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.
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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
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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 \(=\frac{3.5}{28}=0.125\)
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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.
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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 ΔHvap = 40.8KJ⋅mol¯1, fr water Tb=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 = -Pex ( 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: ⇒ \(\mathrm{C}_2 \mathrm{H}_2(g)+\mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{C}_2 \mathrm{H}_3 \mathrm{CO}_2 \mathrm{H}(l)\)
⇒ \(\begin{aligned}
& 2 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^0=-2600 \mathrm{~kJ} \\
& 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-566 \mathrm{~kJ} \text { at } 298 \mathrm{~K}
\end{aligned}\)
⇒ \(2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^0=-566 \mathrm{~kJ}\)
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Multiplying each of the equations [2] and [3] by \(\frac{1}{2}\) and then adding them, we have
\(\begin{array}{r}\mathrm{C}_2 \mathrm{H}_2(g)+3 \mathrm{O}_2(g)+\mathrm{CO}(g) \rightarrow 3 \mathrm{CO}_2(g)+\mathrm{H}_2 \mathrm{O}(l) ; \\
\Delta H^0=\left[\frac{1}{2}(-2600)+\frac{1}{2}(-566)\right] \mathrm{kJ}
\end{array}\)
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 ΔHf– = latent heat of fusion of a substance and Tf = its melting point.
Now, ΔHf = 80 cal⋅g-1 x 18g = 1440 cal
∴ \(\Delta S=\frac{1440}{273} \mathrm{cal} \cdot \mathrm{K}^{-1}=5.27 \mathrm{cal} \cdot \mathrm{K}^{-1}\)
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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 ΔHfº 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 ΔHfO[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