# Given that bond energies of H - H and Cl - Cl are 430  and  respectively and ${\mathrm{\Delta }}_{\mathrm{f}}\mathrm{H}$ for HCI is $-90\mathrm{kJ}{\mathrm{mol}}^{-1}$. Bond enthalpy of HCI is

1. A

290 kJ $mo{l}^{-1}$

2. B

380 kJ $mo{l}^{-1}$

3. C

425 kJ $mo{l}^{-1}$

4. D

245 kJ $mo{l}^{-1}$

FREE Lve Classes, PDFs, Solved Questions, PYQ's, Mock Tests, Practice Tests, and Test Series!

+91

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.

### Solution:

$\begin{array}{l}\frac{1}{2}{\mathrm{H}}_{2}+\frac{1}{2}{\mathrm{Cl}}_{2}\to \mathrm{HCl}\\ \mathrm{\Delta }{\text{H}}_{\mathrm{f}}=\left[\frac{1}{2}\mathrm{\Delta }{\text{H}}_{{\text{H}}_{2}}+\frac{1}{2}\mathrm{\Delta }{\text{H}}_{{\text{Cl}}_{2}}\right]-\left[{\mathrm{\Delta H}}_{\text{HCl}}\right]\\ -90=\left[\frac{1}{2}×430+\frac{1}{2}×240\right]-\mathrm{\Delta }{\text{H}}_{\text{HCl}}\\ -90=215+120-\mathrm{\Delta }{\text{H}}_{\mathrm{HCl}}\\ {\mathrm{\Delta H}}_{\text{HCl}}=335+90=425\mathrm{kJ}/\mathrm{mol}\end{array}$

## Related content

Join Infinity Learn Regular Class Program!

Sign up & Get instant access to FREE PDF's, solved questions, Previous Year Papers, Quizzes and Puzzles!

+91

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.