At 700 K, K_p for the following equilibrium is
5.6 × 10^-3. 2HgO(s) $\rightleftharpoons$ 2Hg(l) + O₂(g)
Suppose 43.1 g of mercury(II) oxide is placed in a sealed 4.00-L vessel at 700 K. What is the partial pressure of oxygen gas at equilibrium?
(R = 0.0821 L · atm/(K · mol) )

Question

At 700 K, K_p for the following equilibrium is
5.6 × 10^-3. 2HgO(s)

\rightleftharpoons

2Hg(l) + O₂(g)
Suppose 43.1 g of mercury(II) oxide is placed in a sealed 4.00-L vessel at 700 K. What is the partial pressure of oxygen gas at equilibrium?
(R = 0.0821 L · atm/(K · mol) )

Quizplus · Accepted Answer

The equilibrium constant Kp is given for the reaction, and since the reaction involves only gases, Kp = P(O2). Therefore, P(O2) = 5.6 × 10^-3 atm at equilibrium.

At 700 K, Kp for the Following Equilibrium Is ⇌\rightleftharpoons⇌ 2Hg(l) + O2(g) Suppose 43

At 700 K, K_p for the Following Equilibrium Is $\rightleftharpoons$ 2Hg(l) + O₂(g)
Suppose 43