At 700 K,K_p for the following equilibrium is 5.6 $\times$ 10^-3.2HgO(s) 2Hg(l)+ O₂(g) Suppose 51.2 g of mercury(II)oxide is placed in a sealed 3.00-L vessel at 700 K.What is the partial pressure of oxygen gas at equilibrium? (R = 0.0821 L · atm/(K · mol))

Question

Quizplus · Accepted Answer

1. Calculate the number of moles of HgO:
n(HgO) = m/M = 51.2 g / 216.59 g/mol = 0.236 mol

2. Use the equilibrium constant expression to find the equilibrium concentrations of Hg and O:
KS = [Hg]^2 × [O]/[HgO]
At equilibrium, let x be the partial pressure of O, then:
KS = (2x)^2 × x / (0.236 - x)
Simplifying this equation gives:
KS(0.236 - x) = 4x^3
Rearranging and solving for x gives: 
x = 0.00556 atm

3. Check the units to make sure they match the given unit of atm. The units of KS are atm^-5, while the equilibrium concentrations are in atm. The units of KS may be checked by calculating KS in terms of the concentrations of Hg, O, and HgO at equilibrium. The units will cancel out, and we find that KS = 2.809 × 10^-25 atm^-5. So the units match the given unit of atm.

Therefore, Choice B (0.0056 atm) is the correct answer.

At 700 K,Kp for the Following Equilibrium Is 5 ×\times×

At 700 K,K_p for the Following Equilibrium Is
5 $\times$