Question

A sealed reaction vessel initially contains 1.113×10-2 moles of water vapor and 1.490×10-2 moles of CO(g).

H20(g) +CO(g) <==> H2(g) + CO2(g)

After the above reaction has come to equilibrium, the vessel contains 8.326×10-3 moles of CO2(g).

What is the value of the equilibrium constant Kc of the reaction at the temperature of the vessel?

Answer 1

Kc =
`([8.326x10-3]^(1) )/([1.113x10-2]^(1)[1.490x10-2]^(1) )`

Kc = 50.2059

Step-by-step explanation:

1. Balance the equation

2. Use the Kc formula

Remember that pure substances, like H2 are not included on the Kc formula

Answer 2

The equilibrium constant for the reaction at the given temperature is approximately 2.0, derived from the concentrations of the substances at equilibrium.

Certainly! Let's denote the change in moles of CO2 as x at equilibrium. The initial concentrations are given as:

[H2O]0 = 1.113 × 10^(-2) mol

[CO]0 = 1.490 × 10^(-2) mol

[H2]0 = 0 mol

[CO2]0 = 0 mol

At equilibrium, the concentrations become:

[H2O]eq = [H2O]0 - x

[CO]eq = [CO]0 - x

[H2]eq = [H2]0 + x

[CO2]eq = [CO2]0 + x

Given that [CO2]eq = 8.326 × 10^(-3) mol, we can express x in terms of this value:

x = [CO2]eq - [CO2]0

x = 8.326 × 10^(-3) - 0

x = 8.326 × 10^(-3) mol

Now, substitute the values into the equilibrium expression for the chemical reaction:

Kc = [H2]eq * [CO2]eq / [H2O]eq * [CO]eq

Kc = (0 + 8.326 × 10^(-3)) * (8.326 × 10^(-3)) / (1.113 × 10^(-2) - 8.326 × 10^(-3)) * (1.490 × 10^(-2) - 8.326 × 10^(-3))

After performing the calculations using a calculator it is found that

Kc ≈ 2.