Question

At 250°C an equilibrium mixture of SbCl3(g), Cl2(g), and SbCl5(g) has the partial pressures 0.670 bar, 0.438 bar, and 0.228 bar, respectively. Calculate the new equilibrium pressures if the volume of the reaction vessel is doubled.

Answer 1

Explanation:

As the given reaction will be $SbCl_{3}(g) + Cl_{2}(g) \rightarrow SbCl_{5}$

Since, it is given that volume of reaction vessel is doubled. Hence, moles of species involved will also increase.

Therefore, the reaction equation will become as follows.

             $2SbCl_{3}(g) + 2Cl_{2}(g) \rightarrow 2SbCl_{5}$

As it is given that partial pressure of $SbCl_{3}$ is 0.670 bar, $Cl_{2}$ is 0.438 bar and $SbCl_{5}$ is 0.228 bar.

Expression to calculate new equilibrium pressure if the volume of reaction vessel is double is as follows.

                  $K_{p} = \frac{[P_{SbCl_{5}}]^{2}}{[P_{SbCl_{3}}]^{2}[P_{Cl_{2}}]^{2}}$

                                   = $\frac{(0.228)^{2}}{(0.670)^{2}(0.438)^{2}}$

                                   = $6.05 \times 10^{3}$

Thus, we can conclude that new equilibrium pressures if the volume of the reaction vessel is doubled is $6.05 \times 10^{3}$.