For the equilibrium at 2.0 bar N_(2(g)) + 3H_(2(g)) harr 2NH_(3(g)) [(del)/(del p) log K_(x)] = ?

For the equilibrium at 2.0 bar N_(2(g)) + 3H_(2(g)) harr 2NH_(3(g)) [(

1.	For the equilibrium at 2.0 bar N_(2(g)) + 3H_(2(g)) harr 2NH_(3(g)) [(del)/(del p) log K_(x)] = ?
Answer» Solution :Kp related by `K_(x)` by equation `Kp=K_(x)(p)^(Delta n_(g))` `log Kp=Delta n_(g)log+log K_(x)` `logK_(x)=logKp-Delta ng logp` `(DEL)/(del p)logK_(x)=0-(Delta ng)/(p)=(0-(-2))/(2)=1`