For the reaction $\mathrm{A_2 + B_2 \rightarrow 2AB}$, the enthalpy change of the reaction as written is $\Delta \mathrm{H}^0 = -200\ \mathrm{kJ\,mol^{-1}}$. AB, $\mathrm{A_2}$ and $\mathrm{B_2}$ are diatomic molecules. If the bond enthalpies of $\mathrm{A_2}$, $\mathrm{B_2}$ and AB are in the ratio $1:0.5:1$, then the bond enthalpy of $\mathrm{A_2}$ is ______ $\mathrm{kJ\,mol^{-1}}$ (Nearest integer)

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Chemical Thermodynamics

For the reaction

A_{2} + B_{2} \to 2AB

, the enthalpy change of the reaction as written is

Δ H^{0} = - 200 kJ mo l^{- 1}

. AB,

A_{2}

and

B_{2}

are diatomic molecules. If the bond enthalpies of

A_{2}

B_{2}

and AB are in the ratio

1 : 0.5 : 1

, then the bond enthalpy of

A_{2}

is ______

kJ mo l^{- 1}

(Nearest integer)

Answer: 400

For the reaction

A_{2} + B_{2} \to 2AB

the given enthalpy change is

- 200 kJ mo l^{- 1}

. Taking the bond enthalpy of

A_{2}

x

, that of

B_{2}

0.5 x

and that of

AB

x

. The reaction enthalpy equals the energy of bonds broken minus the energy of bonds formed,

(x + 0.5 x) - 2 x = - 0.5 x

. Setting

- 0.5 x = - 200

gives

x = 400

. The bond enthalpy of

A_{2}

400 kJ mo l^{- 1}

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer