Question

Consider the following reactions. Reaction 1: A → B, Ea = 28.31 kJ Reaction 2CD; Ea = 20 kJ If the initial temperature is 400 K for both reaction and the rate constants are equal then find the ratio of Arrhenius factor A1:A2 for reaction 1 and 2​

Answer 1

Answer:

It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator; butter goes rancid more quickly in the summer than in the winter; and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be more lethargic on cold days.

The reason for this is not hard to understand. Thermal energy relates direction to motion at the molecular level. As the temperature rises, molecules move faster and collide more vigorously, greatly increasing the likelihood of bond cleavages and rearrangements. Whether it is through the collision theory, transition state theory, or just common sense, chemical reactions are typically expected to proceed faster at higher temperatures and slower at lower temperatures.

Answer 2

Answer:

$A_1:A_2 = 12.18$   for reactions 1 and 2

Concept:

Chemical Kinetics: Arrhenius Equation

Arrhenius equation provides a relationship between the rate constant (k) of a reaction and the temperature of the system. Arrhenius equation can be expressed as follows -

$k = Ae^{\frac{-E_a}{RT} }$

where k = rate constant

           A = Arrhenius factor or  frequency factor

           $E_a$ = Energy of activation

            R = Gas constant = 8.31 $JK^{-1}mol^{-1}$

Given:

For rection 1,A → B, $E_{a_1}$ = 28.31 kJ = 28.31×$10^3 J$

                     Let Arrhenius factor for this reaction be $A_1$.

For reaction 2,C → D, $E_{a_2}$ = 20 kJ = 20×$10^3 J$

                     Let Arrhenius factor for this reaction be $A_2$

Initial temperature, T = 400 K

Solution:

For the first reaction i.e., reaction 1, let the rate constant be $k_1$

Using the Arrhenius equation, we get

$k_1 = A_1e^{\frac{-E_{a_1}}{RT}$              ---------- (i)

Similarly, for the second reaction i.e., reaction 2, let the rate constant be $k_2$

Using the Arrhenius equation, we get

$k_2 = A_2e^{\frac{-E_{a_2}}{RT}$              ----------- (ii)

According to the question, the rate constants for both the reactions are equal, that is,

$k_1 = k_2$

∴ Equating equations (i) and (ii), we get

$A_1e^{\frac{-E_a_1}{RT} } = A_2e^{\frac{-E_a_2}{RT} }$

$\frac{A_1}{A_2} = \frac{e^{\frac{-E_a_2}{RT} }}{e^{\frac{-E_a_1}{RT} }}$

$\frac{A_1}{A_2} = e^{\frac{E_a_1}{RT}- \frac{E_a_2}{RT}}$

$\frac{A_1}{A_2} = e^{\frac{E_a_1-E_a_2}{RT} }$      ----------- (iii)

Putting all values of R, T, $E_{a_1}$ and $E_{a_2}$ in equation (iii), we get

$\frac{A_1}{A_2} = e^{\frac{(28.31X10^3 J)-(20X10^3 J)}{(8.31)(400)} }$

$\frac{A_1}{A_2} = e^{\frac{8.31X10^3}{(8.31)(400)} }$

$\frac{A_1}{A_2} = e^{\frac{10^3}{400} } = e^{\frac{1000}{400} }$

$\frac{A_1}{A_2} = e^{2.5}$

$\frac{A_1}{A_2} = 12.18$

$A_1:A_2 = 12.18$

Hence, the ratio of Arrhenius factors ($A_1:A_2$) for reactions 1 and 2 is 12.18.

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