To calculate the rate constant at \(20^\circ C\) given the rate constant at \(0^\circ C\) and the activation energy, we can use the Arrhenius equation in the form of the following relationship:
\[
\ln \left(\frac{K_2}{K_1}\right) = \frac{E_A}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)
\]
Where:
- \(K_1\) is the rate constant at temperature \(T_1\)
- \(K_2\) is the rate constant at temperature \(T_2\)
- \(E_A\) is the activation energy
- \(R\) is the universal gas constant
- \(T_1\) and \(T_2\) are the absolute temperatures in Kelvin
### Step 1: Convert temperatures to Kelvin
- \(T_1 = 0^\circ C = 273 \, K\)
- \(T_2 = 20^\circ C = 293 \, K\)
### Step 2: Identify given values
- \(K_1 = 7.87 \times 10^{-7} \, s^{-1}\)
- \(E_A = 103 \, kJ \, mol^{-1} = 103 \times 10^3 \, J \, mol^{-1}\)
- \(R = 8.314 \, J \, mol^{-1} \, K^{-1}\)
### Step 3: Substitute values into the equation
Substituting the values into the equation:
\[
\ln \left(\frac{K_2}{K_1}\right) = \frac{103 \times 10^3}{8.314} \left(\frac{1}{273} - \frac{1}{293}\right)
\]
### Step 4: Calculate the right-hand side
First, calculate \(\frac{1}{273} - \frac{1}{293}\):
\[
\frac{1}{273} \approx 0.003663 \, K^{-1}
\]
\[
\frac{1}{293} \approx 0.003414 \, K^{-1}
\]
\[
\frac{1}{273} - \frac{1}{293} \approx 0.003663 - 0.003414 = 0.000249 \, K^{-1}
\]
Now substitute this back into the equation:
\[
\ln \left(\frac{K_2}{K_1}\right) = \frac{103 \times 10^3}{8.314} \times 0.000249
\]
Calculating this gives:
\[
\frac{103000}{8.314} \approx 12313.2
\]
\[
\ln \left(\frac{K_2}{K_1}\right) \approx 12313.2 \times 0.000249 \approx 3.067
\]
### Step 5: Solve for \(K_2\)
Now, we can find \(K_2\):
\[
\frac{K_2}{K_1} = e^{3.067}
\]
Calculating \(e^{3.067}\):
\[
e^{3.067} \approx 21.4
\]
Thus,
\[
K_2 = 21.4 \times K_1 = 21.4 \times 7.87 \times 10^{-7}
\]
Calculating \(K_2\):
\[
K_2 \approx 1.684 \times 10^{-5} \, s^{-1}
\]
### Final Answer:
The rate constant at \(20^\circ C\) is approximately \(1.684 \times 10^{-5} \, s^{-1}\).
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