Enthalpy
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Numerical problems of Enthalpy-
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Q 1- The enthalpy changes for the following reactions at 298 K & 1 atmosphere pressure are given below-
(i) CH 3COOH (l) + 2O 2 (g) ———-> 2CO 2 (g) + 2H 2O(l) Δ H = -874KJ
(ii) CH 3CH 2OH (l) + 3O 2 (g) ———> 2CO 2 (g) + 3H 2O(l) Δ H = -1363KJ
Calculate the internal energy changes for these reactions ?
Solution –
(i) Δ H = Δ U + Δ n g RT
Δ n g=2-2=0
R=8.314 JK -1 mole -1
T = 298 k
Δ H = -874 KJ = -874 X 1000 = -874000 J
-874000 = ΔU + 0 X 8.314 X298
Δ U = -874000 J
Δ U = -874KJ Ans.
(ii)
Δ H = Δ U + Δ n g RT
Δ n g= 2-3 = -1
R=8.314 JK -1 mole -1
T = 298 k
Δ H = -1363 KJ = -1363 X 1000 = -1363000 J
-1363000 = Δ U + (-1) X 8.314 X298
Δ U =- 1363000 + 2477.572
Δ U = – 1360522.43 J
Δ U = – 1360.52 KJ Ans.
Q 2- The enthalpy of combustion of benzoic acid [C 6H 5COOH] at 298 K & 1 atmosphere pressure is -2546 KJ/mole. What is Δ U for the reaction ?
Solution-
Equation for the combustion of benzoic acid is,
2C 6H 5COOH (l) + 13 O 2 (g) ———-> 12 CO 2 (g) + 6H 2O(l)
Δ H = Δ U + Δn g RT
Δ n g= 12 – 13 = -1
R=8.314 JK -1 mole -1
T = 298 k
Δ H = – 2546 KJ = -2546 X 1000 J = -2546000 J
– 2546000= Δ U + (-1) X 8.314 X298
Δ U = – 2546000 + 2477.572
Δ U = – 2543522.43 J
Δ U = – 2543.522 KJ Ans.
Q 3 –
The enthalpy changes for the reaction at 298 K & 1 atmosphere pressure is -92.38 KJ .
N 2 (g) + 3H 2 (g) ———> 2 NH 3 (g) Δ H = – 92.38 KJ
Calculate the internal energy changes for the reaction ?
Solution –
Δ H = Δ U + Δn g RT
Δ n g = 2 – (1+3) = – 2
R=8.314 JK -1 mole -1
T = 298 k
Δ H = – 92.38 KJ = -92.38 X 1000 = – 92380 J
– 92380 = Δ U + ( -2) X 8.314 X 298
Δ U = – 87424.86 J
Δ U = – 87.425 = – 87.43 KJ Ans.
