Question

A certain reaction has a rate constant of 8.0 x 10 -3 M/s at 22.0 0C. It is determined that the activation energy of the reaction is 115 kJ/mol. What is the rate constant of the reaction at 27.00C?

Answer 1

Answer:

The problem can be solved using the Arrhenius equation:

$\boxed{ \bf{k = e^{-E_a/RT}}}$

This can also be written as:

$\text{ln k} = \dfrac{\text{-E}_a}{\text{RT}}$

According to the question, the value of E(a) is 115 kJ/mol, K = 8 × 10⁻³ M and the temperature is 22°C. We are required to find the value of the rate constant (K') at a temperature of 27°C with same activation energy E(a).

Formula:

$\boxed{ln \left[\dfrac{K1}{K2}\right] = \dfrac{E_a}{R} \left[ \dfrac{1}{T_2} - \dfrac{1}{T_1} \right] }$

Based on our question,

• k1 =  8 × 10⁻³ M
• E(a) = 115 kJ/mol
• R = 8.314 J/K.mol
• T1 = 22°C = 295 K
• T2 = 27°C = 300 K

Substituting the values we get:

$\implies ln \left[ \dfrac{8 \times 10^3}{K_2}\right] = -\dfrac{115 \times 10^3}{8.314} \times \left[ \dfrac{1}{300} - \dfrac{1}{295} \right]\\\\\\\implies ln \left[ \dfrac{8 \times 10^3}{K_2}\right] = -13.83 \times 10^3 \times \left[ \dfrac{295 - 300}{300\times 295} \right]\\\\\\\implies ln \left[ \dfrac{8 \times 10^3}{K_2}\right] = (-13.83 \times 10^3) \times (-5.64 \times 10^{-5})\\\\\\\implies ln \left[ \dfrac{8 \times 10^3}{K_2}\right] = 0.78\\\\\\\implies \dfrac{8 \times 10^3}{K_2} = e^{0.78}$

$\implies \dfrac{8 \times 10^3}{K_2} = 2.1814\\\\\\\implies k_2 = \dfrac{8 \times 10^3}{2.1814}\\\\\\\boxed{ \bf{ k_2 = 3.667 \times 10^3\: M}}$

Hence the rate constant at 27°C is 3.667 × 10³ M.