JEE Main 2022 · 26 Jul · Shift 2 · Chemistry Q83

ChemistryInteger+4 / -1Question 83 of 90

For the reaction

$\mathrm{H_2F_2(g)\rightarrow H_2(g)+F_2(g)}$

$\Delta U=-59.6$ kJ mol $^{-1}$ at 27°C.

The enthalpy change for the above reaction is (−) ______ kJ mol $^{-1}$ [nearest integer]

Given: R = 8.314 J K $^{-1}$ mol $^{-1}$ .

Your answer

Answer and worked solution

Official answer: 57

$\Delta H=\Delta U+\Delta n_gRT$ with $\Delta n_g=1$ : $-59.6+\frac{8.314\times300}{1000}=-57.1$ kJ mol $^{-1}$ .

Answer: 57

JEE Main 2022 · 26 Jul · Shift 2, Chemistry question 83