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 $\mathrm{kJ} \mathrm{mol}^{-1}: \mathrm{C}-\mathrm{C}: 347, \mathrm{C}=\mathrm{C}: 611 ; \mathrm{C}-\mathrm{H}: 414, \mathrm{H}-\mathrm{H}: 436$ ]

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Chemical Thermodynamics

For the reaction

C_{2} H_{6} \to C_{2} H_{4} + H_{2}

the reaction enthalpy

Δ_{r} 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

]

Answer: 128

Only the bonds that change need to be counted. Going from

C_{2} H_{6}

C_{2} H_{4} + H_{2}

, one

C - C

bond and two

C - H

bonds are broken while one

C = C

bond and one

H - H

bond are formed.

Δ_{r} H = [347 + 2 (414)] - [611 + 436]

Δ_{r} H = 1175 - 1047 = 128

The reaction enthalpy is

128 kJ mo l^{- 1}

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer