Question

use the references to access important values if needed for this question.
consider the following system at equilibrium where δh = 111 kj, and kc = 6.30, at 723 k:
2nh₃(g) ⇌ n₂(g) + 3h₂(g)
when 0.25 moles of n₂(g) are added to the equilibrium system at constant temperature:
the value of kc
o increases
o decreases
o remains the same
the value of qc
o is greater than kc
o is equal to kc
o is less than kc
the reaction must
o run in the forward direction to reestablish equilibrium
o run in the reverse direction to reestablish equilibrium
o remain in the current position, since it is already at equilibrium
the concentration of n₂ will
o increase
o decrease
o remain the same

Explanation:

Step1: Effect of adding \(N_2\) on \(K_c\)

The equilibrium constant \(K_c\) depends only on temperature. Since the temperature is constant, \(K_c\) remains the same.

Step2: Effect of adding \(N_2\) on \(Q_c\)

The reaction quotient \(Q_c=\frac{[N_2][H_2]^3}{[NH_3]^2}\). When \(N_2\) is added, the numerator increases, so \(Q_c\) is greater than \(K_c\).

Step3: Direction of reaction

When \(Q_c > K_c\), the reaction shifts in the reverse direction to re - establish equilibrium (Le Chatelier's principle).

Step4: Concentration of \(N_2\)

Although the reaction shifts in the reverse direction, the amount of \(N_2\) added is more than the amount that is consumed in the reverse reaction. So the concentration of \(N_2\) increases.

Answer:

The value of \(K_c\): remains the same
The value of \(Q_c\): is greater than \(K_c\)
The reaction must: run in the reverse direction to reestablish equilibrium
The concentration of \(N_2\) will: increase