Reaction kinetics worksheet with answers

Free printable Reaction kinetics worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
t (min)C (mol/L)10 minC = ?

A first-order reaction A → products has a rate constant of 0.2 min⁻¹. The initial concentration of A is 1.5 mol/L. Find the concentration CA after 10 min.

k = 0.2 min⁻¹CA0 = 1.5 mol/Lt = 10 min
CA = mol/L
2.Easy
CA​ (mol/L)r2r = ?CB​ = 0.5 mol/L

The reaction A + B → C follows the rate law r = k·CA², with k = 0.005 L/(mol·s). Find the reaction rate r when CA = 2 mol/L and CB = 0.5 mol/L.

r = k·CA²k = 0.005 L/(mol·s)CA = 2 mol/LCB = 0.5 mol/L
r = mol/(L·s)
3.Medium
t (min)X (%)t = ?95%

A second-order reaction (−rA = k·CA²) in a batch reactor has k = 0.8 L/(mol·min) and starts at 1 mol/L. How long does it take to reach 95% conversion? Find t.

k = 0.8 L/(mol·min)CA0 = 1 mol/LX = 95 %
t = min
4.Medium
t (min)C (mol/L)40 min0.7

In a first-order reaction the concentration of A falls from 1 mol/L to 0.7 mol/L in 40 min. Find the rate constant k.

CA0 = 1 mol/LCA = 0.7 mol/Lt = 40 min
k = min⁻¹
5.Hard
Tk25 °C0.00265 °C0.035

The rate constant of a reaction is 0.002 min⁻¹ at 25 °C and 0.035 min⁻¹ at 65 °C. Find the activation energy Ea.

k1 = 0.002 min⁻¹T1 = 25 °Ck2 = 0.035 min⁻¹T2 = 65 °CR = 8.314 J/(mol·K)
Ea = kJ/mol
6.Hard
t (min)X (%)15 minX = ?

A first-order reaction has k = 0.002 min⁻¹ at 30 °C and an activation energy of 60 kJ/mol. The batch reactor is now run at 60 °C for 15 min. Find the conversion X reached.

k1 = 0.002 min⁻¹T1 = 30 °CEa = 60 kJ/molT2 = 60 °Ct = 15 minR = 8.314 J/(mol·K)
X = %

Answer key

  1. 1. CA = 0.203 mol/L
  2. 2. r = 0.02 mol/(L·s)
  3. 3. t = 23.75 min
  4. 4. k = 0.008917 min⁻¹
  5. 5. Ea = 59.98 kJ/mol
  6. 6. X = 22.58 %

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