Question

The value of enthalpy change (DH) for the reaction

C2H5OH(l) + 3O2(g)$\rightarrow$ 2CO2(g) + 3H2O(I)

at 27⁰C is –1366.5 kJ mol–1. The value of internal energy change for the above reaction at this temperature will be :

• A. –1369.0 kJ
• B. –1364.0 kJ
• C. –1361.5 kJ
• D. –1371.5 kJ

Answer 1

Answer: (B)

–1364.0 kJ

Answer 2

C2H5OH ($\mathcal{l}$) + 3O2 (g)¾® 2CO2(g) + 3H2O($\mathcal{l}$)

$\Delta$ng = 2 – 3 = – 1

$\Delta$U = $\Delta$H – $\Delta$ng RT

$= - 1366.5 - ( - 1) \times \frac{8.314}{10^{3}} \times 300$

= – 1366.5 + 0.8314 × 3 = – 1364 KJ