In a closed flask, exactly of gas and of gas are mixed and allowed to react according to the equatio — Chemical Equilibrium Chemistry Question
Question
In a $1\text{ L}$ closed flask, exactly $3.0\text{ moles}$ of gas $A$ and $3.0\text{ moles}$ of gas $B$ are mixed and allowed to react according to the equation $A(g) + B(g) \rightleftharpoons C(g) + D(g)$. At equilibrium, it is determined that $2.0\text{ moles}$ of $C$ have been formed. What is the numerical value of the equilibrium constant $K_c$?
💡 Solution & Explanation
Initial moles: $A=3$, $B=3$, $C=0$, $D=0$. Let $x$ moles react. At equilibrium: $A=3-x$, $B=3-x$, $C=x$, $D=x$. We are given that equilibrium moles of $C = 2.0$, so $x = 2.0$. Thus, equilibrium moles are: $A = 3 - 2 = 1\text{ mol}$, $B = 3 - 2 = 1\text{ mol}$, $C = 2\text{ mol}$, $D = 2\text{ mol}$. Since volume is $1\text{ L}$, concentrations equal moles. $K_c = \frac{[C][D]}{[A][B]} = \frac{(2)(2)}{(1)(1)} = 4$.