
The rate constants of a reaction at $300K$ & $280K$ respectively are ${K_1}$&${K_2}$. Then:
A.${K_1} = 20{K_2}$
B.${K_2} = 4{K_1}$
C.${K_1} = 4{K_2}$
D.${K_1} = 0.5{K_2}$
Answer
541.3k+ views
Hint: We have to know that when there is an increase in temperature, the portion of molecules which contains kinetic energies higher than the activation energy of the reaction increases. So, the total activation energy of the reaction reduces. Hence, the reaction rate increases.
Complete answer:
We have to know that the rate of a chemical reaction rises with rise in temperature. The rise in temperatures increases the particle’s average kinetic energy thereby increasing the collision frequency of the constituent particles.
With increase in temperature, the reaction rate and the rate constant raise. As an overview, the rate of the reaction and the rate constant becomes two times for every ten degree rise in temperature. This is also known as temperature coefficient. It is the ratio of rate constants of the reaction present at two temperatures by variation in ten degree.
We have seen that the value of rate constant gets doubled for every ten degree increase in temperature. So, for an increase in twenty degree temperature, the rate constant of the reaction could become four times (quadrat times).
Now, coming back to the question
The given temperatures are $300K$ & $280K$.
The difference between temperatures is $20K$.
At $280K$, the rate constant is ${K_2}$.
At $290K$, the rate constant is $2 \times {K_2}$.
At $300K$, the rate constant is $2\left( {2 \times {K_2}} \right)$.
So, at $300K$, the rate constant is $4 \times {K_2}$.
Therefore, at $300K$ the rate constant is ${K_1} = 4 \times {K_2}$
At $300K$ & $280K$ are ${K_1}$ & ${K_2}$ respectively. Then, ${K_1} = 4 \times {K_2}$.
Therefore, the option (C) is correct.
Note: We have to remember that the other factors that influence the rate of a reaction are nature of reactants, concentration of the reactant, and presence of catalyst. Generally, the rate could be faster when both the reactants are present in liquid phase as collision is more frequent in liquid phase.
Complete answer:
We have to know that the rate of a chemical reaction rises with rise in temperature. The rise in temperatures increases the particle’s average kinetic energy thereby increasing the collision frequency of the constituent particles.
With increase in temperature, the reaction rate and the rate constant raise. As an overview, the rate of the reaction and the rate constant becomes two times for every ten degree rise in temperature. This is also known as temperature coefficient. It is the ratio of rate constants of the reaction present at two temperatures by variation in ten degree.
We have seen that the value of rate constant gets doubled for every ten degree increase in temperature. So, for an increase in twenty degree temperature, the rate constant of the reaction could become four times (quadrat times).
Now, coming back to the question
The given temperatures are $300K$ & $280K$.
The difference between temperatures is $20K$.
At $280K$, the rate constant is ${K_2}$.
At $290K$, the rate constant is $2 \times {K_2}$.
At $300K$, the rate constant is $2\left( {2 \times {K_2}} \right)$.
So, at $300K$, the rate constant is $4 \times {K_2}$.
Therefore, at $300K$ the rate constant is ${K_1} = 4 \times {K_2}$
At $300K$ & $280K$ are ${K_1}$ & ${K_2}$ respectively. Then, ${K_1} = 4 \times {K_2}$.
Therefore, the option (C) is correct.
Note: We have to remember that the other factors that influence the rate of a reaction are nature of reactants, concentration of the reactant, and presence of catalyst. Generally, the rate could be faster when both the reactants are present in liquid phase as collision is more frequent in liquid phase.
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