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?

Explanation:

A first order reaction follows the law:

       , where [A] is the concentraion of the reactant A.

Equivalently:

Integrating:

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

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.

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