# Chapter 15 Principles of Chemical Reactivity: Equilibria - Study Questions - Page 583e: 47

#### 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.

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