Question

Heat of formation of $H_2O(g)$ at $1$ atm and 25°C is
$-243 kJ. \Delta E$ for the reaction $H_2(g) + 30 JK^{-1} mol^{-1}\frac{1}{2} O(g)\to H_2O(g)$ at 25°C is

• A. $241.8\, kJ$
• B. $-241.8\, kJ$
• C. $-243\, kJ$
• D. $243\, kJ$

Answer 1

Answer: (B)

$\Delta H = \Delta E + \Delta nRT$
$\Delta n = - 1 /2$
$\therefore -43= \Delta E+\left(-1/2\right)\times8.314\times298\times10^{-3}$
$\therefore \Delta E=-241.76$