PYQ NEET- Chemical Thermodynamics L-6

Question: At standard conditions, if the change in the enthalpy for the following reaction is $-109 \mathrm{~kJ} \mathrm{~mol}^{-1}$.#

$$ \mathrm{H}_2(\mathrm{~g})+\mathrm{Br}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{HBr}(\mathrm{g}) $$

Given that, bond energy of $\mathrm{H}_2$ and $\mathrm{Br}_2$ is $435 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $192 \mathrm{~kJ} \mathrm{~mol}^2$ 1 respectively, what is the bond energy (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of $\mathrm{HBr}$ ?

A) 368

B) 736

C) 518

D) 259

Answer: 368#

Solution:#

$\begin{aligned} & \mathrm{H}2(g)+\mathrm{Br}2(g) \longrightarrow 2 \mathrm{HBr}(g) \ & {[\mathrm{H}-\mathrm{H}] \quad[\mathrm{Br}-\mathrm{Br}] \quad[\mathrm{H}-\mathrm{Br}]} \ & \Rightarrow \quad \Delta_r H=(\Sigma B E){\text {Reactants }}-(\Sigma B E){\text {Products }} \ & {[\because \mathrm{BE}=\text { bond energy }]} \ & \Rightarrow-109=\left[(\mathrm{BE}){\mathrm{H}2}+(\mathrm{BE}){\mathrm{Br}2}\right]-(\mathrm{BE}){\mathrm{HBr}} \times 2 \ & =(435+192)-(\mathrm{BE}){\mathrm{HBr}} \times 2 \ & \Rightarrow(\mathrm{BE})_{\mathrm{HBr}}=368 \mathrm{~kJ} \mathrm{~mol}^{-1} \ & \end{aligned}$