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At a particular temperature k=2.0×10-⁶ mol/L for the reaction. 2CO2 (g) <> 2CO (g) + O2 (g)

If 2.0 mol C02 is initially placed into a 5.0L vessel. Calculate the equilibrium concentration of all species using the ICE table of data given
2CO2 (g) <> 2CO (g) + O2 (g)
Initially partial pressure (mol/L) 0.4. 0 0
Charge in partial pressure (mol/L) _2× +2× +1
Equilibrium partial pressure (mol/L) 0.4_2 2× 1×

1 answer

To analyze the equilibrium concentrations for the reaction

\[ 2 , \text{CO}_2 (g) \rightleftharpoons 2 , \text{CO} (g) + \text{O}_2 (g) \]

we will use an ICE table (Initial, Change, Equilibrium).

Initial Conditions:

The reaction starts with 2.0 moles of \(\text{CO}_2\) in a 5.0 L vessel.

\[ \text{Initial concentrations}: \]

  • \([\text{CO}_2] = \frac{2.0 , \text{mol}}{5.0 , \text{L}} = 0.4 , \text{mol/L}\)
  • \([\text{CO}] = 0 , \text{mol/L}\)
  • \([\text{O}_2] = 0 , \text{mol/L}\)

Change:

Let \( x \) be the amount of \(\text{CO}_2\) that reacts (in mol/L). According to stoichiometry:

  • For every \( 2 \) moles of \(\text{CO}_2\) that react, \( 2 \) moles of \(\text{CO}\) are produced, along with \( 1 \) mole of \(\text{O}_2\).

Thus, the changes in concentrations will be:

  • Change in \([\text{CO}_2] = -x\)
  • Change in \([\text{CO}] = +x\)
  • Change in \([\text{O}_2] = +0.5x\) (since \( 2 \) moles of \(\text{CO}_2\) produce \( 1 \) mole of \(\text{O}_2\))

Equilibrium:

At equilibrium, the concentrations will be:

  • \([\text{CO}_2] = 0.4 - x\)
  • \([\text{CO}] = x\)
  • \([\text{O}_2] = 0.5x\)

Equilibrium Constant:

The equilibrium constant \( K \) has been given as \( 2.0 \times 10^{-6} , \text{mol/L} \).

For the reaction,

\[ K = \frac{[\text{CO}]^2 [\text{O}_2]}{[\text{CO}_2]^2} \]

Substituting in our expressions in terms of \( x \):

\[ K = \frac{x^2 (0.5x)}{(0.4 - x)^2} = \frac{0.5x^3}{(0.4 - x)^2} \]

Substituting \( K = 2.0 \times 10^{-6} \):

\[ 2.0 \times 10^{-6} = \frac{0.5x^3}{(0.4 - x)^2} \]

Solving for \( x \)

For simplification, we can assume \( x \) is small compared to \( 0.4 \), allowing us to approximate \( (0.4 - x) \approx 0.4 \):

\[ 2.0 \times 10^{-6} \approx \frac{0.5x^3}{(0.4)^2} \]

Calculating \( (0.4)^2 = 0.16 \):

\[ 2.0 \times 10^{-6} \approx \frac{0.5x^3}{0.16} \]

Solving for \( x^3 \):

\[ 0.5x^3 \approx 2.0 \times 10^{-6} \cdot 0.16 \] \[ 0.5x^3 \approx 3.2 \times 10^{-7} \] \[ x^3 \approx \frac{3.2 \times 10^{-7}}{0.5} = 6.4 \times 10^{-7} \] \[ x \approx (6.4 \times 10^{-7})^{1/3} \approx 8.64 \times 10^{-3} , \text{mol/L} \]

Equilibrium Concentrations:

Now, we substitute \( x \approx 8.64 \times 10^{-3} \) back into our expressions for equilibrium concentrations:

  • \([\text{CO}_2] = 0.4 - x \approx 0.4 - 0.00864 \approx 0.39136 , \text{mol/L}\)
  • \([\text{CO}] = x \approx 0.00864 , \text{mol/L}\)
  • \([\text{O}_2] = 0.5x \approx 0.5 \times 0.00864 \approx 0.00432 , \text{mol/L}\)

Summary of Equilibrium Concentrations:

  • \([\text{CO}_2] \approx 0.39136 , \text{mol/L}\)
  • \([\text{CO}] \approx 0.00864 , \text{mol/L}\)
  • \([\text{O}_2] \approx 0.00432 , \text{mol/L}\)

These are the equilibrium concentrations for all species in the reaction.