Question

At \(800 \mathrm{~K}\), hydrogen iodide can decompose into hydrogen and iodine gases. $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{H}_{2}(g) $$ At this temperature, \(K=0.0169 .\) What are the partial pressures at equilibrium of the hydrogen and iodine if initially a sealed flask at \(800 \mathrm{~K}\) contains only HI at a pressure of 0.200 atm?

Short answer

Answer: The partial pressures of hydrogen (H2) and iodine (I2) at equilibrium are both 0.00161 atm.

Step-by-step solution

Write the expression for the equilibrium constant K in terms of partial pressures

For the given reaction: $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{H}_{2}(g) $$ The expression for the equilibrium constant (K) in terms of partial pressures is given by: $$ K = \frac{P_{\mathrm{I}_{2}} \times P_{\mathrm{H}_{2}}}{P_{\mathrm{HI}}^2} $$ where \(P_{\mathrm{I}_{2}}\), \(P_{\mathrm{H}_{2}}\), and \(P_{\mathrm{HI}}\) are the partial pressures of iodine, hydrogen, and hydrogen iodide respectively at equilibrium.

Define the changes in partial pressures at equilibrium

Let x be the change in partial pressure of HI at equilibrium. Thus, at equilibrium, the partial pressure of HI is \((0.200 - 2x)\). Since for every 2 moles of HI that decompose, 1 mole of I2 and 1 mole of H2 are formed, the partial pressures of I2 and H2 at equilibrium will be x and x respectively. Now we can rewrite the equilibrium expression in terms of x: $$ K = \frac{x^2}{(0.200 - 2x)^2} $$

Substitute the given value of K and solve for x

Substitute the given value of K (0.0169) into the equilibrium expression and solve for x: $$ 0.0169 = \frac{x^2}{(0.200 - 2x)^2} $$ Solving for x is a quadratic equation (\(12.7104x^2 - 0.0676x + 0.000169 = 0\)), which gives two possible solutions for x. However, only the solution that results in a positive value for the partial pressures is physically meaningful. So we discard the negative solution and find that the positive value of x is approximately 0.00161 atm.

Calculate the partial pressures of H2 and I2 at equilibrium

Now we can calculate the partial pressures of H2 and I2 at equilibrium using the value of x we found: $$ P_{\mathrm{H}_{2}} = x = 0.00161 \mathrm{~atm} $$ $$ P_{\mathrm{I}_{2}} = x = 0.00161 \mathrm{~atm} $$ So, the partial pressures at equilibrium of the hydrogen and iodine are both 0.00161 atm.