Given that, for the reaction, (1)/(2) H_2+(1)/(2) Cl_2 HCl aligned & Δ H_ f (HCl)=-93 kJ / mol & B.E (H_2)=434 kJ / mol & B.E (Cl_2)=242 kJ / mol aligned The bond dissociation energy of HCl is

Correct answer: C. 431 ~kJ / mol Given, (1)/(2) H_2+(1)/(2) Cl_2 HCl, Δ H_ f =-93 ~kJ / mol In this reaction, one mole of HCl is formed Δ H_ f of the reaction will remain same, i.e. -93 ~kJ / mol Let the bond dissociation energy of H-Cl bond be x For the given reaction, aligned Δ H_ f & = B · E_ (reactants) - B · E_ (products) -93 & =(1)/(2)(434)+(1)/(2)(242)-x -93 & =+338-x X & =431 ~kJ / mol aligned

JIPMER 2018

Exam: JIPMER
Year: 2018
Subject: Chemistry
Chapter: Thermodynamics (C)

JIPMER 2018 · Chemistry · Thermodynamics (C)

Given that, for the reaction, (1)/(2) H_2+(1)/(2) Cl_2 HCl aligned...

Given that, for the reaction,

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

\begin{aligned} & \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HCl})=-93 \mathrm{ kJ} / \mathrm{mol} \\ & \text{ B.E }\left(\mathrm{H}_2\right)=434 \mathrm{ kJ} / \mathrm{mol} \\ & \text{ B.E }\left(\mathrm{Cl}_2\right)=242 \mathrm{ kJ} / \mathrm{mol}\end{aligned}

$$ The bond dissociation energy of HCl is

Options

A $232 \mathrm{~kJ} / \mathrm{mol}$
B $331 \mathrm{~kJ} / \mathrm{mol}$
C $431 \mathrm{~kJ} / \mathrm{mol}$
D $530 \mathrm{~kJ} / \mathrm{mol}$

Correct answer

C. $431 \mathrm{~kJ} / \mathrm{mol}$

Step-by-step solution

Given,

\frac{1}{2} \mathrm{H}_2+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow \mathrm{HCl}, \Delta \mathrm{H}_{\mathrm{f}}=-93 \mathrm{~kJ} / \mathrm{mol} \because

In this reaction, one mole of HCl is formed

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

of the reaction will remain same, i.e.

-93 \mathrm{~kJ} / \mathrm{mol}

Let the bond dissociation energy of

\mathrm{H}-\mathrm{Cl}

bond be x For the given reaction,

\begin{aligned} \Delta \mathrm{H}_{\mathrm{f}} & =\Sigma \mathrm{B} \cdot \mathrm{E}_{\text{(reactants) }}-\Sigma \mathrm{B} \cdot \mathrm{E}_{\text{(products) }} \\ -93 & =\frac{1}{2}(434)+\frac{1}{2}(242)-\mathrm{x} \\ -93 & =+338-\mathrm{x} \\ \mathrm{X} & =431 \mathrm{~kJ} / \mathrm{mol}\end{aligned}

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