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If S(g)+e^(-) rarr S^(1-)(g), Delta H = ...

If `S(g)+e^(-) rarr S^(1-)(g), Delta H = -207.6 KJ`
`S(g) + 2e^(-) rarr S^(2-)(g), Delta H = +335.2 KJ`.
The enthalpy for the reaction
`S^(1-)(g)+e^(-) rarr S^(2-)(g)`:

A

`+127.6 KJ`

B

`-127.6 KJ`

C

`+542.8 KJ`

D

`-641.8 KJ`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta H = H_(P-H_(R))`
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