Respuesta :
The rate constant at 298K is 8.175*10^-7 per second.
Data;
- Activation Energy = 102kJ/mol = 102,000J/mol
- Rate constant (k1) = 1.35*10^-4 s^-1
- Rate constant (k2) = ?
- Temperature (T1) = 308K
- Temperature (T2) = 273K
- gas constant (R) = 8.314 J/mol.K
Arrhenius Equation
Using Arrhenius equation two point equation to calculate rate constant at 273K.
[tex]\ln [\frac{k_2}{k_1}] = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2}]\\[/tex]
substituting the values into the equation above,
[tex]\ln [\frac{k_2}{1*35*10^-^4}] = \frac{102000}{8.314} [\frac{1}{308} - \frac{1}{273} ]\\ \ln[\frac{k_2}{1.35*10^-4}] = 12268.8[-0.0004162]\\ \ln[\frac{k_2}{1.35*10^-^4}] = -5.10677\\[/tex]
We can take the antilog of both sides
[tex]e^\ln[\frac{k_2}{1.35*10^-^4}] = e^-5.10677\\[/tex]
[tex]\frac{k_2}{1.35*10^-^4} = 6.0556*10^-^3\\\\k_2 = 1.35*10^-^4 * 6.0556*10^-^3\\k_2= 8.175*10^-^7 s^-^1\\[/tex]
The rate constant at 298K is 8.175*10^-7 per second.
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The rate constant at 273K is [tex]8.18\times 10^{-7} \ s^{-1}[/tex]
Arrhenius Equation
From the question, we are to determine the rate constant at 273 K.
Using the formula,
[tex]ln(\frac{k_{2} }{k_{1} })= \frac{E_{a} }{R} (\frac{1}{T_{1} }- \frac{1}{T_{2} })[/tex]
Where
[tex]k_{1}[/tex] is the initial rate constant
[tex]k_{2}[/tex] is the final rate constant
[tex]E_{a}[/tex] is the activation energy
[tex]R[/tex] is the gas constant
[tex]T_{1}[/tex] is the initial temperature
[tex]T_{2}[/tex] is the final temperature
From the given information
[tex]E_{a} = 102\ KJ/mol = 102000\ J/mol[/tex]
[tex]k_{1} = 1.35 \times 10^{-4} \ s^{-1}[/tex]
[tex]T_{1}= 308 K[/tex]
[tex]T_{2}= 273K[/tex]
[tex]k_{2} = ?[/tex]
and
[tex]R = 8.314 \ J.mol^{-1}.K^{-1}[/tex]
[tex]ln(\frac{k_{2} }{1.35\times 10^{-4} })= \frac{102000}{8.314} (\frac{1}{308}- \frac{1}{273})[/tex]
[tex]ln(\frac{k_{2} }{1.35\times 10^{-4} })= 12268.4628(-0.00041625)[/tex]
[tex]ln(\frac{k_{2} }{1.35\times 10^{-4} })= -5.1067[/tex]
[tex]\frac{k_{2} }{1.35\times 10^{-4} }=e^{-5.1067}[/tex]
[tex]\frac{k_{2} }{1.35\times 10^{-4} }=0.006056[/tex]
[tex]{k_{2} =0.006056 \times 1.35\times 10^{-4}[/tex]
[tex]{k_{2} =0.0081756\times 10^{-4}[/tex]
[tex]{k_{2} =8.1756\times 10^{-7} \ s^{-1}[/tex]
[tex]{k_{2} \approx8.18\times 10^{-7} \ s^{-1}[/tex]
Hence, the rate constant at 273K is [tex]8.18\times 10^{-7} \ s^{-1}[/tex].
Learn more on Arrhenius equation here: https://brainly.com/question/8813467
