Question

Consider the follig system at equilibrium where AH = -198 K dK. = 34.5, at 1.15X10 K. 2802(g) + O2(8) 2250() If the VOLUME of the equilibrium system is suddenly decreased at constant temperature: The value of Ke A. increases. B. decreases. C. remains the same. The value of Qc A. is greater than K B. is equal to Ke C. is less than K 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 number of moles of O2 will: A. increase. B. decrease. C. remain the same. Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

1)
Since temperature is constant, Kc will not change
Answer: C

2)
Since we are decreasing the volume, concentration will increase
Since we have more reactant than product and
Qc = [product]/reactant,
Increase in denominator is more and hence Qc would decrease.
Thus Qc becomes less than Kc
Answer: C

3)
Since Qc is less than Kc, the reaction moves in forward direction
Answer: A

4)
Since reaction runs in forward direction and O2 is reactant, its concentration would decrease.
Answer: B