For the reaction|Thermodynamics and Thermochemistry|JEE MAIN

JEE MAIN 2021

18-03-21 S1

Question

For the reaction
$\mathrm{C}_2 \mathrm{H}_6 \longrightarrow \mathrm{C}_2 \mathrm{H}_4+\mathrm{H}_2$
the reaction enthalpy $\Delta_{\mathrm{r}} \mathrm{H}=$ $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$.
(Round off to the Nearest Integer).
[Given : Bond enthalpies in $\mathrm{kJ} \mathrm{mol}^{-1}: \mathrm{C}-\mathrm{C}: 347$, $\mathrm{C}=\mathrm{C}: 611 ; \mathrm{C}-\mathrm{H}: 414, \mathrm{H}-\mathrm{H}: 436]$

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Solution

$\begin{aligned} & \Delta_{\mathrm{r}} \mathrm{H}=\left[\epsilon_{\mathrm{C}-\mathrm{C}}+2 \epsilon_{\mathrm{C}-\mathrm{H}}\right]-\left[\epsilon_{\mathrm{C}=\mathrm{C}}+\epsilon_{\mathrm{H}-\mathrm{H}}\right] \\ & =[347+2 \times 414]-[611+436]=128\end{aligned}$

Question Tags

JEE Main

Chemistry

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