Question

The equilibrium constant Kc for the following reaction is 8.30. H2(g) + CO2(g) ↔ H2O(g) + CO(g) Initially, 0.40 moles each of H2 and CO2 are injected into a 9.00 L flask. Calculate the concentration of the CO(g) at equilibrium.

Answer 1

alculate the concentration of CO(g) at equilibrium, we can use the equilibrium constant expression and an ICE (Initial-Change-Equilibrium) table.

Given the reaction: H2(g) + CO2(g) ↔ H2O(g) + CO(g)

The equilibrium constant expression for the reaction is:

Kc = [CO(g)] * [H2O(g)] / [H2(g)] * [CO2(g)]

Initially, 0.40 moles each of H2 and CO2 are injected into a 9.00 L flask. Let's assume that at equilibrium, x moles of CO(g) and x moles of H2O(g) are formed. The initial concentrations of H2 and CO2 are 0.40 moles / 9.00 L = 0.0444 M (since the volume is given in liters).

Now, we can set up the ICE table:

scssCopy code        H2(g)   +   CO2(g)   ↔   H2O(g)   +   CO(g)
-------------------------------------------------------
Initial   0.0444 M     0.0444 M        0 M           0 M
Change    -x          -x             +x            +x
Equilibrium 0.0444 - x   0.0444 - x     x           x

We know that the equilibrium constant (Kc) is given as 8.30, so we can write the expression for Kc using the equilibrium concentrations:

Kc = [CO(g)] * [H2O(g)] / [H2(g)] * [CO2(g)] 8.30 = x * x / (0.0444 - x) * (0.0444 - x)

Now, we can solve for x:

8.30 = x^2 / (0.0444 - x)^2 8.30 * (0.0444 - x)^2 = x^2 8.30 * (0.0019744 - 0.0888x + x^2) = x^2 0.01638432 - 0.36864x + 8.3x^2 = x^2 7.3x^2 + 0.36864x - 0.01638432 = 0

Using the quadratic formula, we can solve for x:

x = [ -0.36864 ± √(0.36864^2 - 4 * 7.3 * (-0.01638432)) ] / (2 * 7.3)

x ≈ 0.0157 M (approximately)

Now, we can calculate the concentration of CO(g) at equilibrium:

[CO(g)] (at equilibrium) = x ≈ 0.0157 M

Therefore, the concentration of CO(g) at equilibrium is approximately 0.0157 M.