JEE Mains 26-July-2022 Shift 2 Solved Paper

Section: Chemistry

Question : 83 of 90

Marks: +1, -0

For the reaction
H2F2(g)→H2(g)+F2(g)
∆U=−59.6‌kJ‌mol−1‌ at ‌27∘C
The enthalpy change for the above reaction is (−) _______ kJ‌mol−1 [nearest integer]
Given: R=8.314JK−1mol−1

[26-Jul-2022-Shift-2]

Your Answer:

Validate

Go to Question:

Prev Question

Next Question