Question

The value of $\Delta H^o$ for the reaction ${Cu}^{+}+I^{-}\left(g\right)\to CuI\left(g\right)$ is $-446kJ$ ${mol}^{-1}$. If the ionisation energy of $Cu\left(g\right)$ is $745kJ$ ${mol}^{-1}$ and the electron affinity of $I\left(g\right)$ is $-295kJ$ ${mol}^{-1}$, then the value of $\Delta H^{+}$ for the formation of one mole of $CuI\left(g\right)$ from $Cu\left(g\right)$ and $I\left(g\right)$ is:

• A. $-446kJ$
• B. $450kJ$
• C. $594kJ$
• D. $4kJ$

Answer 1

The correct option is D $4kJ$

Solution:- (D) $4KJ$

${Cu}_{\left(g\right)}⟶{{Cu}^{+}}_{\left(g\right)}+e^{-};\Delta H_1=745KJ/mol$

$\frac{1}{2}{I_2}_{\left(g\right)}+e^{-}⟶{I^{-}}_{\left(g\right)};\Delta H_1=-295KJ/mol$

${{Cu}^{+}}_{\left(g\right)}+{I^{-}}_{\left(g\right)}⟶{CuI}_{\left(g\right)};{\Delta H^0}_{reaction}=-446KJ/mol$

According to Hess's law-

${\Delta H^0}_{reaction}={\Delta H_f}_{product}-{\Delta H_r}_{reactant}$

$\therefore {\Delta H^0}_{reaction}={\Delta H_f}_{CuI}-(\Delta H_1+\Delta H_2)$

$-446={\Delta H_f}_{CuI}-(745+(-295))$

${\Delta H_f}_{CuI}=-446+450=4KJ$

Hence $\Delta H$ for the formation of $CuI$ will be $4KJ$.