an Intensive property.

⇒ \(\mathrm{cal} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} \text { or, cal } \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)

⇒ \(\text { or, } \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} \text { or, } \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \text {. }\)

Molar heated panacea substance(Cm)

=Specify Lear capacity of me substance(c) x Molar mass (M)

Heat absorbed or released by substance (q)

PSEB Class 11 Chemistry Chapter 6 Chemical Thermodynamics Notes

= mass of the substance (m)>specificheat capacity (c) x increase or decrease in temperature <Ar> thus, q= mxcxT

[where AT =final temperature- initial temperature] Using fee equation [1], we can calculate c if we know the values of, m and AT. The heat released or absorbed (q) can be calculated by using equation [1] if we know the values of c, m, and AT. An increase or decrease in temperature (AT) can be calculated by using equation[1] if we know the values of q,c, and m.

Heat Capacity Of A Substance At Constant Volume And Pressure

Comparison of the values of molar heat capacities at constant pressure and constant volume

Comparison for gaseous substances: For gaseous substances, the molar heat capacity at constant pressure is greater than die molar heat capacity at constant volume.

Addition of heat at constant volume: As the volume of the system is constant, no external work by the system is possible. So, all die heat added to the system will be used for increasing the internal energy of the system, which in turn increases the temperature of the die system.

Suppose, the addition of 8a amount of heat to 1 mole of gas causes an increase in the temperature of the gas by dT. Therefore, by definition,

⇒ \(C_{V, m}=\left(\frac{\delta q}{d T}\right)_V\)

Addition of heat at constant pressure: When heat is added to a gas at constant pressure the added heat is used up in two ways. One part of it is expended for the external work done by the system, and the other part of it goes to increase the internal energy of the system.

As a result of an increase in internal energy, the temperature of the system also increases Suppose, when 5q amount of heat is added to l mole of a gas at constant pressure, the temperature of the gas is increased by dT’.

Class 11 Chemistry Chemical Thermodynamics Notes

So, by definition \(C_{P, m}=\left(\frac{\delta q}{d T^{\prime}}\right)_P\) Therefore, if Ihe same amount of heat Is added to I mol of a given gas separately at constant volume and constant pressure, then the Increase in temperature will be small at constant pressure than at constant volume so dt’,dt and cp, m. Cv, m. As a result of this, the molar heat capacity of a gas at constant pressure(cp,m) will be greater than its molar heat capacity at constant volume (cv,m).

Comparison in the case of solids and liquids: when heat is added to a solid or liquid, there occurs no significant change in their volumes. Thus, the work involved in the process of heating a solid or liquid at constant pressure is negligible. This is why the values of Cp m and Cv m are found to be almost the same in the case of a liquid or solid.

For a gas, the ratio of the molar heat capacity at constant pressure (Cp m) to the molar heat capacity at constant volume (Cp,m) is termed as heat capacity ratio (7). Therefore, 7 =CPf m/C V, m

Some relations of Cp and Cv for an ideal gas 1) Internal energy ( U) of an ideal gas depends only on the temperature (T), neither on the pressure (P) nor the volume (V). When an ideal gas undergoes a process involving only P-V work, the heat absorbed by the gas is equal to its internal energy change. Therefore, for an ideal gas

⇒ \(C_V=\left(\frac{\delta q}{d T}\right)_V=\frac{d U}{d T} \text { or, } d \boldsymbol{d}=\boldsymbol{C}_V \boldsymbol{d} T\)

Enthalpy (H) of an ideal gas depends only on temperature (T). It does not depend either on pressure (P) or on volume (V). When an ideal gas undergoes a process involving only P-V work, the heat absorbed by the gas is equal to its enthalpy change. Therefore, for an ideal gas, \(C_P=\left(\frac{\delta q}{d T}\right)_P=\frac{d H}{d T} \text { or, } d H=C_P d T\)

For an ideal gas, the difference between the moon’s heat capacities at constant pressure and constant volume is equal to the universal gas constant (R). Thus, CP, m -CV, m = R.

For 1 mole of an ideal gas, PV = RT

Substitution RT for PV in the relation H = U+ PV gives the enthalpy for1 mol of an ideal gas, i.e., H = U+ RT Differentiating both sides, we have dH = dU + RdT.

Dividing both sides by dT gives

⇒ \(\frac{d H}{d T}=\frac{d U}{d T}+R \quad \text { or, } C_{P, m}=C_{V, m}+R\) Or, \(C_{P, m}-C_{V, m}=R\)

[For an ideal gas, dU = CydT and dH = CpdT}

The change in internal energy (A) with the change in temperature of an ideal gas: Suppose, the ‘n’ mole of an ideal gas undergoes a process in which its temperature changes from T1 to T2 and so does its internal energy from U1 to U2. Therefore, the change in internal energy ofthe gas,

⇒ \(\int_{U_1}^{U_2} d U=\int_{T_1}^{T_2} n C_{V, m} d T\)

⇒ \(\text { or, } U_2-U_1=\int_{T_1}^{T_2} n C_{V, m} d T \text { or, } \Delta U=\int_{T_1}^{T_2} n C_{V, m} d T\)

Class 11 Chemistry Chemical Thermodynamics Notes

If CV,m is considered to be independent of temperature within the temperature range to T2, then | AU = nCV/Therefore, the change in internal energy (AI7) with the temperature change can be calculated by using equation [1], If T2> T1, then AU = positive, ie., with increasing temperature, the internal energy of the system increases. If T2<Ty then A U = negative i.e., with decreasing temperature, the internal energy of the system decreases.

The change in enthalpy (AH) with the change in temperature for an ideal gas: Suppose, an ideal gas undergoes a process in which its temperature changes from T1 to T2.

