Question

A particular first-order reaction has a rate constant of 1.35 × 10^2 s−1 at 25.0 ∘C. What is the magnitude of k at 65.0∘C if Ea = 55.5 kJ/mol.

Answer 1

The magnitude of k at 65.0∘C : 1912.7 /s

Further explanation

Given

k at 25 °C = 1.35 x 10² /s k1

T₁=25 + 273 = 298 K

T₂=65 + 273 = 338 K

Required

the magnitude of k at 65.0∘C

Solution

Arrhenius Equation :

$\tt ln(\dfrac{k_1}{k_2})=(\dfrac{1}{T_2}-\dfrac{1}{T_1})\dfrac{Ea}{R}$

R : gas constant= 8.314 J/molK

Input the value :

$\tt ln(\dfrac{1.35\times 10^2}{k_2})=(\dfrac{1}{338}-\dfrac{1}{298})\dfrac{55.5.10^3}{8.314}\\\\ln(\dfrac{135}{k_2})=-2.651\rightarrow \dfrac{135}{k_2}=e^{-2.651}\rightarrow k_2=1912.7$

Answer 2

The magnitude of k will be "1912.7".

Temperature,

• $T_1 = 25^{\circ} C$ or, $298 \ K$
• $T_2 = 65^{\circ} C$ or, $338 \ K$

Rate constant,

• $1.35\times 10^2 \ s^{-1}$

Gas constant,

• $R = 8.314 \ J/mol$

By using the Arrhenius equation, we get

→ $ln (\frac{k_1}{k_2} ) = (\frac{1}{T_2} - \frac{1}{T_1} )\frac{Ea}{R}$

By substituting the values, we get

→ $ln (\frac{1.35\times 10^2}{k_2} ) = (\frac{1}{338} - \frac{1}{298} )\frac{55.5\times 10^3}{8.314}$

→        $ln (\frac{135}{k_2} ) = -2.651$

                $\frac{135}{k_2} = e^{-2.651}$

                 $k_2 = 1912.7$

Thus the response above is right.

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