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The rate constants for a first-order reaction are found to be 2.76 × 10-5 s-1 at 25° and 6.65 × 10-4 s-1 at 50°C, respectively.

(a) Calculate the activation energy Ea in kilojoules per mole.
(b) Calculate the rate constant at 75°C.

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Answer: (a) To calculate the activation energy (Ea) for a first-order reaction, we can use the Arrhenius equation:

k = Ae^(-Ea/RT)

where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and e is the base of the natural logarithm.

We can rearrange the equation to solve for Ea:

Ea = -RT ln(k/A)

We'll use the rate constants at 25°C and 50°C:

k1 = 2.76 × 10-5 s-1 at 25°C

k2 = 6.65 × 10-4 s-1 at 50°C

T1 = 25 + 273.15 = 298.15 K

T2 = 50 + 273.15 = 323.15 K

Substituting these values into the equation for Ea:

Ea = -R * (T1) * ln(k1/A) = -R * (T2) * ln(k2/A)

Dividing both sides by -R:

T1 * ln(k1/A) = T2 * ln(k2/A)

Dividing both sides by T1 and T2:

ln(k1/A) / T1 = ln(k2/A) / T2

Taking the natural logarithm of both sides:

ln(k2/k1) = (T2/T1) * (ln(k1/A) / T1)

Substituting the values for k1, k2, T1, and T2:

ln(6.65 × 10-4 / 2.76 × 10-5) = (323.15 / 298.15) * ln(2.76 × 10-5 / A)

Solving for ln(k1/A):

ln(k1/A) = ln(6.65 × 10-4 / 2.76 × 10-5) / (323.15 / 298.15)

Using a logarithm calculator, we find that ln(k1/A) = 13.80.

Finally, substituting back into the equation for Ea:

Ea = -R * (T1) * ln(k1/A) = -R * 298.15 * 13.80 = 9,920 J/mol

So, the activation energy for this reaction is approximately 9.92 kJ/mol.

(b) To find the rate constant at 75°C, we can use the Arrhenius equation:

k = Ae^(-Ea/RT)

where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy (9,920 J/mol), R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin (75 + 273.15 = 348.15 K), and e is the base of the natural logarithm.

Substituting these values into the equation:

k = Ae^(-Ea/RT) = Ae^(-99

User Haseeb Rehman
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