AP Chemistry · Free-response skill

An equilibrium calculation from start to finish

The move. Turn lab data into equilibrium concentrations with an ICE table, calculate K, and compare Q with K to predict which way a mixture reacts.

How to do it

  1. Write K first, with products over reactants and coefficients as exponents. Leave out any pure solid or liquid.
  2. Convert every amount to molarity before the ICE table: divide moles by the volume in liters.
  3. In the Change row, tie every species to one unknown with the coefficients: if a reactant with coefficient 3 falls by 3x, a product with coefficient 2 rises by 2x.
  4. Put equilibrium molarities, not starting ones, into K. Report K to the significant figures of the data.
  5. For a direction question, first compute Q from the mixture you are given, then compare it with K, then turn the comparison into a direction for the species asked about.

Practice prompt Try it

A student studies the gas-phase reaction 2NOCl(g)⇌2NO(g)+Cl2(g)2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + \text{Cl}_{2}(g) at a constant temperature. She places 1.000 mol of NOCl(g)\text{NOCl}(g) in an empty, rigid 2.00 L flask. When the flask reaches equilibrium, it contains 0.200 mol of NO(g)\text{NO}(g).

Work through parts a to e with the student’s data, showing each step of your arithmetic.

  1. Write KcK_c for this reaction in terms of the concentrations of the three gases (an expression, not a number).
  2. Work out the equilibrium concentration of Cl2\text{Cl}_{2} and of NOCl\text{NOCl} in the flask.
  3. Put your part b results to work: what is the value of KcK_c at this temperature?
  4. A second flask at the same temperature is filled so that [NOCl]=0.20[\text{NOCl}] = 0.20 M, [NO]=0.020[\text{NO}] = 0.020 M and [Cl2]=0.010[\text{Cl}_{2}] = 0.010 M. What is Q for this mixture?
  5. Is Q from part d larger or smaller than your KcK_c, and so does the amount of NO in the second flask go up or down as it settles to equilibrium?

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 7 points

Model answer, annotated

(a) Kc=[NO]2[Cl2][NOCl]2K_c = \frac{[\text{NO}]^2[\text{Cl}_{2}]}{[\text{NOCl}]^2}

  • Point 1 Products over the reactant, with the coefficients 2, 2 and 1 as exponents.

(b) Initial: [NOCl]=1.000/2.00=0.500[\text{NOCl}] = 1.000/2.00 = 0.500 M; no products. At equilibrium [NO]=0.200/2.00=0.100[\text{NO}] = 0.200/2.00 = 0.100 M, so 2x = 0.100 M and x = 0.0500 M. Then [Cl2]=x=0.0500[\text{Cl}_{2}] = x = 0.0500 M and [NOCl]=0.500−0.100=0.400[\text{NOCl}] = 0.500 - 0.100 = 0.400 M.

  • Point 2 Chlorine has coefficient 1, so it rises by x, half of the NO change.
  • Point 3 NOCl falls by 2x, the same amount that NO rises, because both have coefficient 2.

(c) Kc=(0.100)2(0.0500)(0.400)2=5.00×10−40.160=3.125×10−3K_c = \frac{(0.100)^2(0.0500)}{(0.400)^2} = \frac{5.00 \times 10^{-4}}{0.160} = 3.125 \times 10^{-3}, or 3.1×10−33.1 \times 10^{-3} to the precision of the data.

  • Point 4 Uses the equilibrium row of the ICE table, not the starting amount of NOCl.

(d) Q=(0.020)2(0.010)(0.20)2=4.0×10−60.040=1.0×10−4Q = \frac{(0.020)^2(0.010)}{(0.20)^2} = \frac{4.0 \times 10^{-6}}{0.040} = 1.0 \times 10^{-4}.

  • Point 5 Same expression as K, filled in with the second mixture’s concentrations.

(e) Q (1.0×10−41.0 \times 10^{-4}) is smaller than KcK_c (3.1×10−33.1 \times 10^{-3}). So the net reaction runs forward: NOCl breaks down and the amount of NO goes up until Q rises to KcK_c.

  • Point 6 States the comparison of Q with K plainly.
  • Point 7 Turns the comparison into a direction for NO.