"For the reaction, 2NO2(g) $\rightleftharpoons$ N2O4(g), when $\Delta$S = $-$176.0 JK$-$1 and $\Delta$H = $-$57.8 kJ mol$-$1, the magnitude of $\Delta$G at 298 K for the reaction is ___________ kJ mol$-$1. (Nearest integer)"

Question

Accepted Answer

"Given, $\Delta$H = $-$ 57.8 kJ mol^$-$1

$\Delta$S = $-$ 176 JK^$-$1 mol^$-$1

T = 298 K

Using Gibb's free energy relation

$\Delta$G = $\Delta$H $-$ T$\Delta$S

where, $\Delta$G = change in Gibb's free energy

$\Delta$H = change in enthalpy

T = temperature

$\Delta$S = change in entropy

$\Delta$G = 57.8 kJ/mol $-$ [298 K $\times$ ($-$ 176 Jk^$-$1 mol^$-$1)]

= 57.8 kJ/mol $-$ $\left( {298 \times {{ - 176} \over {1000}}kJ} \right)$ [$\therefore$ 1 kJ = 1000 J]

= $-$ 5.352 kJ/mol

| $\Delta$G | = 5.352

Hence, answer is 5."

For the reaction, 2NO₂(g) $\rightleftharpoons$ N₂O₄(g), when $\Delta$S = $-$176.0 JK^$-$1 and $\Delta$H = $-$57.8 kJ mol^$-$1, the magnitude of $\Delta$G at 298 K for the reaction is ___________ kJ mol^$-$1. (Nearest integer)

Solution

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For the reaction, 2NO2(g) $\rightleftharpoons$ N2O4(g), when $\Delta$S = $-$176.0 JK$-$1 and $\Delta$H = $-$57.8 kJ mol$-$1, the magnitude of $\Delta$G at 298 K for the reaction is ___________ kJ mol$-$1. (Nearest integer)

Solution

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For the reaction, 2NO₂(g) $\rightleftharpoons$ N₂O₄(g), when $\Delta$S = $-$176.0 JK^$-$1 and $\Delta$H = $-$57.8 kJ mol^$-$1, the magnitude of $\Delta$G at 298 K for the reaction is ___________ kJ mol^$-$1. (Nearest integer)