As a result of which its enthalpy changes from H1 to H2. Therefore, the change in enthalpy of the gas in this process is:

⇒ \(\int_{H_1}^{H_2} d H=\int_{T_1}^{T_2} n C_{P, m} d T\)

⇒ \(\text { or, } H_2-H_1=\int_{T_1}^{T_2} n C_{P, m} d T \text { or, } \Delta H=\int_{T_1}^{T_2} n C_{P, m} d T\)

If Cp m is considered to be independent of temperature within the temperature range T1 to T2, then, \(\Delta H=n C_{P, m}\left(T_2-T_1\right)\) So, the change in enthalpy (AH) with the change in temperature can be calculated by using equation [1],

Numerical Examples

Question 1. How much 90g of water from 30°C to 100°C? [Molar heat capacity of water at constant pressure = 75.3 J.mol-1 K-1]
Answer: The specific heat of water at constant pressure,

⇒ \(c_P=\frac{C_{P, m}}{M}=\frac{75.3}{18}=4.18 \mathrm{~J} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

where, Cp m = molar heat capacity at constant pressure, M = molar mass ofthe substance Mass of water, m = 90 g and AT = (373-303)K =70K

∴ q =mx cpx AT = 90 x 4.18 x 70 =26334 J =26.33 kj

∴ 26.33 kl of heat is required to raise the temperature of 90g of water from 30 °C to 100 °C.

Question 2. How much heat will be released when the temperature of1 mol of water changes from 90 °C to 80 °C? Given: Specific heat of water 4.18 J.g-1.K-1
Answer: We know, q = mx cx AT The amount of water = lmol. Therefore, m = 18g, c = 4.18 J.g-1.K-1 and \(\Delta T=[(273+80)-(273+90)] \mathrm{K}=-10 \mathrm{~K}\)

∴ q = 18 X 4.18 X (-10) J = —752.40J

So, the amount of heat that will be liberated when the temperature of1 mol water changes from 90 °C to 80 °C is 752.40 J.

Question 3. Specific heats of an ideal gas at constant volume & constant pressure are 0.015 and 0.025 cal. g-1 .K-1 respectively. Determine the molar mass of the gas.
Answer: \(c_V=0.015 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}, c_P=0.025 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

∴ Molar heat capacity at constant volume, Cy m=M x Cy, and that at constant pressure, Cp m = M x cp[Af= molar mass] Again we know, for an ideal gas, Cp m- Cy m = R.

∴ M(Cp-Cy) = R

⇒ \(\text { or, } \quad M(0.025-0.015) \mathrm{cal}^{-1} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}=1.987 \mathrm{cal}^{-1} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)

or, Af = 198.7g-mol-1 \(\left[ R=1.987 \mathrm{cal} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\right]\)

∴ The molar mass of the gas = 198.7 g-mol-1.

Class 11 Chemistry Chemical Thermodynamics Notes

Question 3. Specific heats of an ideal gas at constant volume And constant pressure are 0.015 and 0.025 cal. g-1.K-1 respectively. Determine the molar mass of the gas.
Answer: \(c_V=0.015 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}, c_P=0.025 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot \mathrm{K}^{-1}\)

∴ Molar heat capacity at constant volume, Cy m=M x Cy and that at constant pressure, Cp m = M x cp[Af= molar mass]

Again we know, for ideal gas, \(C_{P, m}-C_{V, m}=R.\)

∴ M(Cp-Cy) = R

Or, \(M=198.7 \mathrm{~g} \cdot \mathrm{mol}^{-1}\)

Since \(R=1.987 \mathrm{cal} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)

∴ Molar mass ofthe gas = 198.7 g-mol-1.

Question 1. The standard enthalpy of formation of C7H5N3O6 is -x kj.mol-1 at 25c. write the thermochemical equation for the formation reaction of the compound.
Answer:

The constituent elements of the compound C₇H₅N₃O₆(s) whose standard states at 25C are carbon, hydrogen, oxygen and nitrogen C (graphite, s), H₂(g), O₂(g) and N₂(g) respectively. Therefore, the equation representing the formation reaction of the compound should contain C (graphite, s), H₂(g), O₂(g) and N₂(g) as the sources of carbon, hydrogen, oxygen and nitrogen respectively. are Hence, the thermochemical equation for the formation reaction of the given compound is—

⇒ \(\begin{aligned}
& 7 \mathrm{C} \text { (graphite,s) }+\frac{5}{2} \mathrm{H}_2(\mathrm{~g})+3 \mathrm{O}_2(\mathrm{~g})+\frac{3}{2} \mathrm{~N}_2(\mathrm{~g}) \rightarrow \mathrm{C}_7 \mathrm{H}_5 \mathrm{~N}_3 \mathrm{O}_6(\mathrm{~s}) ; \\
& \Delta H_f^0\left[\mathrm{C}_7 \mathrm{H}_5 \mathrm{~N}_2 \mathrm{O}_6(s)\right]=-x \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
\end{aligned}\)

Question 2. Under what conditions the heat of reaction at fixed pressure is equal to that at fixed volume?
Answer: we know \(\Delta H=\Delta U+P \Delta V\)

In case of a reaction involving gaseous substances, the equation [1] can be written as, AH = AU + AnRT [Assuming ideal behaviour of the gases]

If a reaction involves only solid or liquid substances (i.e., no gaseous substance), then the change in volume of the reaction system is negligible i.e., ΔV≈O. According to equation (1), for such type of reaction ΔH ≈ AU.

For example: \(\mathrm{NaOH}(a q)+\mathrm{HCl}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)

Again, if in a gaseous reaction, the difference between the total no. of moles of gaseous products and the total no. of moles ofthe gaseous reactants is zero (i.e., Δn = 0 ), then AH = A17.

Again, if in a gaseous reaction, the difference between the total no. of moles of gaseous products and the total no. of moles ofthe gaseous reactants is zero (i.e., Δn = 0 ), then AH = A17.

Punjab State Board Class 11 Chemistry Solutions Chapter 6 Thermodynamics

Question 3. Calculate AU and AH in calories if one mole of a monoatomic ideal gas is heated at a constant pressure of 1 atm from 25°C to 50°C.
Answer:

For an ideal gas, the changes in internal energy (A U) and enthalpy (AH) due to the change in its temperature are given by the relations

⇒ \(\Delta U=n \mathrm{C}_{V, m} \Delta T \text { and } \Delta H=n \mathrm{C}_{P, m} \Delta T\)

For a monoatomic ideal gas \(C_{V, m}=\frac{3}{2} R \text { and } C_{P, m}=\frac{5}{2} R \text {. }\)

The number of mole of the gas, n = 1 ; and the change in temperature (AT) = 25 K

Therefore \(\Delta U=1 \times \frac{3}{2} R \times 25=1 \times 1.5 \times 1.987 \times 25 \mathrm{cal}\) =74.5cal and \(\Delta H=1 \times \frac{5}{2} R \times 25=1 \times 2.5 \times 1.987 \times 25 \mathrm{cal}=124.18 \mathrm{cal}\)

