Chemistry, 02.08.2019 23:00 silly82

The rate constant of a reaction is 1.1 × 10-4 s-1 at 470 °c, and the activation energy is 264 kj/mol. what is the temperature of the reaction when the rate constant is equal to 4.36 × 10-3?

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Answer from: lmcginnis2003

The temperature of the reaction when rate constant is $4.36 \times {10^{ - 3}}{\text{ }}{{\text{s}}^{ - 1}}$ comes out to be $\boxed{540{\text{ }}^\circ {\text{C}}}$ .

Further Explanation:

Chemical kinetics:

The branch of chemistry that deals with rates of chemical reactions is called chemicalkinetics. This study includes speed of reactions, reaction mechanism, transition states, and characteristics of chemical reaction.

Arrhenius equation:

This equation describes quantitative relationship between rate constant, activation energy and temperatures. This is used for comparison of two rate constants at two different temperatures. Mathematical expression for Arrhenius equation is mentioned below.

$\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{2.303{\text{ R}}}}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$ …… (1)

Here,

${k_2}$ is rate constant at temperature ${T_2}$ .

${k_1}$ is rate constant temperature ${T_1}$ .

${{\text{E}}_{\text{a}}}$ is activation energy.

${T_2}$ isfinal temperature.

${T_1}$ isinitial temperature.

R is universal gas constant.

The activation energy is to be converted from kilojoule to joule. The conversion factor for this is,

$1{\text{ kJ}} = {10^3}{\text{ J}}$

Therefore activation energy can be calculated as follows:

$\begin{aligned}{{\text{E}}_{\text{a}}} &= \left( {264{\text{ kJ/mol}}} \right)\left( {\frac{{{{10}^3}{\text{ J}}}}{{1{\text{ kJ}}}}} \right) \\&= 264000{\text{ J/mol}}\\\end{aligned}$

The temperature is to be converted into Kelvin. The conversion factor for this is,

$0{\text{ }}^\circ {\text{C}} = 273.15{\text{ K}}$

Therefore temperature can be calculated as follows:

$\begin{aligned}{T_1} &= \left( {470 + 273.14} \right){\text{ K}} \\&= {\text{743}}{\text{.15 K}} \\\end{aligned}$

The value of ${k_2}$ is $4.36 \times {10^{ - 3}}{\text{ }}{{\text{s}}^{ - 1}}$ .

The value of ${k_1}$ is $1.1 \times {10^{ - 4}}{\text{ }}{{\text{s}}^{ - 1}}$ .

The value of ${{\text{E}}_{\text{a}}}$ is 264000 J/mol.

The value of R is 8314 J/K.mol.

The value of ${T_1}$ is 743.15 K.

Substitute these values in equation (1).

$\text{log}\left(\dfrac{4.36\times10^{-3}\text{s}^{-1}}{1.1\times10^{-4}\text{s}^{-1}}\right)=\dfrac{264000\:\text{J/mol}}{2.303(8.314\:\text{J/K mol})}\left(\dfrac{1}{743.15\:\text{K}}-\dfrac{1}{T_2}\right)$

Solving for ${T_2}$ ,

${T_2} = {\text{813 K}}$

This temperature is to be converted into degrees Celsius. Therefore ${T_2}$ can be calculated as follows:

$\begin{aligned}{T_2} &= \left( {813 - 273.15} \right){\text{ }}^\circ {\text{C}} \\&= {\text{540 }}^\circ {\text{C}} \\\end{aligned}$

Learn more:

Rate of chemical reaction: The main purpose of conducting experiments:

Answer Details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical Kinetics

Keywords: k2, k1, T1, T2, Ea, R, joule, kilojoule, 540 degrees Celsius, Kelvin, 813 K, 273.15 K, 264000 J/mol, 8.314 J/K.mol.

Answer from: Quest

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Answer from: Quest

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

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