Reaction Quotient Q vs K: Which Way Does the Reaction Go?
Class 11 equilibrium problems often give you a mixture of reactants and products and ask a simple question: will the reaction go forward, go backward, or stay as it is? The tool for this is the reaction quotient, Q. Compare it with the equilibrium constant K, and the direction becomes clear in one line. This page shows the method with worked numbers.
The idea
K is the value of the concentration ratio at equilibrium. Q is the same ratio calculated with the concentrations you have right now, at any moment. If Q differs from K, the system is not at equilibrium and it will move until Q becomes equal to K.
Qc = [C]c [D]d / ( [A]a [B]b ) (using the present concentrations)
Q < K → reaction goes forward (more products form)
Q = K → at equilibrium, no net change
Q > K → reaction goes backward (more reactants form)
How to remember it
Think of Q as "the products-to-reactants ratio now" and K as "the ratio it wants to reach". If the ratio now is too small (Q < K), there are too few products, so products are made. If it is too large (Q > K), there are too many products, so they turn back into reactants.
Rules for writing Q
- Products go in the numerator and reactants in the denominator, each raised to its coefficient.
- Pure solids and pure liquids are left out (their activity is taken as 1).
- Use molar concentrations for Qc. For gases you may use partial pressures to get Qp, but then compare with Kp, not Kc.
- Q and K must be for the same equation written the same way, at the same temperature.
Worked examples
Step 1: Qc = [HI]2 / ( [H2][I2] ).
Step 2: [HI]2 = 0.60 × 0.60 = 0.36. [H2][I2] = 0.20 × 0.10 = 0.020.
Step 3: Qc = 0.36 / 0.020 = 18.
Step 4: 18 < 50, so Q < K.
Answer: The reaction proceeds forward and more HI forms.
Step 1: Let x mol/L of H2 react. Then H2 = 0.20 − x, I2 = 0.10 − x, HI = 0.60 + 2x.
Step 2: (0.60 + 2x)2 = 50 (0.20 − x)(0.10 − x).
Step 3: Expand. Left: 0.36 + 2.4x + 4x2. Right: 50(0.02 − 0.30x + x2) = 1.0 − 15x + 50x2.
Step 4: Move all terms to one side: −46x2 + 17.4x − 0.64 = 0, or 46x2 − 17.4x + 0.64 = 0.
Step 5: x = [17.4 ± √(17.42 − 4 × 46 × 0.64)] / (2 × 46) = [17.4 ± √(302.76 − 117.76)] / 92 = [17.4 ± 13.60] / 92. The two roots are 0.0413 and 0.337.
Step 6: x cannot exceed 0.10 (I2 would go negative), so x = 0.0413.
Step 7: [H2] = 0.1587 M, [I2] = 0.0587 M, [HI] = 0.6826 M.
Check: Q = 0.68262 / (0.1587 × 0.0587) = 0.4659 / 0.009317 = 50.0 ✓.
Answer: At equilibrium [H2] = 0.159 M, [I2] = 0.059 M, [HI] = 0.683 M. Reject the second root because it gives negative concentrations.
Step 1: Qc = [PCl3][Cl2] / [PCl5] = (0.20 × 0.10) / 0.10 = 0.20.
Step 2: 0.20 > 0.040, so Q > K.
Step 3: The reaction goes backward. PCl3 and Cl2 combine to form more PCl5.
Step 4: To find the new equilibrium, let y mol/L of PCl5 form. Then PCl5 = 0.10 + y, PCl3 = 0.20 − y, Cl2 = 0.10 − y. Set (0.20 − y)(0.10 − y) = 0.040 (0.10 + y).
Step 5: y2 − 0.30y + 0.020 = 0.004 + 0.040y, so y2 − 0.34y + 0.016 = 0. y = [0.34 − √(0.1156 − 0.064)] / 2 = [0.34 − 0.2272] / 2 = 0.0564 (the other root, 0.284, exceeds 0.10 and is rejected).
Step 6: PCl5 = 0.1564 M, PCl3 = 0.1436 M, Cl2 = 0.0436 M.
Check: 0.1436 × 0.0436 / 0.1564 = 0.0400 ✓.
Answer: Backward reaction; the new equilibrium has [PCl5] ≈ 0.156 M.
Step 1: Solids are left out of Q, so Qc = [CO2].
Step 2: Suppose Kc = 0.0050 M at some temperature (given). If the CO2 now present is 0.0020 M, then Q = 0.0020 < 0.0050.
Step 3: The reaction goes forward and CaCO3 decomposes until [CO2] reaches 0.0050 M.
Answer: Forward. Note that the amount of solid does not change the answer, as long as some of each solid is present.
- Swapping the comparison. Q < K means forward, Q > K means backward. Write this on your rough sheet before the exam.
- Using equilibrium values for Q. Q uses the concentrations now. If you put equilibrium values in, you simply get K back.
- Including solids or pure liquids. They do not appear in Q or K.
- Forgetting the coefficients as powers. In 2HI, the HI concentration is squared, not multiplied by 2.
- Keeping a root that makes a concentration negative. In quadratic problems, check each root against the starting amounts.
- Mixing Kc and Kp. If Q is built from pressures, compare it with Kp.
Exam relevance
| Question type | Method |
|---|---|
| "In which direction will the reaction proceed?" | Compute Q, compare with K |
| "Is the mixture at equilibrium?" | Check if Q = K |
| "Find the equilibrium concentrations" | Decide the direction from Q, set up an ICE table, solve for x |
| Solid or liquid present | Leave it out of Q |
Check your own syllabus and sample papers to see how deep the numerical part goes. Do not assume a fixed number of marks for this topic.
No single tool in the suite computes Q, so work it out by hand and use the calculator for the arithmetic.
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