After 79 mins 20.0% of a compound has decomposed. What is the half life of this reaction assuming first-order kinetics?
.2=e^(-.693*79/thl)
solve for thlife
take ln both sides
ln.2=-.693*79/thl
grab you calculator.
To determine the half-life of a reaction, we can use the equation for first-order reactions:
ln(N₀/N) = kt
Where:
N₀ is the initial amount of the compound,
N is the amount of the compound remaining after a specific time,
k is the reaction rate constant, and
t is the time elapsed.
In this case, we are given that 20.0% of the compound has decomposed after 79 minutes. Therefore, the amount remaining is 100% - 20.0% = 80.0% = 0.80.
Now, let's plug in the values into the equation:
ln(N₀/N) = kt
ln(1/0.80) = k * 79 minutes
To simplify further, let's solve for k:
ln(1/0.80) = k * 79 minutes
ln(1.25) = k * 79 minutes
Now, we can solve for k by dividing both sides of the equation by 79:
k = ln(1.25)/79
Using a calculator, we find that k ≈ 0.0016 min⁻¹.
The half-life (t₁/₂) can be calculated using the following equation:
t₁/₂ = ln(2) / k
Substituting the value of k we previously calculated:
t₁/₂ = ln(2) / 0.0016 min⁻¹
Using a calculator, we find that t₁/₂ ≈ 432.9 minutes.
Therefore, the half-life of the reaction, assuming first-order kinetics, would be approximately 432.9 minutes.