Question

Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.

2NH3 (g) ⇆ N2 (g) + 3H2 (g)
When 0.38 moles of NH3 (g) are removed from the equilibrium system at constant temperature:
1) The value of Kc:

A. increases.
B. decreases.
C. remains the same.
2) The value of Qc:

A. is greater than Kc.
B. is equal to Kc.
C. is less than Kc.
3) The reaction must:

A. run in the forward direction to restablish equilibrium.
B. run in the reverse direction to restablish equilibrium.
C. remain the same. It is already at equilibrium.
4) The concentration of N2 will:

A. increase.
B. decrease.
C. remain the same.

Answer 1

1) C. remains the same.

2)A. is greater than Kc.

3) B. run in the reverse direction to restablish equilibrium.

4)B. decrease.

Step-by-step explanation:

Step 1: Data given

H° = 111 kJ/mol

Kc = 6.30

Temperature = 723 K

Step 2: The balanced equation

2NH3 (g) ⇆ N2 (g) + 3H2 (g)

Step 3: When 0.38 moles of NH3 (g) are removed from the equilibrium system at constant temperature:

ΔG = - RT ln K

⇒ with R = the gas constant

⇒ with T = temperature

The value of Kc will remain the same because R and T are constant

Step 4: Calculate the value of Q

Kc = [H2]³[N2] / [NH3]²

We will have less reactant, this means the value of Q will increase

Since Kc remains the same an Q increases:

Q is greater than Kc

Step 5: The reaction must:

Since we remove (O.38 moles) NH3, the equilibrium will shift to the left side, the side of the reactants. This is the reverse direction. By doing this there will be made more NH3, and finally the equilibrium will be restablished.

Step 6: The concentration of N2 will:

Since we remove (O.38 moles) NH3, the equilibrium will shift to the left side, the side of the reactants.

There will be made more reactants, and less products (N2 and H2)

The concentration of N2 will be decreased