Calculating Equilibrium Concentrations Practice Test

In a 1.00 L container at constant temperature, the equilibrium $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ is established. Initially, $\mathrm{H_2} = 0.80\ \mathrm{M}$, $\mathrm{I_2} = 0.80\ \mathrm{M}$, and $\mathrm{HI} = 8.00\ \mathrm{M}$. The value of $K_c$ is 64.0. Using an ICE table (assume only $\mathrm{HI}$ changes appreciably), what is the equilibrium concentration of $\mathrm{HI}$?

In a 1.00 L flask at a constant temperature, the reversible reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ is established. Initially, $\mathrm{H_2} = 1.00\ \mathrm{M}$, $\mathrm{I_2} = 1.00\ \mathrm{M}$, and $\mathrm{HI} = 10.0\ \mathrm{M}$. The equilibrium constant is $K_c = 100$. Using an ICE table (assume only $\mathrm{HI}$ changes appreciably), what is the equilibrium concentration of $\mathrm{HI}$?