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 ?

Calculate DeltaG at 298 K for the following reaction if the reaction m

1.	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 ?
Answer» SOLUTION :`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.