At 800 K,K_c for the following equilibrium is 4.2 × 10^-3.
2HgO(s)
2Hg(l) + O₂(g)
Suppose 82.3 g of mercury (II) oxide is placed in a sealed 2.50-L vessel at 800 K.What is the partial pressure of oxygen gas at equilibrium? (R = 0.0821 L ∙ atm/(K ∙ mol) )

Question

At 800 K,K_c for the following equilibrium is 4.2 × 10^-3.
2HgO(s) At 800 K,Kc for the following equilibrium is 4.2 × 10-3. 2HgO(s) 2Hg(l) + O2(g) Suppose 82.3 g of mercury (II) oxide is placed in a sealed 2.50-L vessel at 800 K.What is the partial pressure of oxygen gas at equilibrium? (R = 0.0821 L ∙ atm/(K ∙ mol) ) A) 8.7 atm B) 4.3 atm C) 0.27 atm D) 0.0042 atm E) 22 atm

At 800 K,K<sub>c</sub> for the following equilibrium is 4.2 × 10<sup>-3</sup>. 2HgO(s) 2Hg(l) + O<sub>2</sub>(g) Suppose 82.3 g of mercury (II) oxide is placed in a sealed 2.50-L vessel at 800 K.What is the partial pressure of oxygen gas at equilibrium? (R = 0.0821 L ∙ atm/(K ∙ mol) ) A) 8.7 atm B) 4.3 atm C) 0.27 atm D) 0.0042 atm E) 22 atm

2Hg(l) + O₂(g)
Suppose 82.3 g of mercury (II) oxide is placed in a sealed 2.50-L vessel at 800 K.What is the partial pressure of oxygen gas at equilibrium? (R = 0.0821 L ∙ atm/(K ∙ mol) )

Quizplus · Accepted Answer

First, we need to find the number of moles of HgO present in the reaction vessel.

Mass of HgO = 82.3 g

Molar mass of HgO = 216.59 g/mol

Number of moles of HgO = 82.3 g / 216.59 g/mol = 0.380 moles

From the balanced equation, we can see that the number of moles of oxygen gas produced is equal to the number of moles of HgO consumed. Thus, the equilibrium concentration of oxygen gas is also 0.380 moles/liter.

We can use the ideal gas law to find the partial pressure of oxygen gas:

PV = nRT

P = nRT/V

P = (0.380 mol) * (0.0821 L atm/(K mol)) * (800 K) / (2.50 L)

P = 0.266 atm

Therefore, the partial pressure of oxygen gas at equilibrium is 0.27 atm (choice C).

At 800 K,Kc for the Following Equilibrium Is 4

At 800 K,K_c for the Following Equilibrium Is 4