Question

∆H for the reaction,

2C(s) + 3H2(g) C2H6(g) is -84.4 kJ at 25 0C. Calculate ∆U for the reaction at 25 0C. (R = 8.314 J K-1 mol-1)​

Answer 1

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Solution :

∆H = ∆U + ∆ng RT∆ng

= (moles of product gases) - (moles of reactant gases)∆ng

= 1 - 3 = -2 mol

∆H = -84.4 kJ, R = 8.314 J K-1 mol-1

= 8.314 × 10-3 kJ K-1 mol-1

Substitution of these in above

-84.4 kJ = ∆U + 8.314 × 10-3 kJ K-1 mol-1 ×

298 K × (-2 mol)

= ∆U - 4.96 kJ

Hence, ∆U = -84.4 kJ + 4.96 kJ = - 79.44 kJ

Answer 2

Answer:

∆H = ∆U + ∆ng RT∆ng

= (moles of product gases) - (moles of reactant gases)∆ng

= 1 - 3 = -2 mol

∆H = -84.4 kJ, R = 8.314 J K-1 mol-1

= 8.314 × 10-3 kJ K-1 mol-1

Substitution of these in above

-84.4 kJ = ∆U + 8.314 × 10-3 kJ K-1 mol-1 ×

298 K × (-2 mol)

= ∆U - 4.96 kJ

Hence, ∆U = -84.4 kJ + 4.96 kJ = - 79.44 kJ