Question

A certain first-order reaction (A→products) has a rate constant of 5.70×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?

Answer 1

If a 1st order reaction is given as A→products , then the integrated rate equation for this will be

k = (1/t) * 2.303 log [A]₀ /[A]

where k = rate constant for the reaction

t = time taken

[A]₀ = initial concentration of the reactant

[A]= concentration of the reactant after time t

Now given k = 5.70 × 10⁻³ s⁻¹  

t = ?

concentration of the reactant, [A], to drop to 6.25% of the original concentration. That means if [A₀]= 100%, then after time t, [A]= 6.25%

Now putting these values in the equation

k = (1/t) * 2.303 log [A]₀ /[A]

t = (1/k) * 2.303 log [A]₀ /[A]

= (1/5.70 × 10⁻³ s⁻¹) * 2.303 log [100%]/[6.25%]

= (0.175 × 10³ s) * 2.303 log [16‬]

= (0.175 × 10³ s) * 2.303 * 1.2

= 0.4836 × 10³ s

= 483.6 s

= 8.06‬ min

Thus it will take 8.06‬ min for the concentration of the reactant, [A], to drop to 6.25% of the original concentration