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Question 4. How much hard coal is required to produce the same amount of heat as is produced by the combustion of 2.0 I, of gasoline (mainly isooctane, C8H,b)? Given: AjJJ0 of C8H18 = -5460 kj-mok1 , density of isooctane = 0.692 g. mL-1 and the calorific value of hard coal is 32.75 kj g-1.
Answer:

The mass of 2.0 L of gasoline = 2000 x 0.692 g = 1384 g For 1384 g of the gasoline, the number of moles =\(\frac{1384}{114}=12.14\) [molar mass of the gasoline = 114 g. mol-1 [ For the combustion of 1 mol of gasoline, the amount of the liberated heat is 5460 kj. Therefore, the combustion of 12.14 mol of gasoline will produce 5460 x 12.14 kJ of heat. The calorific value of the hard coal is 32.75 kJ. g 1. So, the amount ofthe hard coal that will produce \(5760 \times 12.14 \mathrm{~kJ} \text { of heat is } \frac{5760 \times 12.14}{32.75} \mathrm{~g}=2135.16 \mathrm{~g}\)

Punjab State Board Class 11 Chemistry Solutions Chapter 6 Thermodynamics

Question 5. One kg of graphite is burnt in a closed vessel. The same amount of graphite is burnt in an open vessel. Will the heat evolved in the two cases be the same? If not, in which case it would be greater?
Answer: Burning of graphite involves the following reaction \(\mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

In an open vessel, if a reaction is carried out, it occurs at constant atmospheric pressure. So, in the reaction, the heat change is the same as the enthalpy change (ΔH). The heat change in a reaction carried out in a closed vessel at constant volume is the same as the internal energy change ( Δ17).

For the above reaction, An = 1-1 = 0. Thus, according to the relation, ΔH = ΔU+ΔnRT, we have AH = AU. Therefore, the burning of 1kg of graphite will produce the same amount of heat irrespective of whether the reaction is carried out at constant volume or constant pressures.

Question 6. Can ΔH be taken as the sole criterion of the spontaneity of a reaction? Justify with an example.
Answer: In an exothermic reaction, ΔH is negative. This means that the energy of the system decreases in an exothermic reaction. On the other hand, ΔH is positive for an endothermic reaction, indicating the energy of the system increases in such a reaction.

As enthalpy decreases in an exothermic reaction, it was once thought that only exothermic reactions would be spontaneous. However, there are some chemical reactions for which AH is positive although they were found to occur spontaneously. Therefore, AH cannot be regarded as the sole criterion for determining the spontaneity of reaction.

Question 7. An Intimate mixture of Fe₂O₃ and ΔA₁₂O₃ is used in solid fuel for rockets. Calculate the fuel value per gram and fuel value per cm3 of the mixture.

ΔH<Fe₂O₃) = 1669,4 kj- mol-1 ,

ΔH(A₁₂O₃) = 832.6 kJ-mol-1
Answer: The reaction is 2A1 + Fe₂O₃→2Fe + A₁₂O₃

The enthalpy change in this reaction is \(\begin{aligned}
\Delta H^0 & =\Delta H_f^0\left(\mathrm{Al}_2 \mathrm{O}_3\right)-\Delta H_f^0\left(\mathrm{Fe}_2 \mathrm{O}_3\right) \\
& =(1669-832.6) \mathrm{kJ}=836.4 \mathrm{~kJ}
\end{aligned}\)

The total mass ofthe reactants (2A1 + Fe₂O₃)

= [2 X 27 + (2 X 55.85 + 3 X 16)] g =213.7 g

PSEB Class 11 Chemistry Chapter 6 Thermodynamics Solutions

Therefore, the fuel per gram of the mixture

⇒ \(=\frac{836.4}{213.7} \mathrm{~kg} \cdot \mathrm{g}^{-1}=3.91 \mathrm{~kJ} \cdot \mathrm{g}^{-1}\)

⇒ \(=\frac{836.4}{213.7} \mathrm{~kg} \cdot \mathrm{g}^{-1}=3.91 \mathrm{~kJ} \cdot \mathrm{g}^{-1}\)

The volume of the mixture of the reactants

⇒ \(=\left(\frac{2 \times 27}{2.7}+\frac{159.7}{5.2}\right) \mathrm{cm}^3=50.7 \mathrm{~cm}^3\)

Therefore, the fuel value per cm3 ofthe mixture

⇒ \(=\frac{836.4}{50.7} \mathrm{~kJ} \cdot \mathrm{cm}^{-3}=16.49 \mathrm{~kJ} \cdot \mathrm{cm}^{-3}\)

Question 8. In a constant volume calorimeter, 3.5g of gas with molecular mass 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K, due to the combustion process. Given, that the heat capacity of the calorimeter is 2.5 kj – K-1, what will be the value of enthalpy of combustion of the gas?
Answer: Por a combustion reaction carried out in a constant volume calorimeter, the amount of liberated heat is given by the relation, q = Calx AT.

Ccal = 2-5 W’ K_1 = (298.45- 298) K =0.45 K

Thus, q = 2.5 x 0.45 kl = 1.125 kj

For 3.5 g ofthe gas, the number of moles \(=\frac{3.5}{28}=0.125\)

PSEB Class 11 Chemistry Chapter 6 Thermodynamics Solutions

Thus, the burning of 0.125 mol ofthe given gas liberates 1.125 kj of heat Hence, the enthalpy of combustion of the gas is \(\frac{1.125}{0.125} \mathrm{~kJ} \cdot \mathrm{mol}^{-1}=9 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 9. If the bond dissociation energies of XY(g), X₂(g) and Y₂(g) are in the ratio of 1: 1: 0.5 and JH for J the formation of XY(g) is -200 kj.mol-1, then what is will be the bond dissociation energy 7 of X₂(g)?
Answer: \(\text { s. } \frac{1}{2} X_2(g)+\frac{1}{2} Y_2(g) \rightarrow X Y(g)\) Suppose, the bond dissociation energy of XY is x. So, the bond dissociation energies of X2 and Y2 will be x and, respectively. For the reaction

⇒ \(\Delta H^0=\Delta H_f^0(\mathrm{XY})-\frac{1}{2} \Delta H_f^0\left(\mathrm{X}_2\right)-\frac{1}{2} \Delta H_f^0\left(\mathrm{Y}_2\right)=-200 \mathrm{~kJ}\)

In terms of bond energies,

⇒ \(\Delta H^0=\left(\frac{1}{2} x+\frac{1}{2} \times \frac{x}{2}\right)-(x)=\frac{3 x}{4}-x=-\frac{x}{4}\)

Therefore, \(-\frac{x}{4}=-200 \mathrm{~kJ}\)

or, x = 800 kJ

Thus, the bond dissociation energy of X₂(g) is 800 kJ.mol-1.

