JEE CHEMISTRY MCQ: HYDROGEN

1 Easy

1 of 1

Application +4 | -1

Given

(A)

2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g}) \quad \Delta \mathrm{H}_1^\theta=-\mathrm{xkJ} \mathrm{mol}^{-1}

(B)

\mathrm{C}

(graphite

)+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) \quad \Delta \mathrm{H}_2^\theta=-\mathrm{ykmol}^{-1}

The

\Delta \mathrm{H}^\theta

for the reaction

\mathrm{C}(

graphite

)+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})

View answer and explanation

Correct option: C

\frac{x-2 y}{2}

Option C (

\frac{x-2 y}{2}

) is correct because it is derived from Hess's Law by manipulating the given equations to obtain the desired reaction.

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