Chemistry · 2024
JEE Main · 30 January 2024, Shift 2 · Q82
Two reactions are given below: 2 Fe_( s ) +3/2 O_2( g ) → Fe_2 O_3( s ), Δ H^°=-822 kJ / mol; C_( s ) +1/2 O_2( g ) → CO_( g ), Δ H^°=-110 kJ / mol…
Two reactions are given below:
$$\begin{aligned}
& 2 \mathrm{Fe}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}, \Delta \mathrm{H}^{\circ}=-822 \mathrm{~kJ} / \mathrm{mol} \\
& \mathrm{C}_{(\mathrm{s})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}^{\circ}=-110 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$
Then enthalpy change for following reaction $\displaystyle 3 \mathrm{C}_{(\mathrm{s})}+\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{(\mathrm{g})}$ is
$\displaystyle \_\_\_\_$ $\displaystyle \mathrm{kJ} / \mathrm{mol}$.
Official answer
From NTA’s final answer key for this paper.
492
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JEE Main 2024 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.