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Calculate change in enthalpy when $39 \mathrm{~g}$ acetylene is completely burnt with oxygen and enthalpy of combustion of acetylene is $1300 \mathrm{~kJ} / \mathrm{mol}$.
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Verified Answer
The correct answer is:
$-1950 \mathrm{~kJ}$
$$
\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_{2(\mathrm{~g})}+\frac{5}{2} \mathrm{O}_{(2)} \longrightarrow 2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{(\ell)} ; \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-1300 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \text { Acetylene } \\
& \text { When } 26 \mathrm{~g} \text { of acetylene is completely burnt, the change in } \\
& \text { enthalpy }=-1300 \mathrm{~kJ} \\
& \therefore \text { For } 39 \mathrm{~g} \text { of acetylene, the change in enthalpy } \\
& =\frac{-1300 \times 39}{26}=-1950 \mathrm{~kJ}
\end{aligned}
$$
\begin{aligned}
& \mathrm{C}_2 \mathrm{H}_{2(\mathrm{~g})}+\frac{5}{2} \mathrm{O}_{(2)} \longrightarrow 2 \mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{(\ell)} ; \Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-1300 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \text { Acetylene } \\
& \text { When } 26 \mathrm{~g} \text { of acetylene is completely burnt, the change in } \\
& \text { enthalpy }=-1300 \mathrm{~kJ} \\
& \therefore \text { For } 39 \mathrm{~g} \text { of acetylene, the change in enthalpy } \\
& =\frac{-1300 \times 39}{26}=-1950 \mathrm{~kJ}
\end{aligned}
$$
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