JEE MAIN - Chemistry (2021 - 1st September Evening Shift - No. 16)
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)
Answer
5
Explanation
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.
$$\Delta$$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.
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