Question

The activation energies of the two reactions are E₁ and E₂ with E₁ > E₂. If the temperature of the reacting system is increased from T₁ (rate constants are K₁ and K₂) to T₂(rate constants are K₁’ and K₂’ ), predict which of the following alternative is correct.

• A. $\frac{{\mathrm{K}}_1^{\text{'}}}{{\mathrm{K}}_1}<\frac{{\mathrm{K}}_2^{\text{'}}}{{\mathrm{K}}_2}$
• B. $\frac{{\mathrm{K}}_1^{\text{'}}}{{\mathrm{K}}_1}>\frac{{\mathrm{K}}_2^{\text{'}}}{{\mathrm{K}}_2}$
• C. $\frac{{\mathrm{K}}_1^{\text{'}}}{{\mathrm{K}}_1}=\frac{{\mathrm{K}}_2^{\text{'}}}{{\mathrm{K}}_2}$
• D. $\frac{{\mathrm{K}}_1}{{\mathrm{K}}_1^{\text{'}}}<\frac{{\mathrm{K}}_2}{{\mathrm{K}}_2^{\text{'}}}$

Answer 1

The correct option is B

$\frac{{\mathrm{K}}_1^{\text{'}}}{{\mathrm{K}}_1}>\frac{{\mathrm{K}}_2^{\text{'}}}{{\mathrm{K}}_2}$

The explanation for the correct answer

Option(B): The expression of the Arrhenius equation is:$\mathrm{K}={\mathrm{Ae}}^{\frac{-\mathrm{Ea}}{\mathrm{RT}}}$

• k denotes the rate constant of the reaction.
• A denotes the pre-exponential factor which, in terms of the collision theory, is the frequency of correctly oriented collisions between the reacting species.
• e is the base of the natural logarithm (Euler’s number).
• E_a denotes the activation energy of the chemical reaction (in terms of energy per mole).
• R denotes the universal gas constant.
• T denotes the absolute temperature associated with the reaction (in Kelvin).

If we take the first reaction we will consider,

${\mathrm{E}}_1=2.303\mathrm{R}\frac{{\mathrm{T}}_1{\mathrm{T}}_2}{({\mathrm{T}}_1-{\mathrm{T}}_2)}\log \left[\frac{{\mathrm{K}}_1^{\text{'}}}{{\mathrm{K}}_1}\right]$

For the second reaction,

${\mathrm{E}}_2=2.303\mathrm{R}\frac{{\mathrm{T}}_1{\mathrm{T}}_2}{({\mathrm{T}}_1-{\mathrm{T}}_2)}\log \left[\frac{{\mathrm{K}}_2^{\text{'}}}{{\mathrm{K}}_2}\right]$

Since, ${\mathrm{E}}_1>{\mathrm{E}}_2$

$\frac{{\mathrm{K}}_1^{\text{'}}}{{\mathrm{K}}_1}>\frac{{\mathrm{K}}_2^{\text{'}}}{{\mathrm{K}}_2}$

Hence, Option(B) is correct.