anthonylopez1

10.02.2021 •

Chemistry

What is the mass in grams of 1.50 x10^15 atoms N

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Ответ:

lailabirdiemae

14.02.2023

Chemistry

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

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5000}\ s^{-1}$

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

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{1000}\ s^{-1}$

The rate constant, k = 0.0006931 s⁻¹

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol

$k_2=0.0006931\ s^{-1}$

$k_1=0.0001386\ s^{-1}$

$T_1=300\ K$

$T_2=310\ K$

So,

$\ln \dfrac{0.0001386}{0.0006931} =-\dfrac{E_{a}}{8.314} \times (\left (\dfrac{1}{300}-\dfrac{1}{310} \right ))$

$-\frac{1}{9300}\times \frac{E_a}{8.314}=\ln \left(\frac{0.0001386}{0.0006931}\right)$

$-\frac{E_a}{77320.2}=\ln \left(\frac{0.0001386}{0.0006931}\right)$

$E_a=-77320.2\ln \left(\frac{0.0001386}{0.0006931}\right)\ J/mol$

$E_a=124453.22\ J/mol$

The activation energy is:- 124453.22 J/mol

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