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DeltaG^(ɵ) for 1/2 N(2)(g)+3/2H(2)(g) hA...

`DeltaG^(ɵ)` for `1/2 N_(2)(g)+3/2H_(2)(g) hArrNH_(3)(g)` is `-16.5 kJ mol^(-1)`. Find out `K_(p)` for the reaction at `25^(@)C`.

Text Solution

Verified by Experts

`logK_p=-(DeltaG^@)/(2.303RT)=-((-16.5 xx 10^3))/(2.303xx8.314xx298)=2.8917`
`K_p =779.41`
`K_p` For reaction `N_2(g) +3H_2(g) hArr 2NH_3(g)` is equal to `(779.41)^2=6.07xx 10^5`
`DeltaG^@=-2.303 xx 8.314xx 298 log 6.07 xx 10^5 ` joule
`=-32.998 kJ mol^(-1)`
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