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

To calculate the new equilibrium pressures when the volume of the reaction vessel is doubled, divide each partial pressure by 2. Therefore, the new equilibrium pressures will be 0.335 bar for SbCl3, 0.219 bar for Cl2, and 0.114 bar for SbCl5.

Step-by-step explanation:

To calculate the new equilibrium pressures when the volume of the reaction vessel is doubled, we can use the ideal gas law and the concept of partial pressures. According to the equation provided, SbCl5 (g) ⇌ SbCl3 (g) + Cl2 (g), we have the partial pressures of SbCl3, Cl2, and SbCl5 as 0.670 bar, 0.438 bar, and 0.228 bar, respectively, at 250°C. When the volume is doubled, the pressure is halved, so each partial pressure will be divided by 2. Therefore, the new equilibrium pressures will be 0.335 bar for SbCl3, 0.219 bar for Cl2, and 0.114 bar for SbCl5.