Questions: Question (2 points) See section 14.5 A reaction vessel contains PCl3, Cl2, and PCl5 with partial pressures of 0.40, 0.40, and 0.60 atm, respectively. The reaction vessel is heated to 450 K where the equilibrium constant (Kp) for the reaction below is 3.9. PCl3(g) + Cl2(g) ⇌ PCl5(g) 1st attempt Part 1 (1 point) See Periodic Table See Hint What is the value of the reaction quotient (Qp)? Part 2 (1 point) See Hint The reaction would be expected to proceed toward

Question (2 points)
See section 14.5

A reaction vessel contains PCl3, Cl2, and PCl5 with partial pressures of 0.40, 0.40, and 0.60 atm, respectively. The reaction vessel is heated to 450 K where the equilibrium constant (Kp) for the reaction below is 3.9.
PCl3(g) + Cl2(g) ⇌ PCl5(g)

1st attempt

Part 1 (1 point)
See Periodic Table
See Hint

What is the value of the reaction quotient (Qp)?

Part 2 (1 point)
See Hint

The reaction would be expected to proceed toward

Transcript text: 17 Question (2 points) See section 14.5 A reaction vessel contains $\mathrm{PCl}_{3}, \mathrm{Cl}_{2}$, and $\mathrm{PCl}_{5}$ with partial pressures of $0.40,0.40$, and 0.60 atm , respectively. The reaction vessel is heated to 450 K where the equilibrium constant ( $K_{\mathrm{p}}$ ) for the reaction below is 3.9 . \[ \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{5(\mathrm{~g})} \] 1st attempt Part 1 (1 point) See Periodic Table See Hint What is the value of the reaction quotient $\left(Q_{p}\right)$ ? $\square$ Part 2 (1 point) See Hint The reaction would be expected to proceed toward $\square$

Solution

Solution Steps

Step 1: Calculate the Reaction Quotient $Q_p$

The reaction quotient $Q_p$ is calculated using the formula:

\[ Q_p = \frac{P_{\text{PCl}_5}}{P_{\text{PCl}_3} \times P_{\text{Cl}_2}} \]

where $P_{\text{PCl}_5}$, $P_{\text{PCl}_3}$, and $P_{\text{Cl}_2}$ are the partial pressures of $\text{PCl}_5$, $\text{PCl}_3$, and $\text{Cl}_2$, respectively.

Given:

$P_{\text{PCl}_5} = 0.60 \, \text{atm}$
$P_{\text{PCl}_3} = 0.40 \, \text{atm}$
$P_{\text{Cl}_2} = 0.40 \, \text{atm}$

Substitute these values into the formula:

\[ Q_p = \frac{0.60}{0.40 \times 0.40} = \frac{0.60}{0.16} = 3.75 \]

Step 2: Determine the Direction of the Reaction

Compare the calculated $Q_p$ with the given equilibrium constant $K_p$.

$Q_p = 3.75$
$K_p = 3.9$

Since $Q_p < K_p$, the reaction will proceed in the forward direction to reach equilibrium, meaning it will proceed toward the formation of more $\text{PCl}_5$.