Question

$2CaS{O}_4\left(s\right)\rightleftharpoons 2CaO\left(s\right)+2S{O}_2\left(g\right)+O2\left(g\right),\Delta H>0$

Above equilibrium is established by taking sufficient amount of $CaS{O}_4\left(s\right)$ in a closed container at 1600 K. Then, which of the following may be correct options?
(Assume that solid $CaS{O}_4$ is present in the container in each case).

• A. Moles of CaO(s) will increase will the increase in temperature.
• B. If the volume of the container is doubled at equilibrium then partial pressure of $S{O}_2\left(g\right)$ will change at new equilibrium.
• C. If the volume of the container is halved partial pressure of $O_2\left(g\right)$ at new equilibrium will remain same.
• D. If two mole of the He gas is added at constant pressure then the moles of CaO(s) will increases.

Answer 1

Answer: (A), (C), (D)

The correct options are
A

Moles of CaO(s) will increase will the increase in temperature.

C

If the volume of the container is halved partial pressure of $O_2\left(g\right)$ at new equilibrium will remain same.

D

If two mole of the He gas is added at constant pressure then the moles of CaO(s) will increases.

As the reaction is endothermic, it will move in the forward direction when heat is supplied. Hence, number of moles of CaO will increase.

With the increase or decrease of volume partial pressure of the gases will remain same. The individual pressures and the volume change, causing a shift in equilibrium, which then leads to reactions happening to reattain equilibrium.

Addition of an inert gas at constant pressure will reduce the partial pressure of all gases. This shifts equilibrium and causes the reaction to go in a direction to increase the number of moles.