Enthalpy : Numericals

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.