Problem: At a particular temperature, K = 1.00 x 102 for the reactionH2(g) + I2(g) ⇌ 2HI(g)In an experiment, 1.00 mole of H2, 1.00 mole of I2, and 1.00 mole of HI are introduced into a 1.00-L container. Calculate the concentrations of all species when equilibrium is reached.

H₂(g) + I₂(g) ⇌ 2 HI(g)          K = 1.00 x 10²

Step 1: Calculate the concentrations of the reactants and products.

Recall molarity: 

$\boxed{\mathbf{Molarity}\mathbf{ }\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{ }\mathbf{=}\frac{\mathbf{mol}}{\mathbf{L}}}$

$$\begin{gathered} \mathbf{\left[}{\mathbf{H}}_{\mathbf{2}}\mathbf{\right]}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{mol}}{\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{L}} \\ \mathbf{\left[}{\mathbf{H}}_{\mathbf{2}}\mathbf{\right]}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{mol}}{\mathbf{L}} \end{gathered}$$

[H₂] = 1.00 M

[I₂] = 1.00 M

[HI] = 1.00 M

Step 2: Calculate the reaction quotient.

$\boxed{\mathbf{Q}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}}$

$\boxed{\mathbf{Q}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}}$

$$\begin{gathered} \mathbf{Q}\mathbf{=}\frac{{\mathbf{\left[}\mathbf{HI}\mathbf{\right]}}^{\mathbf{2}}}{\mathbf{\left[}{\mathbf{H}}_{\mathbf{2}}\mathbf{\right]}\mathbf{\left[}{\mathbf{I}}_{\mathbf{2}}\mathbf{\right]}} \\ \mathbf{Q}\mathbf{=}\frac{{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{)}}^{\mathbf{2}}}{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{)}} \end{gathered}$$

Q = 1.00

Q < K → the reaction shifts in the forward direction to reach equilibrium

Step 3: Construct an ICE chart for the equilibrium reaction.

Step 4: Calculate the change (x) in the reaction using the equilibrium constant.