Answer
See the answer below.
Work Step by Step
a) $p=p(NO_2)+p(N_2O_4)\rightarrow p(N_2O_4)=1.50\ atm-x, p(NO_2)=x$
$K_p=p(NO_2)^2/p(N_2O_4)$
$0.148=x^2/(1.50-x)$
$x^2+0.148x-0.222=0$
$x=0.403$
$f=2x/(1.50-x)\times100\%=73.5\%$
b) It would increase, because, since more moles of the product are formed than moles of reactants are consumed, decreasing pressure shifts the equilibrium to the right.