PSEB Class 11 Chemistry Chapter 6 Solutions

Question 10. For the process H₂O(l)⇌H₂O(g), ΔH = 40.8 kJ⋅mol‾1 at the boiling point of water. Calculate molar entropy change for vaporization from the liquid phase
Answer: Molar entropy change for vaporisation \(\Delta S=\frac{\Delta H_{v a p}}{T_b}\)

Given ΔH_vap = 40.8KJ⋅mol¯1, fr water T_b=373 K

∴ \(\Delta S=\frac{40.8 \times 10^3}{373}=109.38 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\)

∴ Molar entropy change for vaporization of water = 109.38 I.K-1.mol-1

Question 11. A gas confined in a cylinder with a frictionless piston is made to expand from 1L to 5L under a constant pressure of 1.5 atm. During the process, 800 J of heat is supplied from an external source. Calculate the change in internal energy of the gas. (1L- atm = 101.3 J)
Answer: We know, w = -P_ex ( V₂– V₁)

∴  ω= -1.5(5- 1 ) = -6 L⋅atm = -6 x 101.3 I = -607.8J

It is also given that q = + 800 1

Using the 1st law of thermodynamics, we have ΔU = q + w = (800- 607.8) J = 192.2J

∴ The change in internal energy of the gas Is 192.2 J

Question 12. Calculate AH0 of the following reaction at 298 K:

⇒ \(\mathrm{C}_2 \mathrm{H}_2(g)+\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l)\)

Given: \(\begin{array}{r}
\mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l)+3 \mathrm{O}_2(g) \rightarrow 3 \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta H^0=-1368 \mathrm{~kJ}
\end{array}\)

⇒ \(\begin{aligned}
& 2 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-2600 \mathrm{~kJ} \\
& 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-566 \mathrm{~kJ} \text { at } 298 \mathrm{~K}
\end{aligned}\)

Answer: ⇒ \(\mathrm{C}_2 \mathrm{H}_2(g)+\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l)\)

⇒ \(\begin{aligned}
& 2 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-2600 \mathrm{~kJ} \\
& 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-566 \mathrm{~kJ} \text { at } 298 \mathrm{~K}
\end{aligned}\)

⇒ \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-566 \mathrm{~kJ}\)

PSEB Class 11 Chemistry Chapter 6 Solutions

Multiplying each of the equations [2] and [3] by \(\frac{1}{2}\) and then adding them, we have

Subtracting equation [1] from equation [4], we have \(\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_2(g)+\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l) ; \\
& \Delta H^0=\left[\frac{1}{2}(-2600)+\frac{1}{2}(-566)-(-1368)\right] \mathrm{kJ}=-215 \mathrm{~kJ}
\end{aligned}\)

Question 13. For the reaction, N₂(g) + 3H₂(g)→+2NH₂(g), ΔH and ΔS are -95.4 kj and -198.3 J. K-1 respectively. Assuming ΔH and ΔS are independent of temperature, will the reaction be spontaneous at 500 K? Explain
Answer: we know ΔG-ΔH-TΔS

Given That ΔH = -95.4 Kj and ΔS =-198.3J.K-1

∴ ΔG= AG =-95.4 kJ- 500 K(-198.3 X 10-3 kj.K-1)

= 3.75kj

As ΔG > 0 at 500 K, the reaction will not be spontaneous at 500 K

Question 14. The bond energy of any diatomic molecule is defined to be the change in the internal energy for its dissociation. At 298 K, O2(g)→2O(g) AH = 498.3 kj.mol¯1 . Calculate the bond energy of O₂ molecule R = 8.314 J-K-1.mol-1
Answer: Given: O₂(g)→2O(g); ΔH = 498.3 kj⋅mol-1
For the above reaction, An = 2-l = l

We know, ΔH = ΔU+ ΔnRT

∴ 498.3 kj = ΔU+1 x 8.314 x 10~3 x 298 kj.

∴ ΔU = 495.82 kJ

Therefore, the bond energy of the O₂ molecule = 495.82 kJ

Question 15. State the first law of thermodynamics. An ideal gas of volume 6.0 L was made to expand at constant temperature and pressure of atm by supplying heat. If the final volume of the gas was 12.0 L, calculate the work done and the heat supplied in joule in the process [1L.atm= 101.3J]
Answer: We know, w = -Pex V₂– V₁

∴ w = -2(12-6) L- atm = -12 L’- atm = -1215.6 J

As the process is isothermal and the system is an ideal gas, AU = 0 for this process. According to the 1st law of thermodynamics, ΔU = q + ω

∴ 0 = q- 1215.61 or, q = + 1215.6J

Question 16. Two moles of an ideal gas were expanded isothermally against a constant opposing pressure of 1 atm from 20 l to 60 complete w,q, E, and H for the process in joule (given 1L. atm = 101.3J)
Answer:

We know, w = -Pex(V2-V1)

∴ \(\begin{aligned}
w=-1(60-20)=-40 \mathrm{~L} \cdot \mathrm{atm} & =-40 \times 101.3 \mathrm{~J} \\
& =-4.052 \mathrm{~kJ}
\end{aligned}\)

For this process ΔE – 0 and ΔH = 0 because the process is isothermal and the system is an ideal gas. As per the first law of thermodynamics, ΔE = q+ w or, 0 = q- 4.052 kJ

∴ q = 4.052 kJ.

Question 17. The latent heat of fusion of ice at 0°C is 80cal/g; Calculate the molar entropy change for the fusion process.
Answer:

We know, \(\Delta S=\frac{\Delta H_f}{T_f}\)

where ΔH_f– = latent heat of fusion of a substance and Tf = its melting point.

