Consider the following system at equilibrium where H° = 111 kJ, and Kc = 6.30, a
ID: 1047644 • Letter: C
Question
Consider the following system at equilibrium where H° = 111 kJ, and Kc = 6.30, at 723 K:
2NH3(g)--> N2(g) + 3H2(g)
If the TEMPERATURE on the equilibrium system is suddenly decreased:
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 reestablish equilibrium.
B. Run in the reverse direction to restablish equilibrium.
C. Remain the same. Already at equilibrium.
4- The concentration of H2 will:
A. Increase.
B. Decrease.
C. Remain the same.
Consider the following system at equilibrium where H° = 198 kJ, and Kc = 2.90×10-2, at 1.15×103 K.
2SO3(g) --> 2SO2(g) + O2(g)
If the VOLUME of the equilibrium system is suddenly DECREASED 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 reestablish equilibrium.
B. run in the reverse direction to reestablish equilibrium.
C. remain the same. It is already at equilibrium.
4- The number of moles of O2 will:
A. increase.
B. decrease.
C. remain the same.
Explanation / Answer
1)
reaction is an endothermic reaction
so increasing temperature will favout the forward reaction
so value of Kc will increase
Answer: A
2)
Since Kc value increases, Qc is less than Kc
Answer: C
3)
as explained in 1:
A. Run in the forward direction to reestablish equilibrium.
4)
since reaction goes in forward direction, concentration of H2 will increase
Answer: A
I am allowed to answer only 1 question at a time