Answer:
[tex]t=23.5min[/tex]
Explanation:
Hello,
In this case, since the rate equation turns out as shown below due to the first-order kinetics:
[tex]\frac{dC_A}{dt}=-kC_A[/tex]
Its integration results:
[tex]\int\limits^{C_A}_{C_A^0} { \frac{dC_A}{C_A}} \,=-k\int\limits^t_0 {} \, dt\\ln(\frac{C_A}{C_A^0})=-kt[/tex]
However, the rate constant is computed by considering the given half-life time as follows:
[tex]k=\frac{ln(2)}{t_{1/2}}=\frac{ln(2)}{2.42x10^3s}=2.86x10^{-4}s^{-1}[/tex]
In such a way, the required time in minutes to diminish the concentration by 66.8% of the initial turns out:
[tex]C_A=0.668C_A^0[/tex]
[tex]ln(\frac{0.668C_A^0}{C_A^0})=-kt[/tex]
[tex]t=\frac{-ln(0.668)}{2.86x10^{-4}s^1}=1408.3s*\frac{1min}{60s} \\t=23.5min[/tex]
Best regards.
