Given that bond energies of H-H and Cl-Cl are 430 kJ mol^ -1 and 240 kJ mol^ -1 respectively and Δ H_ f for HCl is -90 kJ mol^ -1 , bond enthalpy of HCl is

Correct answer: B. 425 kJ mol^ -1 (1)/(2) H_2+(1)/(2) Cl_2 HCl aligned & Δ H_ HCl =∑ B.E. of reactant -∑ B.E. of products &-90=(1)/(2) × 430+(1)/(2) × 240- B.E. of HCl & B.E. of HCl=215+120+90=425 kJ mol^ -1 aligned

BITSAT 2023Paper 2

Exam: BITSAT
Year: 2023
Paper: Paper 2
Subject: Chemistry
Chapter: Thermodynamics (C)

BITSAT 2023 · Paper 2 · Chemistry · Thermodynamics (C)

Given that bond energies of H-H and Cl-Cl are 430 kJ mol^ -1 and...

Given that bond energies of

\mathrm{H}-\mathrm{H}

and

\mathrm{Cl}-\mathrm{Cl}

are

430 \mathrm{ kJ} \mathrm{ mol}^{-1}

and

240 \mathrm{ kJ} \mathrm{ mol}^{-1}

respectively and

\Delta \mathrm{H}_{\mathrm{f}}

for

\mathrm{HCl}

-90 \mathrm{ kJ} \mathrm{ mol}^{-1}

, bond enthalpy of

\mathrm{HCl}

Options

A $380 \mathrm{ kJ} \mathrm{ mol}^{-1}$
B $425 \mathrm{ kJ} \mathrm{ mol}^{-1}$
C $245 \mathrm{ kJ} \mathrm{ mol}^{-1}$
D $290 \mathrm{ kJ} \mathrm{ mol}^{-1}$

Correct answer

B. $425 \mathrm{ kJ} \mathrm{ mol}^{-1}$

Step-by-step solution

\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{HCl}

\begin{aligned} \\ & \Delta H_{\mathrm{HCl}}=\sum \text{ B.E. of reactant }-\sum \text{ B.E. of products } \\ \\ &-90=\frac{1}{2} \times 430+\frac{1}{2} \times 240-\text{ B.E. of } \mathrm{HCl} \\ \\ & \therefore \quad \text{ B.E. of } \mathrm{HCl}=215+120+90=425 \mathrm{ kJ} \mathrm{ mol}^{-1} \\ \end{aligned}

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