Question

The thermal decomposition of N2O5 obeys first-order kinetics. At 45°C, a plot of ln[N2O5] versus t gives a slope of −6.40 × 10−4 min−1. What is the half-life of the reaction?

Answer 1

Answer:

• 1,080 min

Explanation:

A first order reaction follows the law:

     $rate=k[A]$  , where [A] is the concentraion of the reactant A.

Equivalently:

       $\dfrac{d[A]}{dt}=-k[A]$

Integrating:

    $\dfrac{d[A]}{[A]}=-kdt$

   $\ln \dfrac{[A]}{[A_o]}=-kt$

Half-life means [A]/[A₀] = 1/2, t = t½:

•    t½ = ln (2) / k

That means that the half-life is constant.

The slope of the plot of ln [N₂O₅]  is -k. Then k is equal to 6.40 × 10⁻⁴ min⁻¹.

Thus, you can calculate t½:

   t½ = ln(2) / 6.40 × 10⁻⁴ min⁻¹

   t½ = 1,083 min.

Rounding to 3 significant figures, that is 1,080 min.