JEE MAIN - Chemistry (2022 - 25th June Morning Shift - No. 17)
The standard entropy change for the reaction
4Fe(s) + 3O2(g) $$\to$$ 2Fe2O3(s) is $$-$$550 J K$$-$$1 at 298 K.
[Given : The standard enthalpy change for the reaction is $$-$$165 kJ mol$$-$$1]. The temperature in K at which the reaction attains equilibrium is _____________. (Nearest Integer)
Answer
300
Explanation
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=0$ at equilibrium
$$ \begin{aligned} &\Rightarrow-165 \times 10^3-\mathrm{T} \times(-505)=0 \\\\ &\Rightarrow \mathrm{T}=300 \mathrm{~K} \end{aligned} $$
$$ \begin{aligned} &\Rightarrow-165 \times 10^3-\mathrm{T} \times(-505)=0 \\\\ &\Rightarrow \mathrm{T}=300 \mathrm{~K} \end{aligned} $$
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