
According to the Arrhenius equation, how does temperature affect the rate constant?
Answer
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Hint: Rate constant of a reaction at a particular temperature is defined as the rate of reaction when concentration of reactant is unity. The rate constant depends only on the temperature of the reaction. The relationship between rate constant and temperature can be deduced from the Arrhenius equation which is $K\,=\,A\,{{e}^{\dfrac{-Ea}{RT}}}$. We will perform basic logarithm operations in this equation to deduce the relationship between rate constant and temperature.
Complete answer:
Arrhenius proposed the following empirical equation relating the rate constant K and temperature T
$K\,=\,A{{e}^{\dfrac{-Ea}{RT}}}$ …(i)
Where A= frequency factor or Pre-exponential factor (Constant)
$Ea$= activation energy of the reaction
R= gas constant, whose value= $8.314\,J{{\left( mol\,K \right)}^{-1}}$
The term $\dfrac{-Ea}{RT}$refers to the number of molecules having energy equal to or greater than the activation energy of that particular reaction.
Both A and $Ea$are characteristics of the reaction.
Taking the natural logarithm of Eq. (i) we get
$In\,K\,=\,In\,A\,-\,\dfrac{Ea}{RT}$
Now since, we know that A is constant therefore its natural logarithm $In\A $will also be constant. Let us assume that A does not depend on the temperature. Then from the above equation when temperature of the reaction is increased then as a result the rate constant (K) of a reaction will also increase.
Therefore, we can say that as the temperature of a reaction increases its rate constant also increases.
Note:It should be noted that in general the rate of a reaction increases for each 10 degrees increase in temperature. Also, the rate constant is a constant and it depends only on the temperature of the reaction. It changes only if the temperature fluctuates. Other quantities do not have any effect in it.
Complete answer:
Arrhenius proposed the following empirical equation relating the rate constant K and temperature T
$K\,=\,A{{e}^{\dfrac{-Ea}{RT}}}$ …(i)
Where A= frequency factor or Pre-exponential factor (Constant)
$Ea$= activation energy of the reaction
R= gas constant, whose value= $8.314\,J{{\left( mol\,K \right)}^{-1}}$
The term $\dfrac{-Ea}{RT}$refers to the number of molecules having energy equal to or greater than the activation energy of that particular reaction.
Both A and $Ea$are characteristics of the reaction.
Taking the natural logarithm of Eq. (i) we get
$In\,K\,=\,In\,A\,-\,\dfrac{Ea}{RT}$
Now since, we know that A is constant therefore its natural logarithm $In\A $will also be constant. Let us assume that A does not depend on the temperature. Then from the above equation when temperature of the reaction is increased then as a result the rate constant (K) of a reaction will also increase.
Therefore, we can say that as the temperature of a reaction increases its rate constant also increases.
Note:It should be noted that in general the rate of a reaction increases for each 10 degrees increase in temperature. Also, the rate constant is a constant and it depends only on the temperature of the reaction. It changes only if the temperature fluctuates. Other quantities do not have any effect in it.
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