If the internal energy change for the reaction
`N_(2(g))+3H_(2(g))hArr2NH_(3(g)`) is `-95(KJ)/(mol)` at `27^(@)` such that equilibrium is achieved at same temperature. Then calculate the `DeltaS_(surroundig)` ?

Options

A. `0`
B. `+(1000)/(3)Jk^(-1)mol^(-1)`
C. `-(1000)/(3)Jk^(-1)mol`
D. `-(1)/(3)Jk^(-1)mol^(-1)`

Answer

B

Explanation

`N//A`