$2 {CaSO}_{4} (s) ⇌ 2 CaO (s) + 2 {SO}_{2} (g) + O_{2} (g), ΔH > 0$ . Above equilibrium is established by taking some amount of ${CaSO}_{4} (s)$ in a closed container at 1000 Then which of the following may be correct option?

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Moles of $CaO (s)$ will increase with temperature

If the volume of the container is doubled at equilibrium then partial pressure of ${SO}_{2} (g)$ will change at new equilibrium

If the volume of the container is halved pressure of $O_{2} (g)$ at new equilibrium will remain same

If two moles of the He gas is added at constant pressure then the moles of Ca O(s) will increase

answer is C, D, A.

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Detailed Solution

On changing volume; $P_{8} O_{2}$ or ${PO}_{2}$ does not change at equilibrium.

For endothermic reaction $(ΔH > 0)$ if temperature increases equilibrium shift towards the right. Therefore option A is correct.
For any change in the volume of the container is not going to change the equilibrium pressure at a given temperature if the ${CaSO}_{4}$ is available.

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