Question

A mixture of CO (g) and excess O2(g) is placed in a 1.0 L reaction vessel at 100.0C and a total pressure of 1.50 atm. The CO is ignited and it burns according to the balanced equation below. After reaction, the vessel is cooled to 100.0C and the pressure of the mixture of product gases (O2(g) , CO2(g)) is 1.40 atm. What is the partial pressure of of CO2 in the product mixture

Answer 1

Partial pressure of of CO₂ in the product mixture is 0,20atm

Step-by-step explanation:

The balance equation is:

2CO(g) + O₂(g) → 2CO₂(g)

Total pressure of CO(g) and O₂(g) gases before reaction at 100,0°C and 1,0L is 1,50 atm. You can say:

X₁ + Y₁ = 1,50atm (1)

Where X₁ is initial partial pressure CO and Y₁ is initial partial pressure of O₂

After reaction partial pressures are:

X₂ = X₁ - 2n = 0; 2n = X₁

Y₂ = Y₁ - n

Z₂ = 2n

Where Z₂ is final partial pressure of CO₂

After reaction pressure at 100,0°C and 1,0L is 1,40 atm, that means:

1,40 atm = (Y₂ + Z₂)

1,40 atm = Y₁ - n + 2n

1,40atm = Y₁ + n

1,40 atm = Y₁ + X₁/2 (2)

Replacing (1) in (2)

1,40 atm = 1,50atm - X₁ + X₁/2

-0,10 atm = - X₁/2

0,20 atm = X₁.

As 2n = X₁; 2n = Z₂ = 0,20 atm

I hope it helps!