Now, ΔH_f = 80 cal⋅g-1 x 18g = 1440 cal

∴ \(\Delta S=\frac{1440}{273} \mathrm{cal} \cdot \mathrm{K}^{-1}=5.27 \mathrm{cal} \cdot \mathrm{K}^{-1}\)

PSEB Class 11 Chemistry Chapter 6 Solutions

Question 18. Calculate AG° for the reaction; \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\) at 298K. Given, at 298 K ΔH_fº for H₂O(1) is -286 kj⋅mol-1 and the molar entropies (S°) for H₂ (g), O₂(g) and H₂O(Z) are 130.7, 69.9 J K-1 . mol-1 respectively.
Answer:

For the given reaction, \(\Delta S^0=S_{\mathrm{H}_2 \mathrm{O}(l)}^0-\left[S_{\mathrm{H}_2(\mathrm{~g})}^0+\frac{1}{2} S_{\mathrm{O}_2(g)}^0\right]\)

⇒ \(=\left[69.9-\left(130.7+\frac{1}{2} \times 205.1\right)\right] \mathrm{J} \cdot \mathrm{K}^{-1}=-163.35 \mathrm{~J} \cdot \mathrm{K}^{-1}\)

Given that ΔH_fO[H₂O(l)] = -286 kj-mol-1

So, AH° = -286 kJ for the given reaction.

We know, ΔG° = ΔHO -TΔS0

∴ AG° = [-286 x 103- 298(-163.35)] J = -237.32 kj.

Question 19. Above what temperature the following reaction will be spontaneous?
Answer:

We know, ΔG = ΔH-TΔS. A reaction at a given temperature and pressure is spontaneous if AG < 0 for the reaction. Therefore, at a given pressure and a temperature of TK, the reaction will be spontaneous if—

ΔG < 0 or, ΔH- TΔS < 0

∴ \(T \Delta S>\Delta H \text { or, } T>\frac{\Delta H}{\Delta S}\)

Given that, ΔH = 144.6 kj and ΔS = 0.116 kl. K¯1

∴ \(T>\frac{144.6}{0.116} \text { or, } T>1246.55 \mathrm{~K}\).

So, the reaction will be spontaneous above 1246.55 K.

Question 20. Calculate AH ofthe following reaction at 25°C. 4Fe(s) + 3O₂(g)→2Fe₂O₃(s) Given: Fe₂O₃(s) + 3C(graphite)→2Fe(s) + 3CO(g), ΔH = 117.30 kcal C(graphite) + O₂(g)→CO₂(g) ; AH = – 94.05 kcal \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) ; \Delta H=67.63 \mathrm{kcal} \text { at } 25^{\circ} \mathrm{C}\)
Answer:

⇒ \(\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) ; \Delta H=67.63 \mathrm{kcal} \quad \ldots[1]\)

⇒ \(\mathrm{C} \text { (graphite) }+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) ; \Delta H=-94.05 \mathrm{kcal} \cdots[2]\)

C(graphite) + O₂(g)→CO₂(g); AH = -94.05 kcal ⋅⋅⋅⋅[2]

Fe₂O₃(s) + 3C(graphite)→2Fe(s) + 3CO(g)

AH = 117.30 kcal

Multiplying equation [3] by 2 and equation [1] by 6 and adding them together, we have

2Fe2O₃(s)+6C(graphite)+3O₂(g)→4Fe(s)+6CO₂(g)

AH = [2(117.30) + 6(67.63)] kcal

Subtracting equation [4] from the equation obtained by multiplying equations [2] by 6, we have

4Fe(s) + 3O₂(g)→2Fe₂O₃(s) ;

ΔH = 6(-94.05)- [2(117.30) + 6(67.63)] =-1204.68kcal

Question 1. Find the number of waves formed by a Bohr electron in one complete revolution in its second orbit.
Answer: Number of waves

⇒ \(=\frac{\text { circumference of the orbit }}{\text { wavelength associated with the moving electron }}\)

⇒ \(=\frac{2 \pi r}{\lambda}=\frac{2 \pi r}{(h / m v)}\)

Since \(\lambda=\frac{h}{m v}\)

⇒ \(=\frac{2 \pi}{h} \times(m v r)=\frac{2 \pi}{h} \times \frac{n h}{2 \pi}=n\)

So the number of waves formed in the second orbit = 2

Question 2. Show that an orbital cannot accommodate more than two electrons in it.
Answer:

Each electron present in an atom is expressed by four quantum numbers. Again, according to Pauli’s exclusion principle, no two electrons in an atom can have the same values for all four quantum numbers (n, l, m, and s).

Now for a certain orbital, the quantum numbers n, l, and m have definite values. So, the electrons present in that orbital must have different values for the spin quantum number. But the spin quantum number ‘s’ has only two possible values.

These are \(+\frac{1}{2} \text { and }-\frac{1}{2}\)

Structure of Atom Class 11 Important Questions

Hence, the maximum that can be accommodated in an orbital is 2. For example, values of the quantum numbers of two electrons in Is -orbital no. of electrons are as follows—

Question 3. Mention the similarity and dissimilarity that exist in the significance, conveyed by the given two sets of quantum numbers
Answer:

In the given sets of quantum numbers n and l have identical values. So, two orbitals as indicated by these two sets of quantum numbers belong to the same principal energy level (shell) and also to the same subshell.

As the orbitals belong to the same sub-shell (viz., d -subshell), they will have the same energy.

However, because of the difference in their values of magnetic quantum numbers, they will have different orientations in space.

Under the influence of a magnetic field, the electrons present in these two orbitals will differ in their energy content.

Read And Learn More Class 11 Chemistry Solutions

Question 4. Write the values of quantum numbers of all the electrons in the 3Li -atom.
Answer:

Number of electrons of 3Li -atom = 3. Therefore, its electronic configuration: ls22s1.

Now, for Is -subshell, n = 1 , l = 0 and m = 0 and for 2s -subshell, n = 2 , l = 0 and m = 0.

Therefore, the values of the quantum numbers of the 3 electrons ofLi-atom are:

Structure of Atom Class 11 Important Questions

Question 5. Identify all the quantum numbers of the electrons in the -subshell of a carbon atom.
Answer:

Atomic number of carbon = 6.

∴ Its electronic configuration: ls22s22p2

In case of 2p -subshell, n = 2, l = 1 and m = -1 , 0 , +1 .

