Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

For the reaction

Question:

For the reaction

$\mathrm{C}_{2} \mathrm{H}_{6} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2}$

the reaction enthalpy $\Delta_{\mathrm{r}} \mathrm{H}=$ $\mathrm{kJ} \mathrm{mol}^{-1}$

(Round off to the Nearest Integer).

[Given : Bond enthalpies in $\mathrm{kJ} \mathrm{mol}^{-1}: \mathrm{C}-\mathrm{C}$ :

$347, \mathrm{C}=\mathrm{C}: 611 ; \mathrm{C}-\mathrm{H}: 414, \mathrm{H}-\mathrm{H}: 436]$

Solution:

$\Delta_{\mathrm{r}} \mathrm{H}=\left[\in_{\mathrm{C}-\mathrm{C}}+2 \in_{\mathrm{C}-\mathrm{H}}\right]-\left[\in_{\mathrm{C}=\mathrm{C}}+\in_{\mathrm{H}-\mathrm{H}}\right]$

$=[347+2 \times 414]-[611+436]$

$=128$

Leave a comment

None
Free Study Material