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 -kJ mol^ -1 , bond enthalpy of HCl is:

Correct answer: B. 425 ~kJ ~mol^ -1 (1)/(2) H_2+(1)/(2) Cl_2 HCl 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

NEET 2007

Exam: NEET
Year: 2007
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2007 · 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}

-\mathrm{kJ}

\mathrm{mol}^{-1}

, bond enthalpy of

\mathrm{HCl}

is:

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}

\Delta \mathrm{H}_{\mathrm{HCl}}=\Sigma

B.E. of reactant

-\Sigma

B.E. of products

-90=\frac{1}{2} \times 430+\frac{1}{2} \times 240-

B.E. of

\mathrm{HCl}

\therefore \quad

B.E. of

\mathrm{HCl}=215+120+90

=425 \mathrm{~kJ} \mathrm{~mol}^{-1}

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