1.

For the reaction, `N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g)` `DeltaH =- 95.0 kJ` and `DeltaS = - 19000 J K^(-1)` Calculate the temperature in centigrade at which it will attain equilibrium.

Answer» At equilibrium, `DeltaG = 0`
`T = (DeltaH)/(DeltaS) = (-95 xx 10^(3))/(-190000) = 0.5K`


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