JEE Main 2017 (2 Apr, offline) Chemistry Q31: Given Based…

Question

Given
$\mathrm{C_{(graphite)}+O_2(g)\rightarrow CO_2(g)}$ ;
$\Delta_rH^\circ=-393.5\ \mathrm{kJ\,mol^{-1}}$
$\mathrm{H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)}$ ;
$\Delta_rH^\circ=-285.8\ \mathrm{kJ\,mol^{-1}}$
$\mathrm{CO_2(g)+2H_2O(l)\rightarrow CH_4(g)+2O_2(g)}$ ;
$\Delta_rH^\circ=+890.3\ \mathrm{kJ\,mol^{-1}}$
Based on the above thermochemical equations, the value of $\Delta_rH^\circ$ at 298 K for the reaction
$\mathrm{C_{(graphite)}+2H_2(g)\rightarrow CH_4(g)}$ will be :

Accepted Answer

Option 1: $-74.8\ \mathrm{kJ\,mol^{-1}}$

Answer

$-74.8\ \mathrm{kJ\,mol^{-1}}$

Answer

$-144.0\ \mathrm{kJ\,mol^{-1}}$

Answer

$+74.8\ \mathrm{kJ\,mol^{-1}}$

Answer

$+144.0\ \mathrm{kJ\,mol^{-1}}$

JEE Main 2 April 2017 (offline), Chemistry Q31

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