Q25 / 90 · JEE Main 2021 (01 Sep Shift 2) · Chemistry

For the reaction $2 {NO}_{2} (g) ⇌ N_{2} O_{4} (g),$ when $ΔS = - 176.0 J K^{- 1}$ and $ΔH = - 57.8 kJ {mol}^{- 1}$ the magnitude of $ΔG$ at $298 K$ for the reaction is _____ ${kJmol}^{- 1}$ . (Nearest integer)

Correct answer: 5

Solution

$ΔG = ΔH - TΔS$

$= - 57.8 - \frac{298 (- 176.0)}{1000} = - 57.8 + 52.448 = - 5.352 KJ {mol}^{- 1}$

The Magnitude of $∆ G to the nearest integer is 5 .$

Hence, the correct answer is 5.

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