Calculate `DeltaG` at 298 K for the following reaction if the reaction mixture consists of 1 atm of N_(2),3 atm` of `H_(2)` and 1 atm of `NH_(3)`.
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g),DeltaG^(0)=-33.32 kJ`
What is the effect of increasing pressure on the reaction ?

Explanation

`Q=(p_(NH_(3))^(2))/(p_(N_(2))*p_(H_(2))^(3))=(1^(2))/(1xx3^(2))=3.7xx10^(-2)`
where Q is the reaction quotient
`:.DeltaG=DeltaG^(@)+2.303RTlogQ`
`DeltaG=(-33.32)+2.303xx8.314xx10^(-3)xx298log(3.7xx10^(-2))`
`DeltaG=-41.49kJ`
Thus, increasing the pressure of `H_(2)` from 1 atm (std. condition) to 3 atm, free energy change, `DeltaG` becomes more negative that is from `-33.32kJ` to `-41.49kJ`. Hence increase in pressure favours the forward reaction.