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At \[300\,K\], 50% of molecules collide with energy greater than or equal to \[{E_0}\]. At what temperature, 25% molecules will have energy greater than or equal to \[{E_0}\].

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Last updated date: 25th Jun 2024
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Answer
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Hint: Arrhenius theory states that acid is the substance which on dissolving in the aqueous solution gives \[{{\text{H}}^{\text{ + }}}\]ion. The base is the substance which ionizes the \[{\text{O}}{{\text{H}}^ - }\] ion by dissolving in the aqueous solution.

Complete step by step answer:
We know that, Arrhenius equation is given as,
\[K = A{e^{ - Ea/RT}}\] ……(1)
Here, \[K\]= constant
\[A\]= Arrhenius pre-exponential factor
\[Ea\]= Arrhenius activation energy
\[R\]= Rydberg constant
\[T\]= Temperature
Also from the above equation we know that,
\[\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{{{T_1}}}{{{T_2}}}\] ……(2)
We are given that,
\[
  {T_1} = 300\,K,\,{K_1} = \,50 \\
  {T_2} = \,?,\,{K_2} = \,25 \\
 \]
Using this value in equation (2) we will calculate the value of \[{T_2}\]
Now,
\[
  \dfrac{{{K_1}}}{{{K_2}}} = \dfrac{{{T_1}}}{{{T_2}}} \\
   \Rightarrow \dfrac{{50}}{{25}} = \dfrac{{300}}{{{T_2}}} \\
   \Rightarrow {T_2} = 150\,K \\
\]
Therefore, we can say that, 25% molecule will have energy greater than or equal to \[{E_0}\] at temperature, \[{T_2} = 150\,K\].

Additional Information:
Arrhenius acid in the aqueous solution increases the protons or \[{{\text{H}}^{\text{ + }}}\] ions concentration. For example, hydrochloric acid in water. \[{\text{HCl}}\] dissociates producing \[{{\text{H}}^{\text{ + }}}\] ion and \[{\text{C}}{{\text{l}}^ - }\] ion. An Arrhenius base is a substance that increases the hydroxide ions concentration in the aqueous solution. An example of Arrhenius base is sodium hydroxide compound in water. It dissociates to giving sodium ion and hydroxide ion

Note:
There are some drawbacks of the Arrhenius theory. The Arrhenius theory is applicable in aqueous solution only. According to the theory, \[{\text{HCl}}\] is an acid in the aqueous solution but it is not an acid in benzene, even if it donates \[{{\text{H}}^{\text{ + }}}\] ion to the benzene. Also, the solution of sodium amide in liquid ammonia is not basic in nature, even though amide ion deprotonates the ammonia.