The reaction given is:
\[
\mathrm{BF}_{3}(aq) + \mathrm{NH}_{3}(aq) \rightarrow \mathrm{BF}_{3}\mathrm{NH}_{3}(aq)
\]
The equilibrium constant \( K_c \) for this reaction is 1.4. The equilibrium constant expression for this reaction is:
\[
K_c = \frac{[\mathrm{BF}_{3}\mathrm{NH}_{3}]}{[\mathrm{BF}_{3}][\mathrm{NH}_{3}]}
\]
For vessel A, the concentrations are:
- \([\mathrm{BF}_{3}] = 0.99 \, \text{M}\)
- \([\mathrm{NH}_{3}] = 0.53 \, \text{M}\)
- \([\mathrm{BF}_{3}\mathrm{NH}_{3}] = 0.76 \, \text{M}\)
Calculate the reaction quotient \( Q_c \):
\[
Q_c = \frac{0.76}{0.99 \times 0.53} = \frac{0.76}{0.5247} \approx 1.448
\]
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:
- \([\mathrm{BF}_{3}] = 0.56 \, \text{M}\)
- \([\mathrm{NH}_{3}] = 0.10 \, \text{M}\)
- \([\mathrm{BF}_{3}\mathrm{NH}_{3}] = 1.19 \, \text{M}\)
Calculate the reaction quotient \( Q_c \):
\[
Q_c = \frac{1.19}{0.56 \times 0.10} = \frac{1.19}{0.056} \approx 21.25
\]
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:
- \([\mathrm{BF}_{3}] = 1.37 \, \text{M}\)
- \([\mathrm{NH}_{3}] = 0.74 \, \text{M}\)
- \([\mathrm{BF}_{3}\mathrm{NH}_{3}] = 1.46 \, \text{M}\)
Calculate the reaction quotient \( Q_c \):
\[
Q_c = \frac{1.46}{1.37 \times 0.74} = \frac{1.46}{1.0138} \approx 1.440
\]
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 \([\mathrm{NH}_{3}]\).
Vessel A: \([\mathrm{BF}_{3}]\) and \([\mathrm{NH}_{3}]\) will increase, \([\mathrm{BF}_{3}\mathrm{NH}_{3}]\) will decrease.
\[
\boxed{\text{BF}_3 \uparrow, \text{NH}_3 \uparrow, \text{BF}_3\text{NH}_3 \downarrow}
\]
Vessel B: \([\mathrm{BF}_{3}]\) and \([\mathrm{NH}_{3}]\) will increase, \([\mathrm{BF}_{3}\mathrm{NH}_{3}]\) will decrease.
\[
\boxed{\text{BF}_3 \uparrow, \text{NH}_3 \uparrow, \text{BF}_3\text{NH}_3 \downarrow}
\]
Vessel C: \([\mathrm{BF}_{3}]\) and \([\mathrm{NH}_{3}]\) will increase, \([\mathrm{BF}_{3}\mathrm{NH}_{3}]\) will decrease.
\[
\boxed{\text{BF}_3 \uparrow, \text{NH}_3 \uparrow, \text{BF}_3\text{NH}_3 \downarrow}
\]