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Chemistry Question on Chemical Kinetics

A first order reaction is half completed in 45 min. How long does it need 99.9% of the reaction to be completed?

A

10 Hours

B

5 Hours

C

20 Hours

D

7.5 Hours

Answer

7.5 Hours

Explanation

Solution

A first-order reaction follows the exponential decay equation:
ln⁡([A]t[A]0)=−kt\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt
Where:
[A]t[A]_t is the concentration of reactant at time t
[A]0[A]_0 is the initial concentration of reactant
k is the rate constant
t is the time
We can rearrange the equation to solve for t:
t=ln⁡([A]t/[A]0)−kt = \frac{\ln([A]_t/[A]_0)}{-k}

Given that the reaction is half completed in 45 minutes, we can use this information to find the rate constant (k) for the reaction. At the half-life of a first-order reaction, [A]t[A]0=0.5:\frac{[A]_t}{[A]_0} = 0.5:
0.5=e−k×45 min0.5 = e^{-k \times 45 \text{ min}}

Taking the natural logarithm of both sides:
ln⁡(0.5)=−k×45 min\ln(0.5) = -k \times 45 \text{ min}

Solving for k:
k=ln⁡(0.5)−45 mink = \frac{\ln(0.5)}{-45 \, \text{min}}

Now, let's find the time needed for 99.9% of the reaction to be completed. We'll assume [A]t/[A]0 is 0.001 (0.1% of the initial concentration):
t=ln⁡([A]t[A]0)−kt = \frac{\ln\left(\frac{[A]_t}{[A]_0}\right)}{-k}

t=ln⁡(0.001)−kt = \frac{\ln(0.001)}{-k}

t=ln⁡(0.001)ln⁡(0.5)−45 mint = \frac{\ln(0.001)}{\frac{\ln(0.5)}{-45 \, \text{min}}}
Using a calculator:
t≈7.5 hourst ≈ 7.5\ hours
Therefore, the time needed for 99.9% of the reaction to be completed is approximately 7.5 hours. Option (D) 7.5 hours is the correct answer.