Given that
`O(g)+e^(-)rarrO^(-)(g)` , `DeltaH=34Kcal mol^(-1)`
`O(g)+2e^(-)(g)rarrO^(2-)(g)` , `DeltaH=168Kcal mol^(-1)`
The enthalply change for the reaction
`O^(-)(g)+e^(-)rarrO^(2-)(g)` is

Options

A. `-202Kcalmol^(-1)`
B. `-134Kcalmol^(-1)`
C. `+134Kcalmol^(-1)`
D. `+202Kcalmol^(-1)`

Answer

C

Explanation

Given,
`O(g) + e rarr O^(-)(g),, Delta H = 34 "kcal" mol^(-1)`
`:. O^(-) (g) rarr O (g) + e^(-),, H = -34 "Kcal" mol^(-1)` ...(i)
and `O(g) + 2e^(-) rarr O^(2-)(g),, Delta H = 168 "kcal" mol^(-1)`...(ii)
(i) + (ii) `O_((g))^(-) + e^(-) rarr O^(2-)(g), Delta H = -34 + 168`
`= 134k cal mol^(-1)`