Calculate $\Delta H$ in $\mathrm{kJ}$ for the following reaction $…
Calculate $\Delta H$ in $\mathrm{kJ}$ for the following reaction
$
\mathrm{C}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})
$
Given that,
$
\begin{gathered}
\mathrm{H}_2 \mathrm{O}(g)+\mathrm{C}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_2(g) ; \\
\Delta H=+131 \mathrm{~kJ} \\
\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \\
\Delta H=-282 \mathrm{~kJ} \\
\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(g) ; \\
\Delta H=-242 \mathrm{~kJ}
\end{gathered}
$
- $-393$
- $+393$
- $+655$
- $-655$
Solution
Given,
$
\begin{aligned}
& \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_2(g) ; \\
& \Delta H=131 \mathrm{~kJ} \\
& \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g) ; \\
& \Delta H=-282 \mathrm{~kJ} \\
& \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \text {; } \\
& \Delta H=-242 \mathrm{~kJ} \\
& \mathrm{C}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta \mathrm{H}=\text { ? } \\
&
\end{aligned}
$
On adding Eqs (i), (ii) and (iii), we get Eq (iv)
$
\begin{aligned}
& \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{g}) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_2(g) ; \\
& \Delta H=+131 \mathrm{~kJ} \\
& \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(g) \\
& \Delta H=-282 \mathrm{~kJ} \\
& \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{H}_2 \mathrm{O}(g) \text {; } \\
& \Delta H=-242 \mathrm{~kJ} \\
& \mathrm{C}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \\
& \Delta H=(131-282-242) \mathrm{kJ} \\
& =-393 \mathrm{~kJ} \\
&
\end{aligned}
$
Asked in: AP EAMCET 2008
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