Question

Problem 4.7 : AH for the reaction, 2C(s) + 3H,(9) → CH (9) is -84.4 kJ at 25 °C. Calculate AU for the reaction at 25 °C. (R = 8.314 J K mol')​

Answer 1

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