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ChemistryRates and equilibrium

Rates and equilibrium calculator

A rates and equilibrium calculator does the number work of chemical kinetics and equilibria: rates from a rate equation, orders from experiments, activation energies, and equilibrium constants. Our rates and equilibrium calculator has four tools. Rate uses rate = k[A]m[B]n to find the rate or k with its units. Orders works out the orders from a table of initial rates. Arrhenius finds k, or the activation energy from two temperatures. Kc builds an ICE table to find Kc, or the equilibrium concentrations from Kc.

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Rates and Equilibrium Calculator

Work with rate equations, the Arrhenius equation and Kc

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How to use the calculator

  1. Open the Rates and Equilibrium Calculator and choose Rate, Orders, Arrhenius or Kc.
  2. Rate: type k, the concentrations of A and B, and their orders, to get the rate; or type the rate to get k.
  3. Orders: type up to 6 experiments, each with [A], [B] and the initial rate. Change one concentration at a time between experiments.
  4. Arrhenius: give A, Ea and T for k, or two temperatures with their rate constants for Ea.
  5. Kc: type a reversible equation with state symbols, using <=> for the double arrow, and the starting concentrations. Then give one equilibrium concentration to find Kc, or Kc to find every equilibrium concentration. Presets include the Haber process and esterification.

Every answer has the working, the units, and a graph or table: a rate curve, an Arrhenius plot of ln k against 1/T, or the ICE table. For practice questions with worked answers, press Random problem.

Rate equations and the units of k

\text{rate} = k[\text{A}]^{m}[\text{B}]^{n}m and n are the orders with respect to A and B; m + n is the overall order

Orders are found by experiment, not from the balanced equation. The units of k depend on the overall order, so that the rate always comes out in mol dm⁻³ s⁻¹:

Overall orderUnits of k
0mol dm⁻³ s⁻¹
1s⁻¹
2dm³ mol⁻¹ s⁻¹
3dm⁶ mol⁻² s⁻¹

mol dm⁻³ is the same as mol/L, or M, in the US. For more on rate laws, see the rate equation guide.

Worked examples you can check in the calculator

Example 1: the rate from the rate equation

The calculator opens on k = 0.2, [A] = 0.5 mol dm⁻³ (second order) and [B] = 0.1 mol dm⁻³ (first order). Rate = 0.2 × 0.5² × 0.1 = 0.005 mol dm⁻³ s⁻¹. The overall order is 3, so k is in dm⁶ mol⁻² s⁻¹.

Example 2: k from a measured rate

A reaction is first order in A and first order in B. With [A] = 0.3 and [B] = 0.2 mol dm⁻³, the rate is 0.0036 mol dm⁻³ s⁻¹. Choose to find k: k = 0.0036 ÷ (0.3 × 0.2) = 0.06 dm³ mol⁻¹ s⁻¹.

Example 3: orders from initial rates

Experiment[A] (mol dm⁻³)[B] (mol dm⁻³)Initial rate (mol dm⁻³ s⁻¹)
10.20.10.0015
20.40.10.0030
30.20.30.0135

From 1 to 2, only [A] doubles and the rate doubles: 2 = 2m, so m = 1. From 1 to 3, only [B] triples and the rate goes up 9 times: 9 = 3n, so n = 2. The rate equation is rate = k[A][B]², and from experiment 1, k = 0.0015 ÷ (0.2 × 0.1²) = 0.75 dm⁶ mol⁻² s⁻¹.

Try a rate equation question, then press Mark to check your answer in the Reaction Kinetics course.

Use rate = k[A] or k[A] squared to work out a rate or the rate constant.

Press Mark to have your answer marked in the Reaction Kinetics course.

Example 4: activation energy from two temperatures

k is 0.012 at 300 K and 0.048 at 320 K. Choose Arrhenius and find Ea:

E_a = \frac{8.314 \times \ln\left(0.048 \div 0.012\right)}{\frac{1}{300} - \frac{1}{320}} = 55\ 323\ \text{J mol}^{-1} = 55.3\ \text{kJ mol}^{-1}

The calculator also gives the gradient of the Arrhenius plot, −Ea/R = −6,654 K, and the pre-exponential factor, A = 5.15 × 10⁷. The Arrhenius equation guide explains the plot.

Example 5: Kc from an ICE table

0.5 mol dm⁻³ of N₂O₄ is sealed in a flask and comes to equilibrium with NO₂: N₂O₄(g) ⇌ 2NO₂(g). At equilibrium [NO₂] = 0.1 mol dm⁻³. Choose Kc, type the equation and 0.5 and 0, and give [NO₂] = 0.1. Making 0.1 of NO₂ used 0.05 of N₂O₄, so [N₂O₄] = 0.45:

K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{0.1^2}{0.45} = 0.0222\ \text{mol dm}^{-3}

Example 6: equilibrium amounts from Kc

Ethanoic acid and ethanol make ethyl ethanoate and water, with Kc = 4. Start with 2 of the acid and 1 of the alcohol. Choose to find the amounts: the calculator writes each concentration in terms of x, solves Kc = x² ÷ ((2 − x)(1 − x)) = 4, and finds x = 0.8453. At equilibrium there is 0.845 of the ester and of water, 1.155 of acid and 0.155 of alcohol. Using extra acid pushed far more of the alcohol into ester than an equal mixture would.

practice
Pick an answer, then check it.
Doubling [A] makes the initial rate 4 times bigger, with everything else the same. What is the order with respect to A?

What the calculator doesn't do

  • Kp, the equilibrium constant in partial pressures. Use partial pressures in the same products-over-reactants expression by hand.
  • More than two reactants in a rate equation or an orders table.
  • Integrated rate laws and half-lives, such as [A] against time for a first order reaction.
  • Le Chatelier predictions. It calculates; it doesn't say which way a change of pressure will shift the position. Compare Qc with Kc to judge direction.

Common mistakes

  • Taking orders from the balanced equation. They come from experiment.
  • Forgetting the coefficients as powers in Kc. 2NO₂ means [NO₂]².
  • Putting pure solids into Kc. They are left out; the calculator does this when you give state symbols.
  • Temperatures in °C in the Arrhenius equation. Use kelvin.
  • kJ and J. R is 8.314 J mol⁻¹ K⁻¹, so Ea comes out in J mol⁻¹ before you divide by 1,000.

Common questions

How do you calculate Kc?
Find the equilibrium concentrations, often with an ICE table, then divide the products by the reactants, each to the power of its number in the balanced equation.
What is an ICE table?
A table of Initial, Change and Equilibrium concentrations. The changes follow the ratio of the balanced equation.
How do you work out the units of k?
They depend on the overall order: s⁻¹ for first order, dm³ mol⁻¹ s⁻¹ for second, dm⁶ mol⁻² s⁻¹ for third.
How do you find orders of reaction from data?
Compare two experiments where only one concentration changes. If doubling it doubles the rate, it is first order; if it quadruples the rate, second order; if the rate doesn't change, zero order.
How do you find activation energy from two temperatures?
Multiply R by the natural log of the ratio of the two rate constants, then divide by the difference between 1/T at the lower temperature and 1/T at the higher one, with T in kelvin.

Revision cards

Read the front, say your answer, then flip to check.

Where to go next

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