Sure, let’s solve this step-by-step.
Given:
- The rate constant [tex]\( k_{800K} \)[/tex] at 800 K is [tex]\( 3.241 \times 10^{-5} \, \text{s}^{-1} \)[/tex].
- The activation energy [tex]\( E_a \)[/tex] is [tex]\( 245 \, \text{kJ/mol} \)[/tex].
- We need to find the rate constant [tex]\( k_{990K} \)[/tex] at 990 K (which is [tex]\( 9.90 \times 10^2 \, \text{K} \)[/tex]).
- The universal gas constant [tex]\( R \)[/tex] is [tex]\( 8.314 \, \text{J/mol·K} \)[/tex] (Note: the activation energy needs to be in the same units as [tex]\( R \)[/tex], so convert [tex]\( E_a \)[/tex] from kJ to J).
1. Convert Activation Energy:
[tex]\[
E_a = 245 \, \text{kJ/mol} = 245 \times 10^3 \, \text{J/mol} = 245000 \, \text{J/mol}
\][/tex]
2. Write the Arrhenius Equation:
The Arrhenius equation for the temperature dependence of the rate constant is:
[tex]\[
k_2 = k_1 \exp \left[ \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \right]
\][/tex]
Where:
- [tex]\( k_1 \)[/tex] and [tex]\( k_2 \)[/tex] are the rate constants at temperatures [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex] respectively.
- [tex]\( T_1 = 800 \, \text{K} \)[/tex]
- [tex]\( T_2 = 990 \, \text{K} \)[/tex]
3. Substitute the Values:
[tex]\[
k_2 = 3.241 \times 10^{-5} \cdot \exp \left[ \frac{245000}{8.314} \left( \frac{1}{800} - \frac{1}{990} \right) \right]
\][/tex]
4. Calculate the Exponential Term:
First, calculate the difference in the reciprocals of the temperatures:
[tex]\[
\frac{1}{800} - \frac{1}{990}
\][/tex]
Calculate each term separately:
[tex]\[
\frac{1}{800} = 0.00125
\][/tex]
[tex]\[
\frac{1}{990} \approx 0.001010101
\][/tex]
Then the difference:
[tex]\[
0.00125 - 0.001010101 \approx 0.000239899
\][/tex]
Now calculate the exponent:
[tex]\[
\frac{245000}{8.314} \approx 29476
\][/tex]
[tex]\[
29476 \times 0.000239899 \approx 7.073
\][/tex]
Thus, the rate constant can be expressed as:
[tex]\[
k_2 = 3.241 \times 10^{-5} \cdot \exp(7.073)
\][/tex]
5. Calculate [tex]\( \exp(7.073) \)[/tex]:
[tex]\[
\exp(7.073) \approx 1176.25
\][/tex]
Therefore:
[tex]\[
k_2 \approx 3.241 \times 10^{-5} \times 1176.25 \approx 0.0381 \, \text{s}^{-1}
\][/tex]
So, the rate constant [tex]\( k_2 \)[/tex] at [tex]\( 990 \, \text{K} \)[/tex] is approximately [tex]\(0.0381 \, \text{s}^{-1}\)[/tex].
