Consider and are two reactions. If the rate constant of the reaction can be expressed by the following equation and activation energy of reaction is th of the reaction , then the value of is ________

Question

Consider $A \overset{k_{1}}{\to} B$ and $C \overset{k_{2}}{\to} D$ are two reactions. If the rate constant $(k_{1})$ of the $A ⟶ B$ reaction can be expressed by the following equation $\log_{10} k = 14.34 - \frac{1.5 \times 10^{4}}{T / K}$ and activation energy of $C ⟶ D$ reaction $(E_{a_{2}})$ is $\frac{1}{5}$ th of the $A ⟶ B$ reaction $(E_{a_{1}})$ , then the value of $(E_{a_{2}})$ is ________ ${kJ mol}^{- 1}$ . (Nearest Integer) [2026]

Smart Achievers · Accepted Answer

(57)

$\frac{E_{a 1}}{2.303 R} = 1.5 \times 10^{4}$

$A ⟶ B$

$E_{a 1} = 28.7207 \times 10^{4} J$

$E_{a 1} = 287.207 kJ$

$E_{a 2} = \frac{E_{a 1}}{5} = \frac{287.207}{5} = 57.44 kJ$

Solution

Ans.
(57)

$\frac{E_{a 1}}{2.303 R} = 1.5 \times 10^{4}$

$E_{a 1} = 1.5 \times 10^{4} \times 2.303 \times 8.314$

$E_{a 1} = 28.7207 \times 10^{4} J$

$E_{a 1} = 287.207 kJ$

$E_{a 2} = \frac{E_{a 1}}{5} = \frac{287.207}{5} = 57.44 kJ$

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Solution

Ans. (57) Ea12.303R=1.5×104 Ea1=1.5×104×2.303×8.314 Ea1=28.7207×104 J Ea1=287.207 kJ Ea2=Ea15=287.2075=57.44 kJ

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Ans.
(57)

$\frac{E_{a 1}}{2.303 R} = 1.5 \times 10^{4}$

$E_{a 1} = 1.5 \times 10^{4} \times 2.303 \times 8.314$

$E_{a 1} = 28.7207 \times 10^{4} J$

$E_{a 1} = 287.207 kJ$

$E_{a 2} = \frac{E_{a 1}}{5} = \frac{287.207}{5} = 57.44 kJ$