Question

The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial concentration of A is 0.163 M, how many minutes will it take for the concentration of A to be 66.8% of the initial concentration?

Answer 1

Answer:

$t=23.5min$

Explanation:

Hello,

In this case, since the rate equation turns out as shown below due to the first-order kinetics:

$\frac{dC_A}{dt}=-kC_A$

Its integration results:

$\int\limits^{C_A}_{C_A^0} { \frac{dC_A}{C_A}} \,=-k\int\limits^t_0 {} \, dt\\ln(\frac{C_A}{C_A^0})=-kt$

However, the rate constant is computed by considering the given half-life time as follows:

$k=\frac{ln(2)}{t_{1/2}}=\frac{ln(2)}{2.42x10^3s}=2.86x10^{-4}s^{-1}$

In such a way, the required time in minutes to diminish the concentration by 66.8% of the initial turns out:

$C_A=0.668C_A^0$

$ln(\frac{0.668C_A^0}{C_A^0})=-kt$

$t=\frac{-ln(0.668)}{2.86x10^{-4}s^1}=1408.3s*\frac{1min}{60s} \\t=23.5min$

Best regards.