
A first order reaction is 50% completed in 30 minutes at 27$^{{}^\circ }C$ and in 10 minutes at 47$^{{}^\circ }C$. Calculate the reaction rate constant at 27$^{{}^\circ }C$ and the energy of activation of the reaction in $kJmo{{l}^{-1}}$.
Answer
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Hint: As temperature rises, molecules gain energy and move faster. Therefore, the higher the temperature, the higher the probability that molecules will be moving with the necessary activation energy for a reaction to occur after collision.
Complete step by step answer: A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.
Half-life (symbol ${{t}_{1/2}}$) is the time required for a quantity to reduce to half of its initial value.
For first order reaction $k=\dfrac{{{t}_{1/2}}}{0.693}$.
For knowing rate constant, we can use the half life formula,
Given in the question, we have 30 minutes of half life at 27$^{{}^\circ }C$ and 10 minutes of half life at 47$^{{}^\circ }C$,
At 27$^{{}^\circ }C$, ${{k}_{{{27}^{o}}C}}=\dfrac{0.693}{30}=0.0231mi{{n}^{-1}}$
At 47$^{{}^\circ }C$, ${{k}_{{{47}^{o}}C}}=\dfrac{0.693}{10}=0.0693mi{{n}^{-1}}$
The relation of of rate constant and temperature is as follows:
\[lo{{g}_{10}}\dfrac{{{k}_{1}}}{{{k}_{2}}}=\dfrac{-{{E}_{a}}}{2.303\times R}(\dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}{{T}_{1}}})\]
Substituting values of rate constant and temperature in the above formula to get value activation energy,
\[\begin{align}
& lo{{g}_{10}}\dfrac{0.0231}{0.0693}=\dfrac{-{{E}_{a}}}{19.14}\left( \dfrac{320-300}{320\times {{300}_{1}}} \right) \\
& lo{{g}_{10}}0.333=\dfrac{-{{E}_{a}}}{19.14}\times \dfrac{20}{96000} \\
& {{E}_{a}}=\dfrac{-19.14\times 96000}{20}\times lo{{g}_{10}}0.333 \\
& {{E}_{a}}=43857Jmo{{l}^{-1}}=43.857kJmo{{l}^{-1}} \\
\end{align}\]
Therefore, the value of activation energy will be 43.857kJmol$^{-1}$.
Note: Activation energy is the minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation. The rate constant, or the specific rate constant, is the proportionality constant in the equation that gives the relationship between the rate of a chemical reaction and the concentrations of the given reacting substances.
Complete step by step answer: A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.
Half-life (symbol ${{t}_{1/2}}$) is the time required for a quantity to reduce to half of its initial value.
For first order reaction $k=\dfrac{{{t}_{1/2}}}{0.693}$.
For knowing rate constant, we can use the half life formula,
Given in the question, we have 30 minutes of half life at 27$^{{}^\circ }C$ and 10 minutes of half life at 47$^{{}^\circ }C$,
At 27$^{{}^\circ }C$, ${{k}_{{{27}^{o}}C}}=\dfrac{0.693}{30}=0.0231mi{{n}^{-1}}$
At 47$^{{}^\circ }C$, ${{k}_{{{47}^{o}}C}}=\dfrac{0.693}{10}=0.0693mi{{n}^{-1}}$
The relation of of rate constant and temperature is as follows:
\[lo{{g}_{10}}\dfrac{{{k}_{1}}}{{{k}_{2}}}=\dfrac{-{{E}_{a}}}{2.303\times R}(\dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}{{T}_{1}}})\]
Substituting values of rate constant and temperature in the above formula to get value activation energy,
\[\begin{align}
& lo{{g}_{10}}\dfrac{0.0231}{0.0693}=\dfrac{-{{E}_{a}}}{19.14}\left( \dfrac{320-300}{320\times {{300}_{1}}} \right) \\
& lo{{g}_{10}}0.333=\dfrac{-{{E}_{a}}}{19.14}\times \dfrac{20}{96000} \\
& {{E}_{a}}=\dfrac{-19.14\times 96000}{20}\times lo{{g}_{10}}0.333 \\
& {{E}_{a}}=43857Jmo{{l}^{-1}}=43.857kJmo{{l}^{-1}} \\
\end{align}\]
Therefore, the value of activation energy will be 43.857kJmol$^{-1}$.
Note: Activation energy is the minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation. The rate constant, or the specific rate constant, is the proportionality constant in the equation that gives the relationship between the rate of a chemical reaction and the concentrations of the given reacting substances.
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