Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.2NH3(g) N2(g) + 3H2(g)If the TEMPERATURE on the equilibrium system is suddenly increased:The value of Kc A. increases.  B. decreases.  C. remains the same. The value of Q A. is greater than Kc.  B. is equal to Kc.  C. is less than Kc. The reaction must: A. run in the forward direction to reestablish equilibrium.  B. run in the reverse direction to reestablish equilibrium.  C. remain the same. It is already at equilibrium. The concentration of H2 will: A. increase.  B. decrease.  C. remain the same.

Asked Sep 24, 2019

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

2NH3(g) N2(g) + 3H2(g)

If the TEMPERATURE on the equilibrium system is suddenly increased:

The value of Kc   A. increases.
    B. decreases.
    C. remains the same.


The value of Q   A. is greater than Kc.
    B. is equal to Kc.
    C. is less than Kc.


The reaction must:   A. run in the forward direction to reestablish equilibrium.
    B. run in the reverse direction to reestablish equilibrium.
    C. remain the same. It is already at equilibrium.


The concentration of H2 will:   A. increase.
    B. decrease.
    C. remain the same.

Expert Answer

Step 1

The given problem can be solved by the use of Le Chatelier’s principle. Le Chatelier’s principle states that “ If a reaction is applied to some external disturbance it attempts to maintain equilibrium by opposing the change.”

Given that,

ΔH0 = 111 kJ/mol.

Kc = 6.30.

Since the given value of the enthalpy is positive,  the reaction is endothermic. To favor the reaction in the forward direction the heat should be added in the system.

On the equilibrium system, if the temperature is increased suddenly, the forward reaction is favored because the heat is added in the system. Kc will increase with an increase in temperature and decrease with a decrease in temperature for endothermic reactions.

Since the forward reaction is favored, the Kc value increases. Consequently, product concentration increases while reactant concentration decreases.

Thus, the option (A) is correct.

Step 2

Since the initial concentration of the reactant was greater than that of the present concentration, the value of Qc is less than Kc.

Thus, option (C) is correct.

Step 3

The reaction must run in the forward direction to re-establis...

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