if the initial concentration of the reactant in the first order reaction A→products is 0.64 mol/L and the half-life is 30 s, calculate the concentration of reactant 120 s after initiation of the reaction
half lives = 120/30 = 4
2^4 = 16 and
0.64/16 = ??
To calculate the concentration of the reactant 120 seconds after initiation of the reaction, we can use the first-order reaction equation and the concept of half-life.
The first-order reaction equation is given by: ln[A] = -kt + ln[A]₀, where [A] is the concentration of the reactant at a given time, t is the time elapsed, k is the rate constant, and [A]₀ is the initial concentration of the reactant.
First, we need to find the rate constant (k) using the half-life (t₁/₂) given:
t₁/₂ = (0.693 / k)
Given that the half-life (t₁/₂) is 30 seconds, we can calculate k:
k = 0.693 / t₁/₂
k = 0.693 / 30
k ≈ 0.0231 s⁻¹
Now, we can substitute the values into the first-order reaction equation to find the concentration of the reactant after 120 seconds (t = 120s):
ln[A] = -kt + ln[A]₀
ln[A] = -0.0231 * 120 + ln(0.64)
ln[A] = -2.772 + ln(0.64)
ln[A] ≈ -2.772 + (-0.446)
ln[A] ≈ -3.218
To find the concentration of the reactant (A), we need to exponentiate both sides of the equation:
ln[A] = -3.218
e^(ln[A]) = e^(-3.218)
[A] = 0.0399 mol/L
Therefore, the concentration of the reactant 120 seconds after initiation of the reaction is approximately 0.0399 mol/L.