Given that `O(g) + e^(-) rarr O^(-)(g) Delta H = - 142 kJ mol^(-1)`
`O(g) + 2e^(-) rarr O^(2-)(g) DeltaH =+712kJmol^(-1)`
The heat change for the reaction
`O^(-)(g) + e^(-) rarr O^(2-)(g)` is :

Options

A. `- 570 kJ`
B. 570 kJ
C. `+ 854 kJ`
D. `- 854 kJ`

Answer

C

Explanation

(i) `O(g) + e^(-) rarr O^(-) Delta H = - 142 kJ`
(ii) `O(g) + 2e^(-) rarr O^(2-) Delta H = 712 kJ`
Subtract eq. (i) from eq.(ii).
`O(g) + e^(-) rarr O^(2-) Delta H = 854 kJ`