Ncert Solutions Chemistry Class 11th
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12. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ΔrH for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
12. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ΔrH for the reaction: N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
12. Given? fH of CO (g) = – 110 kJ mol-1
? fH of CO2 (g) = – 393 kJ mol-1
? fH of N2O (g) = 81 kJ mot-1
Upvote 
11. Enthalpy of combustion of carbon to CO2 is – 393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.
11. Enthalpy of combustion of carbon to CO2 is – 393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.
11. The combustion equation is:
C (s) + O2 (g) → CO2 (g);
Upvote 10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. A, ΔfusH = 6.03 KJ mol-1 at 0°C.
Cp [H2O(l)] = 75.3 J mol-1 K-1;
Cp [H2O(s)] = 36.8 J mol-1 K-1.
10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. A, ΔfusH = 6.03 KJ mol-1 at 0°C.
Cp [H2O(l)] = 75.3 J mol-1 K-1;
Cp [H2O(s)] = 36.8 J mol-1 K-1.
10. Total enthalpy change involved in the transformation is the sum of the following changes:
(i) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C
ΔH= Cp [H2O
Upvote 9. Calculate the number of KJ of heat necessary to raise the temperature of 60 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol-1 K-1.
9. Calculate the number of KJ of heat necessary to raise the temperature of 60 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol-1 K-1.
9. No. of moles of Al, n = (60g)/ (27 g mol-1) = 2.22 mol
Molar heat capacity (C) = 24 J mol-1 K-1.
Rise in temperature (? T) = 55 – 35 = 20°C = 20 K
Heat evolved (q) = C x m x T = (24 J mol-1 K-1) x (2.22 mol) x (20 K) = 1065.6 J = 1.067 KJ.
Upvote 8. The reaction of cyanamide,NH2CN(s) with dioxygen was carried out in a bomb calorimeter and ?U was found to be -742,7 KJ-1 mol-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K.
NH2CN (S) + 3/2O2(g) → N2(g) + CO2(g) + H2O(l)
8. The reaction of cyanamide,NH2CN(s) with dioxygen was carried out in a bomb calorimeter and ?U was found to be -742,7 KJ-1 mol-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K.
NH2CN (S) + 3/2O2(g) → N2(g) + CO2(g) + H2O(l)
8. Given,
? U = – 742.7 KJ-1 mol-1
? ng = 2 – 3/2 = + 1/2 mol.
R = 8.314 x 10-3KJ-1 mol-1
T = 298 K
According to the relation? H =? U+? ng RT
Upvote 7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
7. As per the first law of thermodynamics,
? U= q + W
Heat absorbed by the system, q = 701 J
Work done by the system, W = – 394 J
Change in internal energy? U = q + w = 701 – 394 = 307 J.
Upvote 5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are -890.3 KJ mol-1, – 393.5 KJ mol-1 and – 285.8 KJ mol-1 respectively. Enthalpy of formation of CH4(g) will be
(i) – 74.8 KJ mol-1
(ii) – 52.27 KJ mol-1
(iii) + 74.8 KJ mol-1
(iv) + 52.26 KJ mol-1
5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are -890.3 KJ mol-1, – 393.5 KJ mol-1 and – 285.8 KJ mol-1 respectively. Enthalpy of formation of CH4(g) will be
(i) – 74.8 KJ mol-1
(ii) – 52.27 KJ mol-1
(iii) + 74.8 KJ mol-1
(iv) + 52.26 KJ mol-1
5. As per question:
(i) CH4? +2O2? CO2? +2H2? O ΔH1? =−890kJmol−1
(ii) C+O2? CO2? &
Upvote 4. ΔU? of combustion of methane is – X kJ mol–1. The value of ΔH? is
(i) = ΔU?
(ii) > ΔU?
(iii) < ΔU?
(iv) = 0
4. ΔU? of combustion of methane is – X kJ mol–1. The value of ΔH? is
(i) = ΔU?
(ii) > ΔU?
(iii) < ΔU?
(iv) = 0
4.The chemical equation for the combustion reaction is:
CH4 (g) + 2O2 (g) →CO2 (g) + 2H2O (l)
Δng= 1 – 3 = -2
ΔH? = ΔU? + ΔngRT = ΔU? - 2RT
Therefore, ΔH? <ΔU?
i.e. option (iii) is correct.
Upvote Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.
Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.
This is a Short Answer Type Questions as classified in NCERT Exemplar
At constant volume
q = ΔU + (-w)
-w = pΔ q = AU + pΔV
ΔV = 0 (at constant volume)
Hence, qv = ΔU + 0 = ΔU= change in internal energy at constant pressure, qp = AU + pΔV
Since ΔU + pΔV=ΔH
=> qp = ΔH change in enthalpy
Hence, at constant vo
Upvote