The equilibrium constant ($K_c$) of the reaction $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$

Question

The equilibrium constant ($K_c$) of the reaction $A_{2(g)} + B_{2(g)} ightleftharpoons 2AB_{(g)}$ at $100^\circ C$ is $50$. A $1.0 	ext{ L}$ flask containing $1.0 	ext{ mole}$ of $A_2$ is connected to a $2.0 	ext{ L}$ flask containing $2.0 	ext{ moles}$ of $B_2$ via a valve. When the valve is opened and equilibrium is established at $100^\circ C$, how many moles of $AB$ will be formed? (Give your answer to two decimal places).

Professor PC Thomas Classes · Answer

When the flasks are connected, the total volume becomes $1.0 + 2.0 = 3.0 	ext{ L}$. The initial moles are $A_2 = 1.0$ and $B_2 = 2.0$. Let the moles of $A_2$ that react be $x$. At equilibrium, moles of $A_2 = 1-x$, $B_2 = 2-x$, and $AB = 2x$. Since $\Delta n = 0$, $K_c$ can be calculated using moles directly: $K_c = \frac{(2x)^2}{(1-x)(2-x)} = 50$. Expanding this gives $\frac{4x^2}{x^2 - 3x + 2} = 50 \implies 46x^2 - 150x + 100 = 0 \implies 23x^2 - 75x + 50 = 0$. Solving the quadratic equation gives $x = \frac{75 - \sqrt{5625 - 4600}}{46} = \frac{75 - \sqrt{1025}}{46} = \frac{75 - 32.015}{46} = 0.934 	ext{ mol}$. The moles of $AB$ formed is $2x = 2 	imes 0.934 = 1.868 	ext{ mol}$, which rounds to 1.87.