Question

What is the pressure of \(BrCl\) in an equilibrium mixture of \(C{l_2}\), \(B{r_2}\), and \(BrCl\) if the pressure of \(C{l_2}\) in the mixture is \(0.115atm\)and the pressure of \(B{r_2}\) in the mixture is \(0.450atm\)?

\(C{l_2}(g) + B{r_2}(g) \rightleftharpoons 2BrCl(g)\)

\({K_P} = 4.7 \times 1{0^{ - 2}}\)

Short answer

The pressure ofBrCl is 0.049atm

Step-by-step solution

Given information

\(C{l_2}(g) + B{r_2}(g) \rightleftharpoons 2BrCl(g)\)

1. Value of equilibrium constant \({K_{\rm{P}}} = 4.7 \times 1{0^{ - 2}}\)
2. The pressure of \(C{l_2}\)in mixture is \(0.115\,atm\)
3. The pressure of \(B{r_2}\)in mixture is \(0.450\,atm\)

Determine the pressure of \(BrCl\)in mixture

\(\begin{array}{c}{K_P} = \frac{{{{\left( {{P_{BrCl}}} \right)}^2}}}{{{P_{C{l_2}}} \times {P_{B{r_2}}}}}\\{\left( {{P_{BrCl}}} \right)^2} = {P_{C{l_2}}} \times {P_{B{r_2}}} \times {K_P}\\{\left( {{P_{BrCl}}} \right)^2} = 4.7 \times {10^{ - 2}} \times 0.115 \times 0.450\\{\left( {{P_{BrCl}}} \right)^2} = 2.43 \times {10^{ - 3}}\end{array}\)

\(\begin{array}{l}\left( {{P_{BrCl}}} \right) = \sqrt {2.43 \times {{10}^{ - 3}}} \\\left( {{P_{BrCl}}} \right) = 0.049\,atm\end{array}\)