Question

The temperature dependence of the rate constant for a reaction is tabulated as follows: $$ \begin{array}{ll} \hline \text { Temperature (K) } & k\left(M^{-1} \mathrm{~s}^{-1}\right) \\ \hline 600 & 0.028 \\ 650 & 0.22 \\ 700 & 1.3 \\ 750 & 6.0 \\ 800 & 23 \\ \hline \end{array} $$ Calculate \(E_{a}\) and \(A\).

Short answer

The activation energy, \(E_a\), is 49,591 J/mol, and the pre-exponential factor, \(A\), is 9.65 × 10\(^{-7}\) M\(^{-1}\)s\(^{-1}\).

Step-by-step solution

Write down the Arrhenius equation

The Arrhenius equation relates the rate constant \(k\), activation energy \(E_{a}\), and pre-exponential factor \(A\) of a reaction to the temperature \(T\) as follows: \[ k = A e^{-E_a / RT} \] where R is the gas constant, R = 8.314 J\(K^{-1}mol^{-1}\), and T is the temperature in Kelvin.

Take natural logarithm of Arrhenius equation

To find the activation energy and the pre-exponential factor, take the natural logarithm of the Arrhenius equation to make it linear: \[ \ln k = \ln A - \frac{E_a}{RT} \]

Rearrange Arrhenius equation

Rearrange the linearized Arrhenius equation to the form y = mx + b, with y = ln(k) and x = 1/T: \[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \] In this equation, the slope of the graph ln(k) vs. 1/T is \(m = -E_a/R\).

Calculate ln(k) and 1/T for each data point

Use the given temperature and rate constant values to calculate ln(k) and 1/T for each data point: \[ \begin{array}{|c|c|c|} \hline \text{Temperature (K)} & \text{Rate constant } k\left( M^{-1}s^{-1} \right) & \ln (k) & \frac{1}{T} (K^{-1}) \\ \hline 600 & 0.028 & -3.5755 & 0.0016667 \\ 650 & 0.22 & -1.5141 & 0.0015385 \\ 700 & 1.3 & 0.2624 & 0.0014286 \\ 750 & 6.0 & 1.7918 & 0.0013333 \\ 800 & 23 & 3.1355 & 0.0012500 \\ \hline \end{array} \]

Calculate slope of the graph ln(k) vs. 1/T

Using any two points from the table, calculate the slope of the graph. We'll use the points (0.0016667, -3.5755) and (0.0012500, 3.1355): \[ m = \frac{3.1355 - (-3.5755)}{0.0012500 - 0.0016667} = -5961.223 \]

Calculate the activation energy, Ea

Using the slope of the graph and the gas constant R, calculate the activation energy: \[ E_a = -mR = 5961.223 \times 8.314 \, J/molK = 49,591 \, J/mol \]

Calculate the pre-exponential factor, A

To calculate A, choose one of the given temperature and rate constant values and substitute the values of \(E_a\), \(R\), \(T\), and \(k\) in the linearized Arrhenius equation: \[ \ln A = \ln k + \frac{E_a}{RT} \] Choose the first data point with T = 600 K and k = 0.028 M\(^{-1}\)s\(^{-1}\): \[ \ln A = \ln(0.028) + \frac{49,591 \, J/mol}{8.314 \, J/molK \times 600 \, K} = -13.884 \] Take the exponent of both sides to find the value of A: \[ A = e^{-13.884} = 9.65 \times 10^{-7} \, M^{-1}s^{-1} \] The activation energy, \(E_a\), is 49,591 J/mol, and the pre-exponential factor, \(A\), is 9.65 × 10\(^{-7}\) M\(^{-1}\)s\(^{-1}\).