Question

If one starts with pure NO2(g) at a pressure of 0.500 atm, the total pressure inside the reaction vessel when 2NO2(g) ⇌ 2NO(g) + O2(g) reaches equilibrium is 0.674 atm. Calculate the equilibrium partial pressure of NO2.

a) 0.174 atm
b) 0.152 atm
c) 0.326 atm
d) 0.200 atm

Answer 1

The equilibrium partial pressure of NO2 in the reaction 2NO2(g) ⇌ 2NO(g) + O2(g) is 0.152 atm.The equilibrium partial pressure of NO2 would be 0.384 atm, which is obtained by subtracting the change in partial pressure. So option b is correct.

Step-by-step explanation:

For the reaction 2NO2(g) ⇌ 2NO(g) + O2(g), the equilibrium partial pressure of NO2 can be calculated using the given initial pressure of NO2 and the total pressure at equilibrium.

Let the equilibrium partial pressure of NO2 be x.

At equilibrium, the total pressure inside the reaction vessel is 0.674 atm, and the initial pressure of NO2 is 0.500 atm. Therefore, the change in the partial pressure of NO2 is x - 0.500.

Since there are 2 moles of NO2 in the balanced equation, the change in the partial pressure of NO2 is twice the change in the concentration of NO2. So, the change in the partial pressure of NO2 is 2(x - 0.500).

According to the law of partial pressures, the total pressure at equilibrium is the sum of the initial pressure and the change in the pressure at equilibrium. Therefore, we have:

0.674 atm = 0.500 atm + 2(x - 0.500)

Solving for x, we get x = 0.152 atm.

Therefore, the equilibrium partial pressure of NO2 is 0.152 atm.