After 53.0 min, 14.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?
To determine the half-life of a reaction, we need to use the rate constant (k) for the reaction. From the given information, we know that after 53.0 minutes, 14.0% of the compound has decomposed.
In a first-order reaction, the equation describing the rate of the reaction is:
ln(A₀/A) = kt
Where A₀ is the initial concentration of the compound, A is the concentration at time t, k is the rate constant, and t is the time. In this case, A₀/A represents the extent of decomposition.
Since 14.0% of the compound has decomposed, the extent of decomposition (A₀/A) is 0.14. We can substitute this into the equation:
ln(1/0.14) = k(53.0 min)
Solving for k:
k = ln(1/0.14) / 53.0
Now that we have the rate constant, we can determine the half-life. The half-life (t₁/₂) is the time it takes for A₀/A to decrease by half. Therefore, when A₀/A equals 0.5:
ln(1/0.5) = k(t₁/₂)
Substituting the value of k we found earlier:
ln(1/0.5) = ln(1/0.14) / 53.0 * (t₁/₂)
Solving for t₁/₂:
t₁/₂ = (ln(1/0.5) * 53.0) / ln(1/0.14)
Evaluating this expression will give us the half-life of the reaction assuming first-order kinetics.