Thus it appears that, the maximum number of electrons that may be present in 2p -subshell = 6.

But this subshell of carbon contains only 2 electrons.

According to Hund’s rule, these 2 electrons will remain in two separate orbitals as odd electrons, each of which will have the same spin.

Therefore, the values of n, l, m, and s of these 2 electrons will be

Question 6. Identify the orbitals having the following quantum numbers using the symbols s, p, d, f:

1. n=4, l=2
2. n=3,l=1
3. n=2, l=0
4. n=5, l=3
5. n=1,l=0
6. n=3,l=2

Answer: For s, p, d, and /-orbitals, the respective values of azimuthal, quantum number, l are 0, 1, 2, and 3.

Again the principal energy levels are designated by the values of the principal quantum number.

On the basis of their information, the orbitals characterized by the given quantum numbers have been identified.

1. n = 4 , l = 2: 4d orbital
2. n = 3, l = 1 : 3p orbital
3. n = 2,f = 0:2s orbital
4. n = 5, l = 3: 5f orbital
5. n = 1 , l = 0 : Is orbital
6. n = 3 , Z = 2: 3d orbital

Question 7. Explain why 2d -and 3/-orbitals do not exist.
Answer:

For an orbital designated by 2d, quantum numbers will have the values of n = 2 and 1 = 2.

But for n = 2d can have the values of 0 and 1. So 2D -orbital has no real existence.

For 3f orbital, n = 3 and 1 = 3. But for n = 3, l will have

Question 8. If the uncertainty in position and momentum of a particle of mass ‘m’ are equal, then ascertain the minimum uncertainty in its velocity.
Answer:

⇒ \(\Delta x \cdot \Delta p\frac{h}{4 \pi}\)

Or, \(\Delta p \cdot \Delta p\frac{h}{4 \pi}\)

Or, \(\Delta p\frac{1}{2} \sqrt{\frac{h}{\pi}}\)

Or, \(m \times \Delta v \frac{1}{2} \sqrt{\frac{h}{\pi}}\)

Or, \(\Delta v \frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

So the minimum uncertainty in its velocity

⇒ \(=\frac{1}{2 m} \sqrt{\frac{h}{\pi}}\)

Question 9. Isotopes of a certain element are identical in their chemical properties—why? Atomic spectra of elements are called fingerprints—why? What are isosters? Give example.
Answer:

Isotopes

Isotopes of an element have the same number of protons in their nucleus and the same number of electrons in the extranuclear part.

Thus, they have the same electronic configuration in their valance shell and hence identical chemical properties.

The atomic spectra (line spectra) of each element are unique. The line spectra of the two elements resemble each other. Hence, these are regarded as the fingerprints of the elements and thus help in their identification.

Molecules or ions with the same number of atoms and also the same number of electrons are said to be sisters.

Example: CO and N2 form a pair of isosters (each molecule contains 2 atoms and 14 electrons).

Question 10. Which of the following relationships directly expresses Bohr’s concept of a hydrogen atom? E = hv, \(m v r=\frac{n n}{2 \pi}\) in) E = me2, E2-E1 = hv \(\lambda=\frac{h}{m v}.\) A = A.mv Write the values of Planck’s constant in CGS and SI units. Indicate the mass of the photon. Are all photons associated with the same quantity of energy?
Answer:

The following two relationships directly express Bohr’s concept in the case of hydrogen atoms:

Relation (2) states that an electron revolves around the nucleus only in those permitted orbits for which the angular momentum is an integral multiple of \(\)

Relation (4) expresses the idea that whenever an electron jumps from one stationary orbit to another, there is either an emission or absorption of energy which is equal to the difference of the energies associated with the two orbits.

Planck’s constant, h = 6.626 x 10-27 erg-s (in CGS unit) = 6.626 x 10-34 J-s (in SI unit)

In the case of light radiation, a quantum is called a photon. Again a quantum means a definite small amount of energy. So, the photon has no mass. It denotes only a definite but small amount of energy,

The energy associated with a quantum radiation [i.e., a photon) is given by E = hv (where v = frequency of radiation).

Since different radiations have different characteristic values of their frequencies, the energy of all photons cannot be the same. For example, the energy of a photon of blue light is greater than that of red light, since; Ared > Ablue or Ablue < Ared

Class 11 Chemistry Chapter 2 Structure of Atom Questions

Question 11. Calculate the no. of electrons, protons & neutrons present in ammonium ion phosphate Ion. Name a species that Is isosteric with N20.
Answer:

1 ammonium ion \(\left(\mathrm{NH}_4^{+}\right)\)

= 1 N-atom +4 H-atoms -1 electron.

∴ Total number of electrons present in the ion = 7 + 4 x1-1 =10

Number of protons = 7 + 4×1 =11

Number of neutrons = (14-7) + 4×0 =7

1 phosphate ion \(\left(\mathrm{PO}_4^{3-}\right)\)

= 1 P-atom+4 O-atoms +3 electrons

∴ Total number of electrons associated with P04

ion = 15 + 4×8 + 3=50

Number of protons = 15 + 4 x 8 = 47

Number of neutrons = (31- 15) + 4(16- 8) = 48

A species isosteric with N20 isC02

Question 12. “Electron is an essential constituent of atoms of all elements” —explain. Why was it necessary to consider the existence of neutrons in the nucleus of an atom? Write the nuclear reaction related to the discovery of neutrons.
Answer:

“Electron is an essential constituent of atoms of all elements”

The value of charge (e ) and mass (m ) of the electron always remain the same irrespective of the source of their emission.

In the discharge tube experiment, the e/m ratio for the negatively charged particles (electrons) constituting the cathode rays was found to be the same irrespective of the nature of the cathode or the nature of the gas taken in the discharge tube.

Thus it can be said that electron is the essential constituent of all atoms.

After the discovery of proton and electron as subatomic particles, the mass of a proton on an atomic scale has been taken as 1 unit and the mass of an electron as zero.

Under this condition, with the exception of ordinary hydrogen, the atomic mass of no other element could be explained in terms of the number of protons present in the atom of an element.

For example, the atomic mass of sodium =23(H = 1) i.e., an atom of sodium is 23 times heavier than a proton. But it has been proved that the atom of sodium contains only 11 protons which can only account for 11 units of mass.

In order to explain the difference of (23-11) = 12 units of mass, the assumption of the presence of a neutral sub-atomic particle of unit mass in the atom became a necessity.

