Answer
405.3k+ views
Hint:Activation energy is the minimum amount of energy required so that reactants can convert into products. In this question, the Arrhenius equation is applied, which gives a formula for temperature dependence of reaction rate.
Formula used:
$2.303\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
where, ${k_1}$ and ${k_2}$ are the rate constants, ${E_a}$ is the activation energy, ${T_1}$ and ${T_2}$ are room temperature and temperature at refrigerator respectively, and $R$ is the gas constant.
Complete step by step answer:
Here, it is given that the room temperature is $20^\circ C$ and time about when the orange juice gets spoiled is $64\,hrs$ .
To calculate the activation energy of reaction, the formula used is:
$2.303\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
where, ${k_1}$ and ${k_2}$ are the rate constants, ${E_a}$ is the activation energy, ${T_1}$ and ${T_2}$ are room temperature and temperature at refrigerator respectively, and $R$ is the gas constant.
On substituting the values , we get
\[2.303\log \dfrac{1}{3} = \dfrac{{{E_a}}}{{8.314}}\left( {\dfrac{1}{{293}} - \dfrac{1}{{276}}} \right)\]
On further solving, we get,
\[ \Rightarrow {E_a} = \dfrac{{2.303\log \left( {\dfrac{1}{3}} \right)\left( {8.314 \times 293 \times 276} \right)}}{{276 - 293}}\]
\[ \Rightarrow {E_a} = 43457J/mol = 43.46\,kJ/mol\]
To calculate time that it will take for juice to get spoilt at $40^\circ C$, firstly you need to calculate $\dfrac{{{K_2}}}{{{K_1}}}$ .
\[\log \dfrac{{{K_3}}}{{{K_2}}} = \dfrac{{{E_a}}}{{2.303R}}\left( {\dfrac{{{T_3} - {T_2}}}{{{T_3}{T_2}}}} \right)\]
On substituting the values , we get
\[ \Rightarrow \log \dfrac{{{K_3}}}{{{K_2}}} = \dfrac{{43457}}{{2.303 \times 8.314}}\left( {\dfrac{{313 - 293}}{{293 \times 313}}} \right)\]
On simplifying further, we get,
\[
\Rightarrow \log \dfrac{{{K_3}}}{{{K_2}}} = 0.4949 \\
\Rightarrow \dfrac{{{K_3}}}{{{K_2}}} = 3.126 \\
\]
The time required for juice to get spoil at $40^\circ C$ is,
\[\dfrac{{64}}{{3.126}} = 20.47\,hour\]
Therefore, the correct option is (B), that is, (a) $43.46\,kJ\,mo{l^{ - 1}}$ (b) $20.47\,hour$.
Additional information
> In this equation, the Arrhenius equation gives the dependence of the rate constant of a chemical reaction.
$K = A{e^{ - {E_a}/RT}}$
where, $A$ is the pre exponential factor.
> This equation is used to calculate the rate constant and the calculation of energy of activation.
> Increasing the temperature and decreasing the activation energy with the use of a catalyst, it will result in the increase in rate of reaction.
> Rate constant is defined as the proportionality factor in this equation. It is denoted with a symbol $(K)$ . Its unit depends upon the order of reaction.
> Activation energy is defined as the minimum amount of energy required to convert the reactant into product.
Note:
> In this question, the Arrhenius equation is used to calculate the activation energy.
> Activation energy is the minimum amount of energy that is needed to convert reactant into product.
> Decrease in activation energy will increase the rate of reaction and vice – versa.
Formula used:
$2.303\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
where, ${k_1}$ and ${k_2}$ are the rate constants, ${E_a}$ is the activation energy, ${T_1}$ and ${T_2}$ are room temperature and temperature at refrigerator respectively, and $R$ is the gas constant.
Complete step by step answer:
Here, it is given that the room temperature is $20^\circ C$ and time about when the orange juice gets spoiled is $64\,hrs$ .
To calculate the activation energy of reaction, the formula used is:
$2.303\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{{E_a}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
where, ${k_1}$ and ${k_2}$ are the rate constants, ${E_a}$ is the activation energy, ${T_1}$ and ${T_2}$ are room temperature and temperature at refrigerator respectively, and $R$ is the gas constant.
On substituting the values , we get
\[2.303\log \dfrac{1}{3} = \dfrac{{{E_a}}}{{8.314}}\left( {\dfrac{1}{{293}} - \dfrac{1}{{276}}} \right)\]
On further solving, we get,
\[ \Rightarrow {E_a} = \dfrac{{2.303\log \left( {\dfrac{1}{3}} \right)\left( {8.314 \times 293 \times 276} \right)}}{{276 - 293}}\]
\[ \Rightarrow {E_a} = 43457J/mol = 43.46\,kJ/mol\]
To calculate time that it will take for juice to get spoilt at $40^\circ C$, firstly you need to calculate $\dfrac{{{K_2}}}{{{K_1}}}$ .
\[\log \dfrac{{{K_3}}}{{{K_2}}} = \dfrac{{{E_a}}}{{2.303R}}\left( {\dfrac{{{T_3} - {T_2}}}{{{T_3}{T_2}}}} \right)\]
On substituting the values , we get
\[ \Rightarrow \log \dfrac{{{K_3}}}{{{K_2}}} = \dfrac{{43457}}{{2.303 \times 8.314}}\left( {\dfrac{{313 - 293}}{{293 \times 313}}} \right)\]
On simplifying further, we get,
\[
\Rightarrow \log \dfrac{{{K_3}}}{{{K_2}}} = 0.4949 \\
\Rightarrow \dfrac{{{K_3}}}{{{K_2}}} = 3.126 \\
\]
The time required for juice to get spoil at $40^\circ C$ is,
\[\dfrac{{64}}{{3.126}} = 20.47\,hour\]
Therefore, the correct option is (B), that is, (a) $43.46\,kJ\,mo{l^{ - 1}}$ (b) $20.47\,hour$.
Additional information
> In this equation, the Arrhenius equation gives the dependence of the rate constant of a chemical reaction.
$K = A{e^{ - {E_a}/RT}}$
where, $A$ is the pre exponential factor.
> This equation is used to calculate the rate constant and the calculation of energy of activation.
> Increasing the temperature and decreasing the activation energy with the use of a catalyst, it will result in the increase in rate of reaction.
> Rate constant is defined as the proportionality factor in this equation. It is denoted with a symbol $(K)$ . Its unit depends upon the order of reaction.
> Activation energy is defined as the minimum amount of energy required to convert the reactant into product.
Note:
> In this question, the Arrhenius equation is used to calculate the activation energy.
> Activation energy is the minimum amount of energy that is needed to convert reactant into product.
> Decrease in activation energy will increase the rate of reaction and vice – versa.
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