Question

Energy involved to ionise $O_2\left(g\right)$ to $O^{2-}\left(g\right)$ is $639{\text{kJ mol}}^{-1}$. If the ionisation enthalpy of $O^{-}\left(g\right)$ is $141{\text{kJ mol}}^{-1}$. The second electron gain enthalpy of oxygen would be:

• A. $-780{\text{kJ mol}}^{-1}$
• B. $780{\text{kJ mol}}^{-1}$
• C. $498{\text{kJ mol}}^{-1}$
• D. $-498{\text{kJ mol}}^{-1}$

Answer 1

The correct option is B $780{\text{kJ mol}}^{-1}$

Given:
$O\left(g\right)+2e^{-}\to O^{2-}\left(g\right);\Delta H=639{\text{kJ mol}}^{-1}...\left(i\right)$
$O^{-}\left(g\right)\to O\left(g\right)+e^{-};\Delta H=141{\text{kJ mol}}^{-1}...\left(ii\right)$

Adding equation (i) and (ii), we get
$O^{-}\left(g\right)+e^{-}\to O^{2-}\left(g\right);\Delta H=780{\text{kJ mol}}^{-1}...\left(iii\right)$