Question

The reaction ${\mathrm{PCl}}_3(g)+{\mathrm{Cl}}_2(g)\rightleftharpoons {\mathrm{PCl}}_5(g)$ has $K_p=$ $0.0870$ at ${300}^{\circ }\mathrm{C}$. A flask is charged with $0.50$ atm ${\mathrm{PCl}}_3,0.50$ atm ${\mathrm{Cl}}_2$, and $0.20\mathrm{atm}{\mathrm{PCl}}_5$ at this temperature. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of ${\mathrm{Cl}}_2$ in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of ${\mathrm{Cl}}_2$ in the equilibrium mixture?

Short answer

a) The initial reaction quotient Q is 0.80, which is greater than K_p (0.0870), so the reaction will proceed in the reverse direction to reach equilibrium. b) At equilibrium, the partial pressures are $PC{l}_3=0.54598$ atm, $C{l}_2=0.54598$ atm, and $PC{l}_5=0.15402$ atm. c) According to Le Chatelier's Principle, increasing the volume of the system will shift the reaction towards the side with more moles of gas, which is the reactants' side (left). As a result, the mole fraction of $C{l}_2$ in the equilibrium mixture will increase. d) For an exothermic reaction, increasing the temperature will cause the reaction to shift towards the endothermic reaction, which in this case is the decomposition of $PC{l}_5$ into $PC{l}_3$ and $C{l}_2$. Therefore, increasing the temperature will lead to a higher mole fraction of $C{l}_2$ in the equilibrium mixture.

Step-by-step solution

Initial reaction quotient, Q

We will calculate the initial reaction quotient using the given partial pressures. Q is given by $Q=\frac{[PC{l}_5]}{[PC{l}_3][C{l}_2]}$ Plugging in the given values, $Q=\frac{0.20}{(0.50)(0.50)}=0.80$

Comparing Q with K_p (equilibrium constant)

As $K_p=0.0870$ at ${300}^{\circ }\mathrm{C}$ and the calculated Q is 0.80, we have $Q>K_p$. This implies that the reaction will proceed in the reverse direction to reach equilibrium, meaning that $PC{l}_5$ will decompose into $PC{l}_3$ and $C{l}_2$. Now, let's calculate the equilibrium partial pressures of the gases. b) Find equilibrium partial pressures

Changes in partial pressures at equilibrium

Let's represent the change in partial pressures for each gas at equilibrium with x. So, we get: $PC{l}_3(g)+C{l}_2(g)\rightleftharpoons PC{l}_5(g)$ $0.50+x0.50+x0.20-x$

Equilibrium expression

Now, we can write the equilibrium expression using K_p and the adjusted partial pressures: $K_p=\frac{[PC{l}_5]}{[PC{l}_3][C{l}_2]}$ $0.0870=\frac{(0.20-x)}{((0.50+x)(0.50+x))}$

Solve for x

To solve for x, we can multiply both sides of the equation by the denominator and solve a quadratic equation: $0.0870(0.50+x{)}^2=0.20-x$ Solving the quadratic equation, we find $x=0.04598$.

Calculate equilibrium partial pressures

Now, we can calculate the equilibrium partial pressures for each gas: $PC{l}_3$ equilibrium = $0.50+0.04598=0.54598$ atm $C{l}_2$ equilibrium = $0.50+0.04598=0.54598$ atm $PC{l}_5$ equilibrium = $0.20-0.04598=0.15402$ atm c) Effect of increasing volume on the mole fraction of $C{l}_2$

Le Chatelier's Principle

According to Le Chatelier's Principle, if the volume of the system is increased, the reaction will shift towards the side with more moles of gas to counteract the change. In this case, since the reactants' side (left) has more moles, increasing the volume will shift the reaction to the left. The result of the shift to the left side is that the mole fraction of $C{l}_2$ in the equilibrium mixture will increase. d) Effect of increasing temperature on the mole fraction of $C{l}_2$

Le Chatelier's Principle for exothermic reaction

Since the reaction is exothermic, increasing the temperature of the system will cause the reaction to shift in the direction that absorbs more heat, i.e., towards the endothermic reaction. Here, the endothermic reaction is the reverse of the given reaction (i.e., decomposition of $PC{l}_5$ into $PC{l}_3$ and $C{l}_2$. As a result, increasing the temperature will cause the reaction to shift towards the left, leading to a higher mole fraction of $C{l}_2$ in the equilibrium mixture.

Key concepts

Understanding the Reaction Quotient (Q)

The reaction quotient, denoted as Q, offers a snapshot of a system's status: It tells us which direction a reaction must shift to achieve equilibrium. Given the initial partial pressures, Q is calculated similarly to the equilibrium constant K_p, but without the assumption that the reaction is at equilibrium.

The equation,
$Q=\frac{[PC{l}_5]}{[PC{l}_3][C{l}_2]}$,
requires concentrations or, as in the scenario of gases, partial pressures of reactants and products. Once you derive the value of Q and compare it to K_p, you can predict which way the reaction will shift: If Q is larger than K_p, the system will shift left, converting products into reactants. Conversely, if Q is smaller, the shift will be to the right, forming more products.

Le Chatelier's Principle and Its Impact

Le Chatelier's Principle is pivotal when adjusting to changes in a reaction at equilibrium. This principle states that if a system at equilibrium undergoes a change in concentration, temperature, or pressure, the equilibrium will shift to counteract the imposed change and a new equilibrium will be established.

Increasing volume leads to a decrease in pressure, and per this principle, the equilibrium shifts towards the side with more gaseous moles, to increase the pressure again. In the given problem, this means an increased mole fraction of Cl₂. Furthermore, for an exothermic reaction, raising temperature favors the endothermic direction—again a shift to the left, increasing the mole fraction of Cl₂, balancing the added heat.

The Role of Equilibrium Constant (Kp)

The equilibrium constant, K_p, is integral to predicting the extent of a reaction and its position at equilibrium. It applies to gas-phase reactions where partial pressures measurements are convenient. Defined as the ratio of the partial pressures of products to reactants raised to their stoichiometric coefficients, K_p indicates the favorability of a reaction:
$K_p=\frac{[PC{l}_5]}{[PC{l}_3][C{l}_2]}$.

A large K_p suggests a greater proportion of products at equilibrium, whereas a small value indicates a reaction that favors reactants. By setting up an equilibrium expression using K_p, as done in the given example, you can solve for unknown changes in concentration (or pressure) and calculate the system's new equilibrium state.

Partial Pressures in Reaction Equilibria

Partial pressures play a critical role when dealing with gaseous equilibria. The pressure exerted by each gas component in a mixture is its partial pressure, and it contributes to the total pressure of the system. Dalton's Law dictates that the sum of each gas's partial pressure equates to the overall pressure.

In the context of our equilibrium problem, changes in partial pressures directly correspond to shifts in the equilibrium condition. Calculating the equilibrium partial pressures as shown in the step by step solution allows you to measure not only where the equilibrium lies but also how perturbations, such as volume changes or temperature fluctuations, will shift the equilibrium — critical for comprehending the system's dynamic behavior.