Question

Iodine chloride decomposes at high temperatures to iodine and chlorine gases. $$ 2 \mathrm{ICl}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Cl}_{2}(g) $$ Equilibrium is established at a certain temperature when the partial pressures of \(\mathrm{ICl}, \mathrm{I}_{2},\) and \(\mathrm{Cl}_{2}\) are (in atm) 0.43,0.16 , and \(0.27,\) respectively. (a) Calculate \(K\). (b) If enough iodine condenses to decrease its partial pressure to \(0.10 \mathrm{~atm}\), in which direction will the reaction proceed? What is the partial pressure of iodine when - auilibrium is reestablis]

Short answer

The equilibrium constant K for the reaction is approximately 0.285. (b) In which direction will the reaction proceed when the partial pressure of iodine is decreased, and what are the new partial pressures at equilibrium? When the partial pressure of iodine is decreased, the reaction will shift in the forward direction. The new partial pressure of iodine at equilibrium is approximately 0.137 atm.

Step-by-step solution

(a) Calculate the equilibrium constant K

To calculate the equilibrium constant (K) for this reaction, we can use the equation: $$ K = \frac{[\mathrm{I}_2][\mathrm{Cl}_2]}{[\mathrm{ICl}]^2} $$ We are given the partial pressures of the reactants and products at equilibrium: \([\mathrm{ICl}] = 0.43\,\mathrm{atm}\), \([\mathrm{I}_2] = 0.16\,\mathrm{atm}\), and \([\mathrm{Cl}_2] = 0.27\,\mathrm{atm}\). Substitute these values into the equation: $$ K = \frac{(0.16)(0.27)}{(0.43)^2} $$ Now, calculate K: $$ K ≈ 0.285 $$

(b) Identify the direction of the reaction when Iodine pressure is decreased

When the partial pressure of iodine is decreased to \(0.10\,\mathrm{atm}\), we need to determine if the reaction will shift to re-establish equilibrium. For this, use the reaction quotient Q to compare it to the equilibrium constant K. The equation for the reaction quotient is the same as for K: $$ Q = \frac{[\mathrm{I}_2][\mathrm{Cl}_2]}{[\mathrm{ICl}]^2} $$ Given that Iodine's partial pressure decreases to \(0.10\,\mathrm{atm}\), we get: $$ Q = \frac{(0.10)(0.27)}{(0.43)^2} $$ Now, calculate Q: $$ Q ≈ 0.156 $$ Since \(Q < K\), the reaction will shift in the forward direction, i.e., the reaction proceeds towards the formation of more products (I₂ and Cl₂) to re-establish equilibrium.

Calculate the new partial pressure of iodine

Let the change in the partial pressures be x. Since the forward reaction is favored, the change in partial pressures will be: $$ [\mathrm{ICl}] = 0.43 - 2x $$ $$ [\mathrm{I}_2] = 0.10 + x $$ $$ [\mathrm{Cl}_2] = 0.27 + x $$ At the new equilibrium, the reaction quotient will be equal to the equilibrium constant K. So, we can write: $$ K = \frac{[(0.10+x)(0.27+x)]}{(0.43-2x)^2} $$ We already know that K is approximately equal to 0.285. Now we can solve for x: $$ 0.285 = \frac{[(0.10+x)(0.27+x)]}{(0.43-2x)^2} $$ This is a quadratic equation. Solving for x will give us the change in partial pressures required to re-establish equilibrium. You can use a quadratic formula or a suitable numerical method to find x. After solving the equation, you will get x ≈ 0.037. Since we are interested in finding the new partial pressure of iodine, we can plug this value into the expression for the partial pressure of iodine: $$ [\mathrm{I}_2]_{new} = 0.10 + x = 0.10 + 0.037 ≈ 0.137\,\mathrm{atm} $$ So, the new partial pressure of iodine when equilibrium is reestablished is approximately \(0.137\,\mathrm{atm}\).