Given: A (g) + B (g) ⇋ C (g) + D (g) At equilibrium a 2.00 liter container was found to contain 1.60 moles of C, 1.60 moles of D, 0.50 moles of A and 0.50 moles of B. Calculate Kc. enter a number to 2 decimal places If 0.10 mole of B and 0.10 mole of C are added to this system, what is the value of Q? enter a number to 2 decimal places What will the new equilibrium concentration of A be? enter a number to 3 decimal places

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Chemistry

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5 December, 00:49

Given: A (g) + B (g) ⇋ C (g) + D (g) At equilibrium a 2.00 liter container was found to contain 1.60 moles of C, 1.60 moles of D, 0.50 moles of A and 0.50 moles of B. Calculate Kc. enter a number to 2 decimal places If 0.10 mole of B and 0.10 mole of C are added to this system, what is the value of Q? enter a number to 2 decimal places What will the new equilibrium concentration of A be? enter a number to 3 decimal places

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Answers (1)

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Porter Sweeney · Answer 1

Kc = 10.24

Q = 9.07

[A] = 0.262 mol/L

Explanation:

In a reversible reaction, the equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. When this happens, the concentrations remain constant. The ratio between the multiplication of the concentration of the products by the multiplication of the reactants (each concentration elevated by the substance's coefficient) is called Kc, the equilibrium constant.

The value of the Kc depends on the temperature, and the pure liquids and solids are considered to have concentration equal to 1 (because it's activity is equal to 1, and the activity is aproximated to the concentrantion). So, for the reaction given, the concentrations at the equilibrium are:

[A] = 0.50 moles / 2.00 liter = 0.25 mol/L

[B] = 0.50 moles / 2.00 liter = 0.25 mol/L

[C] = 1.60 moles / 2.00 liter = 0.80 mol/L

[D] = 1.60 moles / 2.00 liter = 0.80 mol/L

Kc = [C]*[D]/[A]*[B]

Kc = 0.8*0.8/0.25*0.25

Kc = 0.64/0.0625

Kc = 10.24

The value of Q, the reaction quotient, is calculated as the value of Kc, but now, with the concentrations at a certain time and not necessariy in equilibrium. The new concentrantions of B and C will be:

[B] = (0.50 + 0.10) / 2.00 = 0.3 mol/L

[C] = (1.60 + 0.10) / 2.00 = 0.85 mol/L

Q = [C]*[D]/[A]*[B]

Q = 0.85*0.8/0.25*0.3

Q = 0.68/0.075

Q = 9.07

Because more product was added, by the Le Chatelier's principle, the reaction will shift in order to consume C and D, and forms more A and B, and so the equilibrium will be achieved again, so, let's do an equilibrium chart:

A (g) + B (g) ⇄ C (g) + D (g)

0.25 0.3 0.85 0.8 Initial

+x + x - x - x Reacts (stoichiometry is 1:1:1:1)

0.25+x 0.3+x 0.85-x 0.8-x Equilibrium

Kc = (0.85-x) * (0.8-x) / (0.25+x) * (0.3+x)

10.24 = (0.68 - 1.65x + x²) / (0.075 + 0.55x + x²)

10.24x² + 5.632x + 0.768 = 0.68 - 1.65x + x²

9.24x² + 7.282x - 0.088 = 0

Solving by a graphic calculator, and knowing that x > 0 and x < 0.8

x = 0.012 mol/L

So, [A] = 0.25 + 0.012 = 0.262 mol/L