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# At room temperature $\left( {20^\circ C} \right)$ orange juice gets spoiled in about $64$ hours. In a refrigerator, at $3^\circ C$ juice can be stored three times as long before it gets spoilt. Estimate (a) the activation energy of the reaction that causes the spoilage of juice. (b) How long should it take for juice to get spoilt at $40^\circ C$ ?A. (a) $21.9\,kJ\,mo{l^{ - 1}}$ (b) $10.47\,hour$ B. (a) $43.46\,kJ\,mo{l^{ - 1}}$ (b) $20.47\,hour$ C. (a) $434.6\,kJ\,mo{l^{ - 1}}$ (b) $10.47\,hour$ D. None of these

Last updated date: 20th Jun 2024
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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.

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) $20.47\,hour$.

> 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.