Question

The concentration of a reactant decreases from 1.2 x10-3 M to 6.0 x10-4 M in 24 hours. Then it decreases to 3.0 x10-4 M in another 48 hours. What is the value of the rate constant?

Answer 1

Answer:

3.47 × 10¹ M⁻¹ h⁻¹

Explanation:

From the question, it's clearly depicted that it is a second order reaction rate.

NOW, rate constant for second order reaction can be given as:

k(t)= $[\frac{1}{[A]_t} - \frac{1}{[A]_o}]$

Given that:

t = 24 hours

Let the initial concentration of reactant $[A]_o=1.20 *10^{-3}M$

Final concentration $[A]_t=6.0 *10^{-4}M$

∴

k(t)= $[\frac{1}{6.0*10^{-4}} - \frac{1}{1.2*10^{-3}}]$

k(24)hours = 833.33 M⁻¹

k = $\frac{833.33M^_{-1}}{24h}$

k = 34.72 M⁻¹ h⁻¹

k = 3.472 × 10¹ M⁻¹ h⁻¹

∴ the value of the rate constant (k) = 3.472 × 10¹ M⁻¹ h⁻¹