Consider the following system at equilibrium where Delta H degree = -111 kJ, and
ID: 505982 • Letter: C
Question
Consider the following system at equilibrium where Delta H degree = -111 kJ, and K_c = 0.159, at 723 K. N_2(g) + 3H_2(g) 2NH_3(g) If the VOLUME of the equilibrium system is suddenly increased at constant temperature: The value of K_c C A. increase. B. decreases. C. remains the same. The value of Q_c C A. is greater than K_c. B. is equal to K_c. C. is less than K_c. The reaction mutt: A 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 number of moles of H_2 will: A A. increase. B. decrease. C. remain the same.Explanation / Answer
We are increasing volume here
In other words we are trying to decrease pressure
so, according to Le Chatellier's principle,
Reaction will try to increase the pressure
Hence it will move in a direction which have more gaseous molecules
Here reactant has more gaseous molecule
So equilibrium will move to left
The value of Kc will be same since Temperature is constant
The value of Qc is greater than Kc, that is why the reaction has to move to left to establish equilibrium again
The reaction must run in the reverse direction to reestablish equilibrium
the number of moles of H2 will increase