Questions: Reaction Quotient (Q) and Equilibrium Comparison

Question 1

For the reaction A(g) ⇌ B(g), K = 10. You measure [A] = 2 M and [B] = 30 M. What happens next?

• A. The reaction shifts forward because products are present in large amounts
• B. The reaction shifts in reverse because Q (= 15) is greater than K (= 10)
• C. The reaction is at equilibrium because both species are present
• D. The reaction shifts forward because more reactant must be consumed

Answer: (B)

Q = [B]/[A] = 30/2 = 15. Since Q > K (15 > 10), the product-to-reactant ratio exceeds the equilibrium ratio — there are too many products relative to where the system needs to end up. The reaction runs in reverse, consuming B and regenerating A, until Q falls back to 10. Options A and D describe the opposite direction; option C confuses 'both species present' with Q = K.

Question 2

A reaction is at equilibrium (Q = K). A student adds excess product, then claims 'the equilibrium constant K has increased because there are more products.' What is the correct analysis?

• A. The student is correct; more products always increase K
• B. K depends only on temperature and is unchanged; adding product raises Q above K, so the reaction shifts in reverse until Q falls back to K
• C. Adding product shifts the reaction forward, so both Q and K increase equally
• D. K increases only when temperature rises; concentration changes increase Q but not K

Answer: (B)

K is a thermodynamic constant that depends only on temperature, not on concentrations. Adding product increases Q (more products in the numerator) above the fixed K value, making Q > K. This drives the reverse reaction until equilibrium is restored at Q = K. Option D gets part of it right (K depends on T) but implies concentration can't change Q — it can and does, which is the whole point of the Q framework.

Question 3

Q and K use the same mathematical expression (products over reactants raised to stoichiometric powers), but K is only meaningful at the equilibrium state.

• T. True
• F. False

Answer: True

Q and K are calculated identically, but Q can be computed at any instant using current concentrations. K is the specific value that Q takes when the system has reached equilibrium. K is fixed for a given temperature; Q changes as concentrations change. The power of Q is precisely that it can be evaluated before equilibrium to predict the direction of change.

Question 4

If Q < K for a reaction, the system is already at equilibrium and no net change occurs.

• T. True
• F. False

Answer: False

Q < K means the product-to-reactant ratio is currently below the equilibrium ratio — there are too many reactants relative to equilibrium. The forward reaction will proceed to generate more products, increasing Q until it equals K. Q = K is the equilibrium condition; Q < K and Q > K both indicate that the system is NOT at equilibrium.

Question 5

A system is at equilibrium. More reactant is added. Using Q and K, explain why the forward reaction now proceeds.

Model answer: Adding reactant increases the denominator in the Q expression, making Q smaller than K. Since Q < K, the product-to-reactant ratio is now below the equilibrium ratio — the system has 'too many' reactants relative to where it needs to end up. The forward reaction runs to consume reactants and produce products, raising Q back toward K.

This question connects Q directly to Le Chatelier's principle. Le Chatelier says the system shifts to relieve the stress of added reactant — but Q vs. K shows why quantitatively. The addition lowers Q below K, and the forward reaction is the only way to restore Q = K. Every Le Chatelier prediction is a Q-vs-K comparison in disguise.