Question

What will be the initial rate of a reaction if its rate constant is 10⁻³ min⁻¹ and concentration of reactant is 0.2 mol dm⁻³. How much of the reactant will be converted into products in 200 minutes?

Answer 1

The reaction's rate constant is 10⁻³ min⁻¹. The provided reaction is first order, according to the rate constant units. Therefore, the reaction's rate law can be expressed as

Rate = k [A]

At [A] = 0.2 mol dm⁻³

Rate = k [A]

= 10⁻³ × 0.2

= 2.0 × 10⁻⁴ mol dm⁻³ min⁻¹

For a first-order reaction,

k = `2.303/t log_10  [A]_0/([A])`

Putting [A]₀ = 100, t = 200 min, we have

10⁻³ = `2.303/200 log_10  100/([A])`

or, `log_10  100/([A]) = (200 xx 10^-3)/2.303` = 0.0868

or, `100/([A])` = antilog₁₀ 0.0868 = 1.221

or, [A] = `100/1.221` = 81.9

Hence, percentage of the substance decomposed = 100 − 81.9 = 18.1%