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Sagot :
Certainly! Let's solve the given question step by step.
Given:
The rate constant for the decomposition of ethylene oxide follows:
[tex]$ \log k \left( \text{in s}^{-1} \right) = 14.34 - \frac{1.25 \times 10^4}{T} $[/tex]
We need to calculate:
(i) The activation energy of the reaction.
(ii) The rate constant at 700 K.
(iii) The frequency factor, A.
### (i) Calculate the Activation Energy of the Reaction
The general form of the Arrhenius equation is:
[tex]$ \log k = \log A - \frac{E_a}{2.303RT} $[/tex]
Comparing with the given equation:
[tex]$ \log k = 14.34 - \frac{1.25 \times 10^4 K}{T} $[/tex]
Here, the term [tex]$\frac{1.25 \times 10^4 K}{T}$[/tex] represents the ratio [tex]$\frac{E_a}{2.303RT}$[/tex].
We need to find the activation energy [tex]\( E_a \)[/tex]:
Given:
[tex]\[ \frac{E_a}{2.303R} = 1.25 \times 10^4 K \][/tex]
[tex]\[ R = 8.314 \text{ J/mol·K} \][/tex]
Solving for [tex]\( E_a \)[/tex]:
[tex]\[ E_a = 1.25 \times 10^4 \times 2.303 \times 8.314 \][/tex]
[tex]\[ E_a = 1.25 \times 10^4 \times 19.14282 \][/tex]
[tex]\[ E_a = 239285.25 \text{ J/mol} \text{ or } 239.285 \text{ kJ/mol} \][/tex]
However, this value can be converted to Joules (as the provided answer is in Joules) simply.
Thus, the activation energy ([tex]\( E_a \)[/tex]) is:
[tex]\[ E_a = 239285.25 \text{ J/mol} \][/tex]
### (ii) Calculate the Rate Constant at 700 K
Now, using the given formula:
[tex]\[ \log k = 14.34 - \frac{1.25 \times 10^4}{700} \][/tex]
First, calculate the exponent:
[tex]\[ \frac{1.25 \times 10^4}{700} = 17.8571 \][/tex]
So,
[tex]\[ \log k = 14.34 - 17.8571 = -3.5171 \][/tex]
From logarithms, we know that:
[tex]\[ k = 10^{\log k} = 10^{-3.5171} \approx 3.843628408177594 \text{ s}^{-1}\][/tex]
Thus, the rate constant at 700 K is approximately:
[tex]\[ k \approx 3.843628 \text{ s}^{-1} \][/tex]
### (iii) Calculate the Frequency Factor, [tex]\( A \)[/tex]
From the provided equation:
[tex]\[ \log A = 14.34 \][/tex]
Taking the inverse logarithm:
[tex]\[ A = 10^{14.34} \approx 218776162394955.2 \text{ s}^{-1} \][/tex]
Thus, the frequency factor [tex]\( A \)[/tex] is:
[tex]\[ A = 2.187761623949552 \times 10^{14} \text{ s}^{-1} \][/tex]
### Summary of Results:
(i) The activation energy ([tex]\( E_a \)[/tex]) is 239285.0 J/mol.
(ii) The rate constant ([tex]\( k \)[/tex]) at 700 K is approximately 3.843628 s[tex]\(^{-1}\)[/tex].
(iii) The frequency factor ([tex]\( A \)[/tex]) is [tex]\( 2.187761623949552 \times 10^{14} \text{ s}^{-1} \)[/tex].
These are the final answers to the given question.
Given:
The rate constant for the decomposition of ethylene oxide follows:
[tex]$ \log k \left( \text{in s}^{-1} \right) = 14.34 - \frac{1.25 \times 10^4}{T} $[/tex]
We need to calculate:
(i) The activation energy of the reaction.
(ii) The rate constant at 700 K.
(iii) The frequency factor, A.
### (i) Calculate the Activation Energy of the Reaction
The general form of the Arrhenius equation is:
[tex]$ \log k = \log A - \frac{E_a}{2.303RT} $[/tex]
Comparing with the given equation:
[tex]$ \log k = 14.34 - \frac{1.25 \times 10^4 K}{T} $[/tex]
Here, the term [tex]$\frac{1.25 \times 10^4 K}{T}$[/tex] represents the ratio [tex]$\frac{E_a}{2.303RT}$[/tex].
We need to find the activation energy [tex]\( E_a \)[/tex]:
Given:
[tex]\[ \frac{E_a}{2.303R} = 1.25 \times 10^4 K \][/tex]
[tex]\[ R = 8.314 \text{ J/mol·K} \][/tex]
Solving for [tex]\( E_a \)[/tex]:
[tex]\[ E_a = 1.25 \times 10^4 \times 2.303 \times 8.314 \][/tex]
[tex]\[ E_a = 1.25 \times 10^4 \times 19.14282 \][/tex]
[tex]\[ E_a = 239285.25 \text{ J/mol} \text{ or } 239.285 \text{ kJ/mol} \][/tex]
However, this value can be converted to Joules (as the provided answer is in Joules) simply.
Thus, the activation energy ([tex]\( E_a \)[/tex]) is:
[tex]\[ E_a = 239285.25 \text{ J/mol} \][/tex]
### (ii) Calculate the Rate Constant at 700 K
Now, using the given formula:
[tex]\[ \log k = 14.34 - \frac{1.25 \times 10^4}{700} \][/tex]
First, calculate the exponent:
[tex]\[ \frac{1.25 \times 10^4}{700} = 17.8571 \][/tex]
So,
[tex]\[ \log k = 14.34 - 17.8571 = -3.5171 \][/tex]
From logarithms, we know that:
[tex]\[ k = 10^{\log k} = 10^{-3.5171} \approx 3.843628408177594 \text{ s}^{-1}\][/tex]
Thus, the rate constant at 700 K is approximately:
[tex]\[ k \approx 3.843628 \text{ s}^{-1} \][/tex]
### (iii) Calculate the Frequency Factor, [tex]\( A \)[/tex]
From the provided equation:
[tex]\[ \log A = 14.34 \][/tex]
Taking the inverse logarithm:
[tex]\[ A = 10^{14.34} \approx 218776162394955.2 \text{ s}^{-1} \][/tex]
Thus, the frequency factor [tex]\( A \)[/tex] is:
[tex]\[ A = 2.187761623949552 \times 10^{14} \text{ s}^{-1} \][/tex]
### Summary of Results:
(i) The activation energy ([tex]\( E_a \)[/tex]) is 239285.0 J/mol.
(ii) The rate constant ([tex]\( k \)[/tex]) at 700 K is approximately 3.843628 s[tex]\(^{-1}\)[/tex].
(iii) The frequency factor ([tex]\( A \)[/tex]) is [tex]\( 2.187761623949552 \times 10^{14} \text{ s}^{-1} \)[/tex].
These are the final answers to the given question.
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