Question

Consider the following hypothetical reaction: $$ \mathrm{X}_{2}(g)+\mathrm{R}(s) \rightleftharpoons \mathrm{X}_{2} \mathrm{R}(g) $$ \(\mathrm{R}\) has a molar mass of \(73 \mathrm{~g} / \mathrm{mol}\). When equilibrium is established, a 2.5-L reaction vessel at \(125^{\circ} \mathrm{C}\) contains \(15.0 \mathrm{~g}\) of \(\mathrm{R}\), 4.3 atm of \(\mathrm{X}_{2}\), and 0.98 atm of \(\mathrm{X}_{2} \mathrm{R}\). (a) Calculate \(K\) for the reaction at \(125^{\circ} \mathrm{C}\). (b) The mass of \(\mathrm{R}\) is doubled. What are the partial pressures of \(\mathrm{X}_{2}\) and \(\mathrm{X}_{2} \mathrm{R}\) when equilibrium is reestablished? (c) The partial pressure of \(\mathrm{X}_{2}\) is decreased to \(2.0 \mathrm{~atm}\). What are the partial pressures of \(\mathrm{X}_{2}\) and \(\mathrm{X}_{2} \mathrm{R}\) when equilibrium is reestablished?

Short answer

Question: Calculate the partial pressures of X2 and X2R at equilibrium when the mass of R is doubled and the partial pressure of X2 is decreased to 2.0 atm. Answer: For the scenario where the mass of R is doubled, the partial pressures of X2 and X2R at equilibrium are 3.943 atm and 1.337 atm, respectively. When the partial pressure of X2 is decreased to 2.0 atm, the new partial pressures of X2 and X2R at equilibrium are 1.377 atm and 0.357 atm, respectively.

Step-by-step solution

Write the equilibrium expression for Kp

For the given reaction, the equilibrium constant Kp in terms of the partial pressures is: $$ K_p = \frac{P_{(\mathrm{X}_2 \mathrm{R})}}{P_{(\mathrm{X}_2)}} $$

Calculate initial moles of R and partial pressures of gases at equilibrium

We are given the mass of R (15 g) and its molar mass (73 g/mol). We can calculate the initial moles of R: $$ \text{moles of R} = \frac{15 \ \mathrm{g}}{73 \ \mathrm{g/mol}} = 0.205 \ \mathrm{mol} $$ The partial pressures of X2 and X2R at equilibrium are given as 4.3 atm and 0.98 atm, respectively.

Calculate Kp at 125°C

Plug the partial pressures into the Kp expression calculated in Step 1: $$ K_p = \frac{0.98 \ \mathrm{atm}}{4.3 \ \mathrm{atm}} = 0.228 $$ So, at \(125^{\circ} \mathrm{C}\), the equilibrium constant \(K_p = 0.228\).

Calculate new partial pressures when mass of R is doubled

When the mass of R is doubled, we have: $$ \text{moles of R} = 2 \times 0.205 \ \mathrm{mol} = 0.410 \ \mathrm{mol} $$ Let \(\Delta P_{X_2}\) be the change in partial pressure of \(X_2\), and \(\Delta P_{X_2R}\) be the change in partial pressure of \(X_2R\). At the new equilibrium, $$ K_p = \frac{P_{(\mathrm{X}_2 \mathrm{R})}+\Delta P_{X_2R}}{P_{(\mathrm{X}_2)}-\Delta P_{X_2}} $$ Use the calculated value of Kp to solve for the change in partial pressures: $$ 0.228 = \frac{0.98 + \Delta P_{X_2R}}{4.3 - \Delta P_{X_2}} $$ Since the stoichiometry is 1:1, the change in partial pressures is equal: \(\Delta P_{X_2} = \Delta P_{X_2R}\). Substituting this into the equation, we obtain: $$ 0.228 = \frac{0.98 + \Delta P}{4.3 - \Delta P} $$ Solving for \(\Delta P\), we get \(\Delta P = 0.357 \ \mathrm{atm}\). The new partial pressures of \(X_2\) and \(X_2R\) are: $$ P_{(\mathrm{X}_2)} = 4.3 - 0.357 = 3.943 \ \mathrm{atm} $$ $$ P_{(\mathrm{X}_2 \mathrm{R})} = 0.98 + 0.357 = 1.337 \ \mathrm{atm} $$

Calculate new partial pressures when X2 is decreased to 2.0 atm

When the partial pressure of X2 is decreased to 2.0 atm, we can write the Kp expression as: $$ K_p = \frac{P_{(\mathrm{X}_2 \mathrm{R})}+\Delta P_{X_2R}}{P_{(\mathrm{X}_2)}+\Delta P_{X_2}} $$ Replacing the values of Kp and the given partial pressure of X2, $$ 0.228 = \frac{0.98 + \Delta P_{X_2R}}{2.0 + \Delta P_{X_2}} $$ Again, since the stoichiometry is 1:1, the change in partial pressures is equal: \(\Delta P_{X_2} = \Delta P_{X_2R}\). Substituting this into the equation, we obtain: $$ 0.228 = \frac{0.98 + \Delta P}{2.0 + \Delta P} $$ Solving for \(\Delta P\), we get \(\Delta P = -0.623 \ \mathrm{atm}\). The new partial pressures of \(X_2\) and \(X_2R\) are: $$ P_{(\mathrm{X}_2)} = 2.0 - 0.623 = 1.377 \ \mathrm{atm} $$ $$ P_{(\mathrm{X}_2 \mathrm{R})} = 0.98 - 0.623 = 0.357 \ \mathrm{atm} $$