Question

A reaction has a rate constant of 0.012s ^ - 1 at 400.0 K and 0.691s ^ - 1 at 450.0 K.

Determine the activation barrier for the reaction.

Answer 1

Answer:

Ea ≈ 64.4 kJ/mol

Step by step Explanation:
We can use the Arrhenius equation to relate the rate constant (k) to the activation energy (Ea), the universal gas constant (R), and the temperature (T):

k = Ae^(-Ea/RT)

where A is the pre-exponential factor and R is the gas constant. Taking the natural logarithm of both sides of the equation, we get:

ln(k) = ln(A) - Ea/RT

We can use the given rate constants and temperatures to form two equations:

ln(0.012) = ln(A) - Ea/(R*400.0 K)

ln(0.691) = ln(A) - Ea/(R*450.0 K)

Subtracting the second equation from the first, we can eliminate ln(A) and solve for Ea:

ln(0.012/0.691) = (Ea/R)(1/400.0 K - 1/450.0 K)

Ea = -ln(0.012/0.691) * R/(1/400.0 K - 1/450.0 K)

Using R = 8.314 J/(mol K), we get:

Ea = -ln(0.012/0.691) * 8.314 J/(mol K) / (1/400.0 K - 1/450.0 K)

Ea ≈ 64.4 kJ/mol

Therefore, the activation barrier for the reaction is approximately 64.4 kJ/mol

Answer 2

The rate constant of a reaction is related to the activation energy (Ea) and the temperature (T) by the Arrhenius equation:

k = A * e^(-Ea/RT)

where k is the rate constant, A is the pre-exponential factor, R is the gas constant, and T is the absolute temperature.

Taking the natural logarithm of both sides of the equation, we get:

ln(k) = ln(A) - (Ea/RT)

We can use this equation to determine the activation energy for the reaction by comparing the rate constants at two different temperatures. For example, at 400.0 K and 450.0 K, we have:

ln(k1) = ln(A) - (Ea/RT1)

ln(k2) = ln(A) - (Ea/RT2)

where k1 is the rate constant at 400.0 K, k2 is the rate constant at 450.0 K, and RT1 and RT2 are the product of the gas constant and temperature at each temperature.

Taking the difference between the two equations, we get:

ln(k2/k1) = Ea/R * (1/RT1 - 1/RT2)

Solving for the activation energy (Ea), we get:

Ea = -R * ln(k2/k1) / (1/RT1 - 1/RT2)

Substituting the given values, we get:

Ea = -8.314 J/mol*K * ln(0.691/0.012) / (1/400.0 K - 1/450.0 K)

Ea = 93.8 kJ/mol

Therefore, the activation energy for the reaction is 93.8 kJ/mol.