For the reaction $CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$, hydrogen gas is introduced

Question

For the reaction $CO_{(g)} + 2H_{2(g)} ightleftharpoons CH_3OH_{(g)}$, hydrogen gas is introduced into a $5.0 	ext{ L}$ flask at $327^\circ C$ containing $0.2 	ext{ mole}$ of $CO_{(g)}$ and a catalyst, until the total equilibrium pressure is $4.92 	ext{ atm}$. At this point, $0.1 	ext{ mole}$ of $CH_3OH_{(g)}$ has formed. Calculate the equilibrium constant $K_c$ for the reaction. ($R = 0.082 	ext{ L atm K}^{-1}	ext{mol}^{-1}$, round to two decimal places).

Professor PC Thomas Classes · Answer

Temperature $T = 327^\circ C = 600 	ext{ K}$. Total moles at equilibrium $n_{tot} = \frac{PV}{RT} = \frac{4.92 	imes 5}{0.082 	imes 600} = 0.5 	ext{ mol}$. At equilibrium, moles of $CH_3OH = 0.1 	ext{ mol}$. Initial $CO = 0.2 	ext{ mol}$, so equilibrium $CO = 0.2 - 0.1 = 0.1 	ext{ mol}$. The equilibrium moles of $H_2 = n_{tot} - (	ext{moles of } CO + 	ext{moles of } CH_3OH) = 0.5 - 0.2 = 0.3 	ext{ mol}$. The volume is $5.0 	ext{ L}$, so the concentrations are $[CO] = 0.1/5 = 0.02 	ext{ M}$, $[H_2] = 0.3/5 = 0.06 	ext{ M}$, $[CH_3OH] = 0.1/5 = 0.02 	ext{ M}$. $K_c = \frac{[CH_3OH]}{[CO][H_2]^2} = \frac{0.02}{(0.02)(0.06)^2} = \frac{1}{0.0036} = 277.777...$, which rounds to 277.78.