Question

Given that:
$O_{\left(g\right)}+e^{-}\to O_g^{-}\Delta H=-142kJmo{l}^{-1}$
$O_{\left(g\right)}+2e^{-}\to O_{\left(g\right)}^{2-}\Delta H=+702kJmo{l}^{-1}$
The enthalpy (heat) change for the reaction is represented by the equation-
$O_{\left(g\right)}^{-}+e^{-}\to O^{2-}$ is:
(correct answer +2, wrong answer -0.5)

• A. $-844kJ$
• B. $-560kJ$
• C. $+560kJ$
• D. $+844kJ$

Answer 1

The correct option is D $+844kJ$

$O_{\left(g\right)}+e^{-}\to O_g^{-}\Delta H_1=-142kJmo{l}^{-1}$............(1)
$O_{\left(g\right)}+2e^{-}\to O_{\left(g\right)}^{2-}\Delta H_2=+702kJmo{l}^{-1}$...........(2)
Applying Hess's law of heat summation,
Substracting equation (1) from (2)
$O_{\left(g\right)}^{-}+e^{-}\to O^{2-}$............(3)

Enthalpy for the reaction (3) is obtained as follows

$\Delta H_3=\Delta H_2-\Delta H_1=702-(-142)=844\text{kJ}$