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  • For the reaction <img alt="\mathrm{C}_{2} \mathrm{H}_{6} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2}" src="/latex-image/?%5Cmathrm%7BC%7D_%7B2%7D%20%5Cmathrm%7BH%7D_%7B6%7D%20%5Crightarrow%20%5Cmathrm%7BC%7D_%7B2%7D%20%5Cmathrm%7BH%7D_%7B4

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}= ___________kJ mol-1. (Round off to the Nearest Integer). [Given : Bond enthalpies in kJ mol-1 : C–C : 347, C=C : 611; C–H : 414, H–H : 436]
 

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Given reaction:

\mathrm{C}_{2} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{H}_{2}

Need to find the reaction enthalpy \Delta_{r} \mathrm{H}.

We know

\Delta_{r} \mathrm{H}= \Delta_{\textup{Product}} \mathrm{H}-\Delta_{\textup{Reactant}} \mathrm{H}

= (6 x EC-H + 1 x EC-C) - (4 x EC-H + 1 x EC=C + 1 x EH-H)

= 2 x EC-H + 1 x EC-C - 1 x EC=C - 1 x EH-H)

[Given : Bond enthalpies in kJ mol-1 : C–C : 347, C=C : 611; C–H : 414, H–H : 436]

After putting the value-

\Delta_{r} \mathrm{H}= 2 × 414 + 347 – 611 – 436 

\Delta_{r} \mathrm{H}= 828 + 347 – 1047

\Delta_{r} \mathrm{H}= 128\textup{ kJ/mol}

Ans = 128

Posted by

Kuldeep Maurya

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