Question

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

Answer 1

Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

A. is greater than Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium.

4) The concentration of N2 will:

B. decrease.

Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

$ln(K)=\frac{\Delta _RG}{RT}$

2) The value of Qc:

A. is greater than Kc, since the reaction quotient is:

$Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}$

Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.

4) The concentration of N2 will:

B. decrease, since less reactant is forming the products.

Best regards.

Answer 2

Answer:

1) C. remains the same.

2)A. is greater than Kc.

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

4)B. decrease.

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