Q26 / 120 · VITEEE 2007 · Chemistry
Given: $2 \mathrm{C}(\mathrm{s})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})$;
$\begin{array}{r}
\Delta \mathrm{H}=-787 \mathrm{~kJ} \\
\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286 \mathrm{~kJ} \\
\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+2 \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \\
; \Delta \mathrm{H}=-1310 \mathrm{~kJ}
\end{array}$
The heat of formation of acetylene is
- A$-1802 \mathrm{~kJ}$
- B$+1802 \mathrm{~kJ}$
- Correct answerC$+237 \mathrm{~kJ}$
- D$-800 \mathrm{~kJ}$
Solution
We have to find
$2 \mathrm{C}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{2(\mathrm{~g})} \quad \Delta \mathrm{H}=?$
This is the equation for formation of acelytene Given
$2 \mathrm{C}_{(\mathrm{s})}+2 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{CO}_{2(\mathrm{~g})} ; \Delta \mathrm{H}=-787 \mathrm{~kJ}$ ...(1)
$\mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_{2} \mathrm{O}_{(\ell)} ; \Delta \mathrm{H}=-286 \mathrm{~kJ}$ ...(2)
$\begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2(\mathrm{~g})}+2 \frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} & \\ & \longrightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}_{(\ell)} \\ & \Delta \mathrm{H}=-1310 \mathrm{~kJ} ...(3) \end{aligned}$
Addeq.(1) and (2)
$\begin{array}{l}
2 \mathrm{C}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})}+2 \frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{CO}_{2(\mathrm{~g})} \\
+\mathrm{H}_{2} \mathrm{O}_{(\ell)} ; \Delta \mathrm{H}=-1073 \mathrm{~kJ} ...(4)
\end{array}$
Subtract eq. (3) form eq. (4) we get
$\begin{array}{l}
2 \mathrm{C}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{2(\mathrm{~g})} \\
\Delta \mathrm{H}=+237 \mathrm{~kJ}
\end{array}$
Hence, the correct answer is option: 'C'.