Nuclear reaction associated with the discovery of neutron \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \longrightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 n.\)

Question 13. What is meant by line spectra of atoms? How many lines are observed in the visible region of the hydrogen spectrum? Calculate the wave number of the first line.
Answer:

line spectra of atoms

When an electric discharge is passed through a gaseous element enclosed in a discharge tube under low pressure and the emitted light is analyzed by a spectroscope, the spectrum consists of a large number of lines (with different frequencies) which are grouped into different series. The complete spectrum obtained is known as line spectra or atomic spectra.

In the visible region of the hydrogen spectrum, four lines are obtained: Ha [red], Hg [bluish green], (blue), and Hg (violet). The series consisting of these lines is called the Balmer series.

The equation for wave number in the hydrogen spectrum is given by,

For the first line in the Balmer series, n1 = 2, n2 = 3

∴ \(\bar{v}=1.09678 \times 10^5\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=1.09678 \times 10^5\left(\frac{1}{4}-\frac{1}{9}\right) \mathrm{cm}^{-1}\)

Question 14. Write three differences between electromagnetic waves and matter waves. Show that the circumference of the ‘ n ‘th electronic orbit is ‘ n’ times the de Broglie wavelength of the wave associated with the motion of the electron i.e., 2nrn = nX.
Answer:

According to Bohr’s theory, the angular momentum of a revolving electron in the nth orbit is given by,

Again according to the de-Broglie equation,

From equations [1 ] and [2] we have \(\frac{n h}{2 \pi r_n}=\frac{h}{\lambda} \text { or, } 2 \pi r_n=n \lambda \text { (Proved) }\)

Structure of Atom Class 11 Questions and Answers

Question 15. Give electronic configurations of the following species and show which of these contain the same number of electrons: Cl-, N3″, P3 “, K+, Na+, Mg2+, Ar, S2-, Ne, 02~, Al3+, Ca2+. In an atom of an inert gas, the difference between the number of p -electrons and s -electrons is equal to the number of d -electrons present in that atom. Identify the inert gas and indicate its atomic number.
Answer:

So, N3-, Na+, Mg2+, O2-, Ne, and Al3+ have the same number of electrons (10), so they are isoelectronic. The electronic configuration of each of them ls22s22p6.

Again, Cl “, P3-, K+, Ar, S2-, and Ca2+ have the same number of electrons (18), so they are isoelectronic. Their electronic configuration is:

The inert gas is krypton (Kr) whose atomic number is 36, and has an electronic configuration:

ls22s22p63s23p63d104s24p6

Total number of s -electrons = 8 and total number of p -electrons = 18

∴ Difference in the number of s and p -electrons = 18- 8 = 10 = number of d -electrons.

Question 16. What is the maximum number of electrons present in subshell (s), for which n + l = 3? Will the p -electrons present in the N -atom have the same energy? What is the effect of the magnetic field on these electrons?
Answer:

According to the given conditions, If n = 3, then 1 = 0 (since n + l = 3). These two values of quantum numbers indicate 3s -subshell. 3s subshell can hold a maximum of 2 electrons.

Again, if n = 2, then l = 1 (since n + l = 3). These two values of quantum numbers denote 2psubshell. 2p -subshell may contain a maximum of 6 electrons.

So, a total of (2 + 6) = 8 electrons will be present in subshells for which n + l = 3.

The electronic configuration of N-atom (atomic number = 7 ) is given by: ls22s22p3. The electrons present in the p -subshell will occupy singly px, py, and pzorbitals respectively, i.e., they remain unpaired.

As these are degenerate orbitals, the energy ofthese three electrons will be the same.

As px, py, and pz -orbitals have different orientations in space, the effect of the magnetic field on the electrons present in these orbitals becomes different.

Consequently, it causes a difference in energy. Because of this difference in energy, fine splitting of the spectral lines occurs.

Hence, it may be stated that under the influence of the applied magnetic field, the energies of the electrons of the N-atom become different.

Structure of Atom Class 11 Questions and Answers

Question 17. How would you show that the maximum number 4 0 +1 +i of electrons that can be accommodated in an orbit with a principal quantum number is 2n2? In an atom, the 4f-subshell is completely filled up with electrons. How many electrons in that sub-shell will have the value zero for their magnetic quantum number?
Answer:

The electrons, revolving around the nucleus in an atom, are described by four quantum numbers [viz., n, l, m, and s ]. For a given value of (n), (l) can have n values [these are / = 0, 1, 2…..(n-1)]. This shows that the number of subshells present in the ‘ n ’ th orbit = n.

Again the number of orbitals present in any subshell =(2/+l). Moreover, each orbital may contain a maximum of 2 electrons. Based on this concept, the maximum number of electrons that a particular principal energy level can accommodate can be calculated.

Thus for the principal quantum number’ n values of l =0,1,2…..(n-1). For 1 = 0, number oforbitals present = 2×0 + 1 = 1 For1=1, number oforbitals present = 2×1 + 1 = 3 For 1 = 2 , number oforbitals present = 2×2 + 1 = 5 For / = (w — 1) , number of orbitals present = 2(n- 1) +1 = (2n- 1).

∴ Total number of orbitals in n -th energy level

=l+3+5+7+ … + (2n- 1)

Thus the total number of electrons in ‘ n ‘th energy level = n2 x 2 = 2n2.

For the f-sub-shell, l = 3, and for l = 3, possible values of m = -3, -2, -1, 0, +1, +2, +3 This shows that the f-subshell consists of 7 orbitals.

Out of these, only one orbital is associated with magnetic quantum number, m = 0. This orbital has the capacity to accommodate a maximum of 2 electrons.

Question 18. One predicts the following sets of quantum numbers for some of the electrons present in an atom. State with reasons, which are permissible and which are not. The mass number of an element is twice its atomic number. 2p-subshell of the atom of that element contains 4 electrons. Find the number of protons, neutrons, and electrons present in that atom. What is the valency of that element?
Answer:

Among the given sets of quantum numbers only and are possible but are not permitted.

n = 2, / = 2, m = +2, s = \(+\frac{1}{2}\)

If n = 2 then 1 has only two possible values 0 and 1.

