Mr Cole Chemistry

Arrhenius Equation Questions

Arrhenius Calculations

Rate Equations Worksheet

1. Calculation Practice

Calculate the following based on the information provided.
a)
Calculate the value of the rate constant at 25oC for a reaction that has an Arrhenius constant of 2.48×108 and an activation energy of 35.5kJmol-1.

T = 298 K
Ea = 35,500 J mol-1
k = Ae-Ea/RT = 2.48×108 × e-(35500)/(8.31 × 298)
k = 2.48×108 × 5.95×10-7
k = 147.5

b)
Calculate the value of the Arrhenius constant at 0oC given that k is 4.2×10-8 and the activation energy is 92.4kJmol-1.

T = 273 K
A = k / e-Ea/RT = 4.2×10-8 / e-(92400)/(8.31 × 273)
A = 4.2×10-8 / 2.04×10-18
A = 2.05 x 1010

c)
Calculate the value of the activation energy for a reaction if at 20oC, the value for k is 8.82×10-4 and A is 3.51×107.

T = 293 K
ln k = ln A – Ea/RT
Ea = RT(ln A – ln k) = 8.31 × 293 × ln(3.51×107 / 8.82×10-4)
Ea = 2434.83 × 24.41
Ea = 59437 J mol-1 = 59.4 kJ mol-1

d)
Calculate the value of the pre-exponential factor for a reaction that has a rate constant of 2.28×10-2 at 298k and an activation energy of 24kJmol-1.

A = k / e-Ea/RT = 2.28×10-2 / e-(24000)/(8.31 × 298)
A = 2.28×10-2 / 6.18×10-5
A = 369

e)
Calculate the temperature of the reaction where k equals 1.10×10-2, 4.00×109 equals Y (Pre-exponential factor), and Ea equals 88.0kJmol-1.

ln(A/k) = Ea/RT
T = Ea / (R × ln(A/k))
T = 88000 / (8.31 × ln(4.00×109 / 1.10×10-2))
T = 88000 / (8.31 × 26.62)
T = 398 K

2. Complex Calculations

2.
A reaction has an activation energy of 62.5kJmol-1 and a pre-exponential factor of 1.66×1010. Calculate the temperature increase needed to double the rate of reaction when compared to reacting at 25oC.

Step 1: Calculate k at 298 K (25oC)
k1 = 1.66×1010 × e-62500/(8.31×298) = 0.182

Step 2: Calculate T for double the rate (k2 = 0.364)
T2 = Ea / (R × ln(A/k2))
T2 = 62500 / (8.31 × ln(1.66×1010/0.364))
T2 = 62500 / (8.31 × 24.54) = 306.4 K

Step 3: Difference
Difference = 306.4 – 298 = 8.4 K (or 8.4oC)

3.
An experiment is completed twice at 298k, once with a catalyst and once without. The uncatalyzed reaction has an activation energy of 120kJmol-1. The pre-exponential factor is 1.5×1010. The reaction is 20 000 times faster with the catalyst. What is the activation energy of the catalysed reaction?

Ratio kcat / kuncat = 20,000
ln(20000) = (Ea,uncat – Ea,cat) / RT
9.903 = (120000 – Ea,cat) / (8.31 × 298)
24524 = 120000 – Ea,cat
Ea,cat = 120000 – 24524 = 95476 J mol-1
Ea,cat = 95.5 kJ mol-1

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