JEE PYQ: Chemical Equilibrium Question 4
Question 4 - 2021 (18 Mar Shift 1)
For the reaction $\text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_4 + \text{H}_2$
the reaction enthalpy $\Delta_r H =$ ____ $\text{kJmol}^{-1}$ (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]
Show Answer
Answer: 128
Solution
Bonds broken: 1 C$-$C + 6 C$-$H = 2831. Bonds formed: 1 C$=$C + 4 C$-$H + 1 H$-$H = 2703. $\Delta_r H = 128$ kJ/mol.
BETA
