Questions: A chemical engineer is studying the following reaction: BF3(aq) + NH3(aq) → BF3NH3(aq) At the temperature the engineer picks, the equilibrium constant Kc for this reaction is 1.6. The engineer charges ("fills") three reaction vessels with boron trifluoride and ammonia, and lets the reaction begin. He then measures the composition of the mixture inside each vessel from time to time. His first set of measurements are shown in the table below. Predict the changes in the compositions the engineer should expect next time he measures the compositions. reaction vessel compound concentration expect change in con entration A BF3 0.93 M increase decrease 0 (no change) NH3 0.60 M increase decrease (no change) BF3NH3 0.90 M increase decrease (no change) B BF3 1.16 M increase decrease (no change) NH3 0.75 M increase decrease (no change) BF3NH3 1.41 M increase decrease (no change) C BF3 0.66 M increase decrease (no change) NH3 0.33 M increase decrease (no change) BF3NH3 1.17 M increase decrease (no change)

Solution Steps

The reaction given is:

\[ \mathrm{BF}_{3}(\mathrm{aq}) + \mathrm{NH}_{3}(\mathrm{aq}) \rightarrow \mathrm{BF}_{3}\mathrm{NH}_{3}(\mathrm{aq}) \]

The equilibrium constant \( K_c \) for this reaction is 1.6. This constant is defined as:

\[ K_c = \frac{[\mathrm{BF}_{3}\mathrm{NH}_{3}]}{[\mathrm{BF}_{3}][\mathrm{NH}_{3}]} \]

For vessel A, the concentrations are:

• \([\mathrm{BF}_{3}] = 0.93 \, \text{M}\)
• \([\mathrm{NH}_{3}] = 0.60 \, \text{M}\)
• \([\mathrm{BF}_{3}\mathrm{NH}_{3}] = 0.90 \, \text{M}\)

Calculate the reaction quotient \( Q_c \):

\[ Q_c = \frac{0.90}{0.93 \times 0.60} = \frac{0.90}{0.558} \approx 1.6122 \]

Since \( Q_c > K_c \), the reaction will shift to the left, decreasing \([\mathrm{BF}_{3}\mathrm{NH}_{3}]\) and increasing \([\mathrm{BF}_{3}]\) and \([\mathrm{NH}_{3}]\).

For vessel B, the concentrations are:

• \([\mathrm{BF}_{3}] = 1.16 \, \text{M}\)
• \([\mathrm{NH}_{3}] = 0.75 \, \text{M}\)
• \([\mathrm{BF}_{3}\mathrm{NH}_{3}] = 1.41 \, \text{M}\)

Calculate the reaction quotient \( Q_c \):

\[ Q_c = \frac{1.41}{1.16 \times 0.75} = \frac{1.41}{0.87} \approx 1.6207 \]

Since \( Q_c > K_c \), the reaction will shift to the left, decreasing \([\mathrm{BF}_{3}\mathrm{NH}_{3}]\) and increasing \([\mathrm{BF}_{3}]\) and \([\mathrm{NH}_{3}]\).

For vessel C, the concentrations are:

• \([\mathrm{BF}_{3}] = 0.66 \, \text{M}\)
• \([\mathrm{NH}_{3}] = 0.33 \, \text{M}\)
• \([\mathrm{BF}_{3}\mathrm{NH}_{3}] = 1.17 \, \text{M}\)

Calculate the reaction quotient \( Q_c \):

\[ Q_c = \frac{1.17}{0.66 \times 0.33} = \frac{1.17}{0.2178} \approx 5.3720 \]

Since \( Q_c > K_c \), the reaction will shift to the left, decreasing \([\mathrm{BF}_{3}\mathrm{NH}_{3}]\) and increasing \([\mathrm{BF}_{3}]\) and \([\mathrm{NH}_{3}]\).

Final Answer