At 600 $\circ$ C,the equilibrium constant for the reaction 2HgO(s) → 2Hg(l)+ O₂(g) Is 2.8.Calculate the equilibrium constant for the reaction 1/2O₂(g)+ Hg(l) → HgO(s)

Question

Quizplus · Accepted Answer

The equilibrium constant for the reverse reaction (the second given reaction) is the inverse of the equilibrium constant for the forward reaction (the first given reaction):

K' = 1/K = 1/2.8 = 0.36

Therefore, choice (D) is the correct answer.

At 600 ∘\circ∘ C,the Equilibrium Constant for the Reaction 2HgO(s) → 2Hg(l)+ O2(g)

At 600^$\circ$C,the Equilibrium Constant for the Reaction 2HgO(s) → 2Hg(l)+ O₂(g)