Question

Given
$\mathrm{C}_{\text {(graphite) }}+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})$;
$\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$;
$\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g})$;
$\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=+890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Based on the above thermochemical equations, the value of $\Delta_{\mathrm{r}} \mathrm{H}^{\circ}$ at 298 K for the reaction
$\mathrm{C}_{\text {(graphite) }}+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})$ will be :

Accepted Answer

(C) $-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Answer

(A) $+74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Answer

(B) $+144.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Answer

(D) $-144.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

JEE Main · 2017 · 02 Apr 2017 · Offline · Chemical Thermodynamics

Solution

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