JEE MAIN - Chemistry (2022 - 30th June Morning Shift - No. 6)
$$\Delta$$G$$^\circ$$ vs T plot for the formation of MgO, involving reaction
2Mg + O2 $$\to$$ 2MgO, will look like :
2Mg + O2 $$\to$$ 2MgO, will look like :
_30th_June_Morning_Shift_en_6_1.png)
_30th_June_Morning_Shift_en_6_2.png)
_30th_June_Morning_Shift_en_6_3.png)
_30th_June_Morning_Shift_en_6_4.png)
Explanation
For the reaction,
$$ \begin{aligned} &2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{MgO}(\mathrm{s}) \\\\ &\Delta S^{\circ}=-v e \\\\ &\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \end{aligned} $$
$-T \Delta S^{\circ}$ is positive,
Hence, $\Delta G^{\circ}$ increase with increase in temperature.
Also, when there is phase change there is an abrupt increase in the slope.
$$ \begin{aligned} &2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{MgO}(\mathrm{s}) \\\\ &\Delta S^{\circ}=-v e \\\\ &\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \end{aligned} $$
$-T \Delta S^{\circ}$ is positive,
Hence, $\Delta G^{\circ}$ increase with increase in temperature.
Also, when there is phase change there is an abrupt increase in the slope.
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