Chemistry, 13.12.2019 06:31 supermimi8078

Areaction is known to exhibit 1st order kinetics. at 300k the concentration of reactant is reduced to one half of its initial value after 5000s. in contrast, at 310k the conc. is halved after 1000s. using this information to calculate: i) the rate constant for the reaction at 300kii) the time required for the reaction to be reduced to halfiii) the activation energy for the reaction

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Answer from: marquetter7689

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

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

Answer from: Quest

C) the increase in atmospheric pressure pushes the walls of the balloon outward thus making it is the answer.hope i you.

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