NEET Chemistry Equilibrium: Q vs K and Common Traps

Master NEET Chemistry equilibrium with worked Q vs K examples, Kp–Kc checks and a focused mock test review routine.

Start with the question, not a memorised shift rule

In equilibrium questions, three different tasks often look similar: writing an equilibrium expression, deciding which direction a mixture will move, and predicting the effect of changing conditions. Keep them separate. A correct formula can still produce a wrong answer if you use concentrations from the wrong moment or treat a catalyst as a change in the equilibrium constant.

This guide uses original practice examples, not reproduced previous-year questions. Read the NCERT Equilibrium chapter alongside your notes, and verify the applicable syllabus and information bulletin through the official NEET website. Chapter weightage and exact future questions are not guaranteed.

Write the expression before substituting numbers

For aA + bB ⇌ cC + dD, the school-level concentration expression is Kc = [C]^c[D]^d / ([A]^a[B]^b). The coefficients become powers; they are not multipliers outside the brackets. At equilibrium, this expression gives Kc. Using the same expression for a mixture at another instant gives Qc, the reaction quotient.

For a specified reaction at a fixed temperature:

  • Qc < Kc: the net reaction proceeds towards products until equilibrium is approached.
  • Qc > Kc: the net reaction proceeds towards reactants.
  • Qc = Kc: the mixture is at equilibrium; forward and reverse reactions continue at equal rates.

Use the same expression and compatible conventions for Q and K. Thermodynamic equilibrium constants are formulated using activities; the concentration and partial-pressure expressions here are the usual school-level treatment.

Worked example 1: reaction direction

For H2(g) + I2(g) ⇌ 2HI(g), suppose Kc = 16 at the stated temperature. At one instant [H2] = 0.20 mol/L, [I2] = 0.20 mol/L and [HI] = 0.40 mol/L.

Qc = (0.40)^2 / (0.20 × 0.20) = 0.16 / 0.04 = 4.

Because 4 < 16, the net reaction moves forward. The mixture is not already at equilibrium merely because all three substances are present. This calculation predicts direction, not the final concentration of HI.

Worked example 2: the coefficient trap

For 2A(g) ⇌ B(g), let Kc = 8 and the instantaneous concentrations be [A] = 0.50 mol/L and [B] = 1.0 mol/L.

Qc = [B]/[A]^2 = 1.0/0.25 = 4. The forward direction is favoured. Using 1.0/(2 × 0.50) would give 1, but that expression is incorrect: the coefficient 2 is an exponent.

Count only gaseous species for Kp and Kc

In the ideal-gas school-level relation Kp = Kc(RT)^Δn, Δn is the total gaseous product coefficient minus the total gaseous reactant coefficient. Match the value of R to the units used and use temperature in kelvin.

For N2(g) + 3H2(g) ⇌ 2NH3(g), Δn = 2 − 4 = −2. Therefore Kp = Kc/(RT)^2. For H2(g) + I2(g) ⇌ 2HI(g), Δn = 0, so Kp = Kc under this convention.

Pure solids and pure liquids are omitted from the usual heterogeneous-equilibrium expression because their activities are taken as constant. For CaCO3(s) ⇌ CaO(s) + CO2(g), Kp is expressed using the CO2 partial pressure. Do not count the solid coefficients in Δn for the Kp–Kc relation.

Distinguish a shift from a change in K

At a fixed temperature, changing concentration or pressure can disturb Q and shift the equilibrium composition; it does not change K for that specified reaction. A catalyst speeds up the approach to equilibrium without changing K or the equilibrium composition.

Temperature is different. For an exothermic forward reaction, raising temperature favours the reverse direction and reduces its equilibrium constant. For an endothermic forward reaction, raising temperature favours products and increases its equilibrium constant.

Pressure statements also need conditions. Increasing pressure by decreasing the volume of an ideal gaseous mixture favours the side with fewer gas molecules if the gaseous coefficient totals differ. Adding inert gas at constant volume does not change the reacting gases' partial pressures. Do not use a bare “pressure increases” shortcut without reading how the change is made.

Try three checks before reading the answers

  1. For A(g) ⇌ B(g), Kc = 3, [A] = 0.20 mol/L and [B] = 0.60 mol/L. What is the net direction?
  2. For 2SO2(g) + O2(g) ⇌ 2SO3(g), what is Δn?
  3. A catalyst is added to a mixture already at equilibrium at fixed temperature. Does K increase?

Answers: (1) Qc = 0.60/0.20 = 3, so there is no net change in composition at equilibrium. (2) Δn = 2 − 3 = −1. (3) No. Both directions become faster, but K and the equilibrium composition are unchanged.

Turn mistakes into a short revision cycle

Use a suggested four-session cycle rather than repeatedly reading the whole chapter. First practise writing ten expressions without calculations. Next solve reaction-direction problems and state Q, K and the comparison explicitly. In the third session mix Kp–Kc, temperature and catalyst questions. Finally take a timed mixed set and reattempt mistakes later without viewing the solution.

In your error log, label each miss as expression, exponent, gaseous-species count, temperature condition, arithmetic or interpretation. Review the label with the largest number of errors before increasing speed. Our mock-test analysis guide explains how to make that review repeatable.

Move from topic practice to mixed papers through Scholareum mock tests and the test-series catalogue. Combine this routine with the NEET Physics graph guide and the NEET subject-wise mock strategy. Use the exam calendar as a planning aid and confirm dates with the official authority.

Frequently asked questions

Is Q calculated only at equilibrium?

No. Q describes the current mixture. It equals K when that mixture is at equilibrium under the stated conditions.

Does a large K mean the reaction happens quickly?

No. K describes equilibrium position, not reaction speed. A product-favoured equilibrium can still be reached slowly.

Should I memorise every numerical answer?

Memorise the expression-writing method and the conditions behind each rule. Recalculate original and official practice problems; recognising a number is not evidence that you can solve a fresh question.