So, the value of 1 cannot be 2.

n = 4, l = 0, m = +1 , s = \(+\frac{1}{2}\)

If l= 0, m = +1 is not possible.

n = 3,l= 2,m = +1,s = 1 ; the value of s may be either +- or . It can never have the value 1.

n = 5, l = 4, m = —3 , s = \(+\frac{1}{2}\)

In the case of this set of quantum numbers, the values of n, /, m, and s are quite consistent.

n = 0, l = 0 , m = 0, s = \(-\frac{1}{2}\) ; The value of the principal quantum number n can never be zero.

n = 1,1-0, m = 0 , s = \(+\frac{1}{2}\)

Values of n,l, m, and s are consistent.

According to the given condition, the electronic configuration ofthe element is lsz2sz2pÿ2pj,2p’.

So, the atomic number of that element is 8. Obviously, the element is oxygen.

O-atom can attain stable electronic configuration by accepting two electrons in its outermost shell. So, its valency is 2.

So, as given in a mass number of the element = 2×8 = 16. Hence, the number of neutrons =16-8 = 8. Number of electrons =8.

PSEB Class 11 Chemistry Structure of Atom Questions

Question 19. Why does 2p -subshell have the capacity to accommodate more electrons than 2s -publicly? Arrange the following sets of electrons in order of decreasing energy.

1. n = 4, l = 0, m = 0, s = \(+\frac{1}{2}\)
2. n = 3, l = 1,
3. m = 1 , s = -i
4. n = 3, / = 2, m = 0, s = \(\frac{1}{2}\)
5. n = 3 , Z = 0, m = 0, s = \(-\frac{1}{2}\)

Which is the lowest principal energy level that permits the existence of the ‘g’ sub-shell?
Answer: For s -sub-shell,l = 0. So, the number of orbitals in this sub-shell =21+1 = 2×0+1 =1.

• An orbital can accommodate a maximum of 2 electrons.
• So, the maximum number of electrons that can be accommodated in 2s -sub-shell = 2.
• Again for p -subshell, 1 = 1. So, the number of orbitals in this sub-shell = 21+ 1 =2x 1 + 1 =3.
• Maximum number of electrons that can be accommodated in 2p -sub-shell =2×3 = 6
• So, 2p -subshell contains more electrons than 2s.

The relative order of energies of various sub-shells can be predicted by the following rules:

A sub-shell with the lower value of (n + /) has lower energy, If two sub-shells have an equal value of (n + l), the sub-shell with a lower value of n has lower energy.

So, we have the following sequence of energies of the electrons: (c) > (a) > (b) > (d) (3)

For the ‘g’ sub-shell, the azimuthal quantum number 1 = 4.

Thus the lowest principal energy’ level that permits the existence of g -sub-shell is given by, n = /+ 1 =4+ 1 = 5.

Question 20. Give sets of quantum numbers for describing all the electrons present In the 3p -subshell. What do you mean by nucleon?
Answer:

3p -subshell may contain a maximum of 6 electrons. For orbitals of the 3p -sub-shell, n = 3 and 1 = 1. Again for 1 = 1, the values of m are -1, 0, and +1. Furthermore, for each value of m, s may have two values \(\left(+\frac{1}{2} \text { and }-\frac{1}{2}\right)\)

The sub-atomic particles present in the nucleus i.e., protons and neutrons are commonly called nucleons.

Question 21. In a neon (Nc) atom, how many electrons are there that spin in hi anti-clockwise direction? For how many electrons of Cl-atom, n + l = 3?
Answer:

Electronic configuration of Ne (atomic no. = 10 )

Each ofthe 5 orbitals (Is, 2s, 2px, 2py, 2pz) present in the Ne atom contains one electron pair each, as displayed by its electronic configuration.

But one of the two electrons in each orbital spins in the clockwise direction while the other spins in the anti-clockwise direction. Out ofthese 10 electrons, 5 electrons spin in the anti-clockwise direction.

Electronic configuration ofCl: ls22s22p63s23p5 In case of electrons in Is -subshell n + l =1 + 0=1 In case of 2 electrons in 2s -subshell, n + l = 2 + 0 = 2 In case of 6 electrons in 2p -subshell, n + l = 2 + 1 = 3 In case of2 electrons in 3s -subshell, n + l = 3 + 0 = 3 In case of5 electrons in 3p -subshell, n + l = 3 +1 = 4 It is observed that in case of 6 electrons in 2p -sub¬ shell and 2 electrons in 3s -sub-shell, the value of n + l will be equal to 3.

Question 22. Electronic configurations of the outermost shell of the atoms of some elements are given below. From these identify the elements and write their atomic numbers.

1. 3s2 ,
2. 3p4 ,
3. 2p4 ,
4. 3p6 ,
5. 5p5

Electronic configuration of the atom: ls22s22p63s2
∴ Total number of electrons = 12 = number of protons = atomic number.

So, the element is magnesium (Mg). It has a valency of 2 because it can donate two electrons from the outermost shell to attain a stable inert gas-like electronic configuration.

Electronic configuration of the atom: ls22s22p63s23p4

Total number of electrons = 16 = number of protons = atomic number.

So, the element is sulfur (S). It has valency 2 because it can gain two electrons in the outermost shell to attain an inert gas like electronic configuration.

Electronic configuration ofthe atom: ls22s22p4

∴ Total number of electrons =8 = number of protons = atomic number.

So, the element is oxygen. It has a valency of 2 because it can gain two electrons in the outermost shell to attain inert gas like electronic configuration.

Electronic configuration: ls22s22p63s23p6

Total number of electrons = 18 = number of protons = atomic number.

So, the element is argon (Ar). Outermost shell of this element is fully filled with electrons. It is an inert gas having valency.

From the outermost electronic configuration of the atom it is clear that it has a total of2(lst shell) +8 (2nd shell) +18 (3rd shell) +18 (4th shell) +7 (5th shell) = 53 electrons.

So it has 53 protons in the nucleus, indicating that its atomic number is 53. The element must be iodine with a valency of 1.

PSEB Class 11 Chemistry Structure of Atom Questions

Question 23. Calculate the difference in radius between the first and third orbit of the hydrogen atom.
Answer:

The radius of n th orbit of H-atom is given by,

rn = 0.529 x n2 A

∴ r3-r1 = 0.529 X (32- 12)A = 0.529 X 8 = 4.232A.

The difference in radius between the first and third orbit of the hydrogen atom = 4.232A.