Question

Initially, 0.600 mol of A is present in a 2.00 L solution. 2A(aq) ⇌ 2B(aq) + C(aq) At equilibrium, 0.160 mol of C is present. Calculate K.

Answer 1

The equilibrium constant (K) for the reaction 2A(aq) ⇌ 2B(aq) + C(aq) is calculated using the initial and equilibrium concentrations of the reactants and products in the reaction. The steps involve determining the changes in concentration for A, B, and C, followed by applying the equilibrium expression to find K, which is 0.032 M.

Step-by-step explanation:

To calculate the equilibrium constant (K) for the equilibrium reaction 2A(aq) ⇌ 2B(aq) + C(aq), we can use the data provided.

Initially, we have 0.600 mol of A in a 2.00 L solution, resulting in an initial concentration [A] of 0.300 M.

• At equilibrium, we know that there are 0.160 mol of C in the same volume, so [C] is 0.080 M.
• Since the reaction shows the formation of 2B and C from 2A, for every mole of C formed, two moles of A are reacted.
• Hence, the change in the concentration of A is double that of C (Δ[A] = -2 × Δ[C]).
• This change in concentration of A is -2 × 0.080 M = -0.160 M.
• The equilibrium concentration of A is 0.300 M - 0.160 M = 0.140 M.
• The change in concentration of B is equal to the formation of C since B and C are formed in a 2:1 ratio:
• Δ[B] = +Δ[C] = 0.080 M, thus the equilibrium concentration of B is also 0.080 M.
• The expression for the equilibrium constant for this reaction is K = [B]²[C]/[A]².
• Substituting the equilibrium concentrations into this expression gives us K = (0.080 M)²(0.080 M)/(0.140 M)².

After calculating, we find that K = 0.032 M.

Answer 2

To calculate the equilibrium constant, K, for the given reaction 2A(aq) ⇌ 2B(aq) + C(aq), we can use an ICE table to track the changes in concentrations. By substituting the equilibrium values of [A], [B], and [C] into the equilibrium expression, we can solve for x, which will give us the equilibrium constant, K.

Step-by-step explanation:

To calculate the equilibrium constant, K, for the given reaction:

2A(aq) ⇌ 2B(aq) + C(aq)

We need to determine the concentrations of A, B, and C at equilibrium.

Given that the initial moles of A is 0.600 mol and the equilibrium moles of C is 0.160 mol, we can set up an ICE table to track the changes in concentrations:

Initial:

[A] = 0.600 mol

[B] = 0 mol

[C] = 0 mol

Change:

[A] decreases by x mol

[B] increases by 2x mol

[C] increases by x mol

Equilibrium:

[A] = 0.600 - x mol

[B] = 2x mol

[C] = 0.160 + x mol

We can plug these values into the equilibrium expression:

K = ([B]²[C]) / [A]²

Simplifying, we get:

K = (4x²(0.160 + x)) / (0.600 - x)²

Finally, we can substitute the equilibrium values of [A], [B], and [C] into the expression and solve for x, which will give us the equilibrium constant, K.