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At 25^{\circ}C and 1 atm pressure, the enthalpy of combustion of benzene (l) and acetylene (g) are \mathrm{-3268\ kJ\ mol^{-1}} and \mathrm{-1300\ kJ\ mol^{-1}}, respectively. The change in enthalpy for the reaction 3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l}), is

Option: 1

+324 \mathrm{~kJ} \mathrm{~mol}^{-1}


Option: 2

+632 \mathrm{~kJ} \mathrm{~mol}^{-1}


Option: 3

-632 \mathrm{~kJ} \mathrm{~mol}^{-1}


Option: 4

-732 \mathrm{~kJ} \mathrm{~mol}^{-1}


Answers (1)

best_answer

Given,

\underset{\text{Benzene}}{\mathrm{C}_{6} \mathrm{H}_{6}(l)} +\left(\frac{15}{2}\right) \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}, \mathrm{\Delta }_{\mathrm{c}} \mathrm{H}=-3268 \mathrm{~kJ} / \mathrm{mol} \quad (1)

\underset{\text { (Acetylene) }}{\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})}+\left(\frac{5}{2}\right) \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}, \mathrm{\Delta _{C}} \mathrm{H}=-1300 \mathrm{~kJ} / \mathrm{mol} \quad (2)

We need to find the enthallpy change of the following reaction

3 \mathrm{C}_{2} \mathrm{H}_{2} \mathrm{(g)} \rightarrow \mathrm{C}_{6} \mathrm{H}_{6} \text { (l) }, \Delta_{\text{r}} \mathrm{H}=\text { ? }

So, by doing 3 \times(2)-(1), we get the required reaction.

Thus, enthalpy change of the reaction is given as 

\begin{aligned} &\mathrm{\Delta_{r} H=3 \times(-1300)-(-3268)} \\ &\mathrm{\Delta_{r} H}=-632 \mathrm{~kJ} / \mathrm{mol} \end{aligned}

Hence, the correct answer is Option (3)

Posted by

Deependra Verma

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