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Chemistry · 2021

JEE Main · 18 March 2021, Shift 1 · Q53

For the reaction C_2 H_6 → C_2 H_4+ H_2 the reaction enthalpy Δ_r H = \ \ \ \ kJ mol^-1. (Round off to the Nearest Integer). [Given: Bond enthalpies…

For the reaction $$\mathrm{C}_2 \mathrm{H}_6 \rightarrow \mathrm{C}_2 \mathrm{H}_4+\mathrm{H}_2 $$ the reaction enthalpy $\displaystyle \Delta_{\mathrm{r}} \mathrm{H}=$ \_\_\_\_ $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$. (Round off to the Nearest Integer). [Given : Bond enthalpies in $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}: \mathrm{C}-\mathrm{C}: 347, \mathrm{C}=\mathrm{C}: 611$; C-H : $\displaystyle 414$, H - H : $\displaystyle 436$]
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.