Answer:
Explanation:
A first order reaction follows the law:
[tex]rate=k[A][/tex] , where [A] is the concentraion of the reactant A.
Equivalently:
[tex]\dfrac{d[A]}{dt}=-k[A][/tex]
Integrating:
[tex]\dfrac{d[A]}{[A]}=-kdt[/tex]
[tex]\ln \dfrac{[A]}{[A_o]}=-kt[/tex]
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.
The half-life of the reaction is 1.080 min. As given the order of the reaction is first order.
The rate of the reaction for a first-order reaction can be given as:
Rate= k[A]
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.
Calculation for t½:
t½ = ln(2) / 6.40 × 10⁻⁴ min⁻¹
t½ = 1.083 min
Thus, the half-life of this reaction is 1.083 min.
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