Question

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

Answer 1

Answer:

For 1: The value of $K_c$ is 10.24

For 2: The value of $Q_c$ is 9.07

For 3: The new equilibrium concentration of A is 0.220 M

Explanation:

We are given:

Volume of the container = 2.00 L

Equilibrium moles of A = 0.50 moles

Equilibrium moles of B = 0.50 moles

Equilibrium moles of C = 1.60 moles

Equilibrium moles of D = 1.60 moles

We know that:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution}}$

For the given chemical reaction:

$A(g)+B(g)\rightleftharpoons C(g)+D(g)$

• For 1:

The expression of $K_c$ for above equation follows:

$K_c=\frac{[C][D]}{[A][B]}$

We are given:

$[A]_{eq}=\frac{0.50}{2.00}=0.25$

$[B]_{eq}=\frac{0.50}{2.00}=0.25$

$[C]_{eq}=\frac{1.60}{2.00}=0.8$

$[D]_{eq}=\frac{1.60}{2.00}=0.8$

Putting values in above equation, we get:

$K_c=\frac{0.8\times 0.8}{0.25\times 0.25}\\\\K_c=10.24$

Hence, the value of $K_c$ is 10.24

• For 2:

Added moles of B = 0.10 moles

Added moles of C = 0.10 moles

$Q_c$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

$Q_c=\frac{[C][D]}{[A][B]}$

Now,

$[A]=\frac{0.50}{2.00}=0.25$

$[B]=\frac{0.60}{2.00}=0.3$

$[C]=\frac{1.70}{2.00}=0.85$

$[D]=\frac{1.60}{2.00}=0.8$

Putting values in above equation, we get:

$Q_c=\frac{0.85\times 0.8}{0.25\times 0.3}\\\\Q_c=9.07$

Hence, the value of $Q_c$ is 9.07

• For 3:

Taking equilibrium constant as 10.24 for calculating the equilibrium concentration of A.

$K_c=10.24$

$[B]_{eq}=\frac{0.60}{2.00}=0.3$

$[C]_{eq}=\frac{1.70}{2.00}=0.85$

$[D]_{eq}=\frac{1.60}{2.00}=0.8$

Putting values in expression 1, we get:

$10.24=\frac{0.85\times 0.8}{[A]\times 0.3}$

$[A]_{eq}=\frac{0.85\times 0.8}{0.3\times 10.24}=0.220$

Hence, the new equilibrium concentration of A is 0.220 M