Chemistry, 12.02.2020 03:19 CoolRahim9090

What is the activation energy (in kJ/mol) of a reaction whose rate constant increases by a factor of 89 upon increasing the temperature from 307 K to 343 K? R = 8.314 J/(mol • K). Only enter the numerical value as an integer in the answer box below. Do NOT type in the unit (kJ/mol).

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The least quantity of energy needed by the reactants to start a chemical reaction and result in products is called activation energy.

Activation energy can be given by the Arrhenius equation:

$\text{k} & = \text{Ae} ^{ \dfrac{\text{- Ea}}{\text{RT}} }$

Where Ea is the activation energy.

The activation energy needed is 15.9 KJ/mol.

This can be calculated by:

Given,

Temperature (T1) = 307 K Temperature (T2) = 343 K Gas constant (R) = 8.314 J/(mol • K) rate constant = $\frac{\text{K}2}{\text{K}1} = 89$

To calculate the activation energy (Ea):

$\text{ln} \;\dfrac{\text{K2}}{\text{K1}} & = \dfrac{\text{Ea}}{\text{R}} \;(\dfrac{1}{\text{T}1} \;- \dfrac{1}{\text{T2}})$

Where, Ea is the activation energy, k is the rate constant, R is the Universal Gas Constant and T is the temperature.

$\text{ln} \;\dfrac{\text{89 K1}}{\text{K1}} & = \dfrac{\text{Ea}}{\text{8.314 J/mol.K }} \;(\dfrac{1}{\text{307 K}} \;- \dfrac{1}{\text{343 K}})$

$\text{ln} 89 = {\frac {\text{Ea}} { 8.314 \text{J/mol.K}} \times (0.0325 - 0.00291)$

$\text{ln} 89 = {\frac {\text{Ea}} { 8.314 \text{J/mol.K}} \times (2.95 \times 10^{2} )$

$4.488= {\frac {\text{Ea}} { 8.314 \text{J/mol.K}} \times (2.95 \times 10^{2} )$

$\text{Ea} = \dfrac{4.488 \times (2.95 \times 10^{2} )}{ 8.314 \text{J/mol.K}}$

$= \dfrac{0.1324}{8.314}$

Ea = 0.0159

Ea = 1.59 x 10² J/mol

Therefore, the required activation energy is 15.9 KJ/mol.

To learn more about activation energy refer to the link:

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Answer from: xbeatdroperzx

15.9 KJ/mol

Explanation:

Given data:

Temperature = T1 = 307 K

Temperature = T2 = 343 K

Gas constant R= 8.314 J/(mol • K)

rate constant = k2/K1 = 89

To find:

Activation energy (in kJ/mol) = Ea = ?

Formula:

The Arrhenius equation gives the relation between temperature and reaction rates:

$ln\frac{K2}{K1} = \frac{Ea}{R} (\frac{1}{T1} - \frac{1}{T2})$

here, in this equation

k = the rate constant

Ea = the activation energy

R = the Universal Gas Constant

T= the temperature

Solution:

$ln\frac{89 K1}{K1} = \frac{Ea}{8.314 J/mol . k\\} (\frac{1}{307 K} - \frac{1}{343 K})$

ln 89 = Ea / 8.314 J/mol.K x (0.0325 - 0.00291)

ln 89 = Ea / 8.314 J/mol.K x (2.95 x 10^2 )

4.488 = Ea / 8.314 J/mol.K x (2.95 x 10^2)

Ea = 4.488 x (2.95 x 10^2) / 8.314 J/mol.K

= 0.1324 / 8.314

Ea = 0.0159

Ea = 1.59 x 10^2 J/mol

= 15.9 KJ/mol

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