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If Delta(f)G^(@) for NH(3)(g) is - 16.4 ...

If `Delta_(f)G^(@)` for `NH_(3)(g)` is `- 16.4 kJ mol^(-1)`, then `Delta G^(@)` for the reaction : `N_(2)(g) + 3 H_(2)(g) rarr 2NH_(3)(g)` is

A

`32.8 kJ mol^(-1)`

B

`16.4 kJ mol^(-1)`

C

`- 16.4 kJ mol^(-1)`

D

`- 32.8 kJ mol^(-1)`

Text Solution

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The correct Answer is:
D
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