A 2.50 L vessel is injected with 5.42 mols of C2H6, 8.90 mols of O2, 6.14 mols of H2O, and 2.67 mols of CO2.

Question

1. A 2.50 L vessel is injected with 5.42 mols of C₂H₆, 8.90 mols of O₂, 6.14 mols of H₂O, and 2.67 mols of CO₂. Set up but do not solve the equation to find equilibrium concentrations if K_c for this reaction is 3.92 x 10² and the unbalanced reaction is C₂H₆ (g) + O₂ (g) ⇌ CO₂ (g) + H₂O (g).

Answer 1

Here are the steps on their own if you would like to try it first with some guidance:

Step 1: Balance the reaction.

Step 2: Use Molarity = moles / Liters to find the initial concentrations.

Step 3: Calculate the reaction quotient Q. This is done the same way that the equilibrium constant is calculated.

Step 4: Compare Q to Kc. This determines the direction of the reaction. It also tells you in the upcoming ICE table which molecules will be increasing in concentration and which ones will be decreasing.

Step 5: Set up an ICE table.

Step 6: Substitute the expressions for the final concentrations into the Kc equation.

Solution:

Step 1: Balance the equation.

C2H6 (g) + O2 (g) ⇌ CO2 (g) + H2O (g) C=2, H=6, O=2 / C=1, H=2, O=3

C2H6 (g) + O2 (g) ⇌ 2CO2 (g) + H2O (g) C=2, H=6, O=2 / C=2, H=2, O=5

C2H6 (g) + O2 (g) ⇌ 2CO2 (g) + 3H2O (g) C=2, H=6, O=2 / C=2, H=6, O=7

C2H6 (g) + 3.5O2 (g) ⇌ 2CO2 (g) + 3H2O(g) C=2, H=6, O=7 / C=2, H=6, O=7

Balanced but not whole: Multiply all coefficients by 2 to get the smallest whole number ratio

2C2H6 (g) + 7O2 (g) ⇌ 4CO2 (g) + 6H2O(g) C=4, H=12, O=14 / C=4, H=12, O=14

2C2H6 (g) + 7O2 (g) ⇌ 4CO2 (g) + 6H2O(g) Balanced Equation

Step 2: Molarity = moles / Liters

Volume of container: 2.50 L

5.42 mols of C₂H₆ --> [C₂H₆] = 5.42 / 2.50 = 2.17 M

8.90 mols of O₂ -->[O₂ ] = 8.90 / 2.50 = 3.56 M

6.14 mols of H₂O -->[H₂O] = 6.14 / 2.50 = 2.46 M

2.67 mols of CO₂ -->[CO₂] = 2.67 / 2.50 = 1.07 M

Step 3: Calculate the reaction quotient.

Q = [products] / [reactants] At concentrations other than equilibrium

Q = [CO₂]⁴ [H₂O]⁶ / [O₂ ]⁷ [C₂H₆]²

Q = [1.07]⁴ [2.46]⁶ / [3.56]⁷ [2.17]²

Q = 0.00851

Step 4: Compare Q to Kc.

This will indicate which direction the reaction must proceed in order to reach equilibrium.

Q is smaller than Kc. On the number line, Q needs to move to the right to reach Kc. The reaction will proceed to the right. Reactants will be used up and products will be formed in order for Q to reach Kc.

<------Q--------------------------Kc------------>

0.00851 3.92 x 10²

-->

Step 5: ICE Table

[C₂H₆] [O₂ ] [CO₂] [H₂O]

Initial 2.17 3.56 1.07 2.46 Starting concentrations, in units of M

Change -2x -7x +4x +6x Step 4 determined that reactants are consumed (-x) and

products are formed (+x). The coefficients are from Step 1.

End 2.17-2x 3.56-7x 1.07+4x 2.46+6x These are the concentrations at equilibrium.

Step 6: Set up Kc:

Kc = [products] / [reactants]

Kc = [CO₂]⁴ [H₂O]⁶ / [O₂ ]⁷ [C₂H₆]² When concentrations are at equilibrium

3.92x10² = [1.07+4x]^4 [2.46+6x]^6 / [3.56 - 7x]^7 [2.17 - 2x]^2

This is the final expression for Kc at equilibrium.

Hope this helps! - Emily

Answer 2

Based on the concentration of reactants and products, the value of Q (reaction quotient) is much less than Kc, so the reaction will proceed to the right.

Balanced equation is…

2C2H6 + 7O2 <==> 4CO2 + 6H2O

Equilibrium concentrations will be

C2H6 = 2.17-2x

O2 = 3.56-7x

CO2 = 1.07+4x

H2O = 2.46+6x

Kc = 3.92x10² = (1.07+4x)⁴(2.46+6x)⁶ / (2.17-2x)²(3.56-7x)⁷