Respuesta :
Answer:
The rate constant at 300 K , k = 0.0001386 s⁻¹
5000 s is the time required.
The activation energy is:- 124453.22 J/mol
Explanation:
(i)
Half life is the time at which the concentration of the reactant reduced to half. So, 5000 s is the half life.
At 300 K
Half life expression for first order kinetic is:
Half life = 5000 s
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{5000}\ s^{-1}[/tex]
The rate constant, k = 0.0001386 s⁻¹
(ii)
As, stated above, Half life is the time at which the concentration of the reactant reduced to half. So, 5000 s is the time required.
(iii)
At 310 K , Half life = 1000 s
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]k=\frac {ln\ 2}{1000}\ s^{-1}[/tex]
The rate constant, k = 0.0006931 s⁻¹
Using the expression,
[tex]\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )[/tex]
Wherem
[tex]k_1\ is\ the\ rate\ constant\ at\ T_1[/tex]
[tex]k_2\ is\ the\ rate\ constant\ at\ T_2[/tex]
[tex]E_a[/tex] is the activation energy
R is Gas constant having value = 8.314 J / K mol
[tex]k_2=0.0006931\ s^{-1}[/tex]
[tex]k_1=0.0001386\ s^{-1}[/tex]
[tex]T_1=300\ K[/tex]
[tex]T_2=310\ K[/tex]
So,
[tex]\ln \dfrac{0.0001386}{0.0006931} =-\dfrac{E_{a}}{8.314} \times (\left (\dfrac{1}{300}-\dfrac{1}{310} \right ))[/tex]
[tex]-\frac{1}{9300}\times \frac{E_a}{8.314}=\ln \left(\frac{0.0001386}{0.0006931}\right)[/tex]
[tex]-\frac{E_a}{77320.2}=\ln \left(\frac{0.0001386}{0.0006931}\right)[/tex]
[tex]E_a=-77320.2\ln \left(\frac{0.0001386}{0.0006931}\right)\ J/mol[/tex]
[tex]E_a=124453.22\ J/mol[/tex]
The activation energy is:- 124453.22 J/mol
