A first-order reaction follows the exponential decay equation:
\(\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt\)
Where:
\([A]_t\) is the concentration of reactant at time t
\([A]_0\) is the initial concentration of reactant
k is the rate constant
t is the time
We can rearrange the equation to solve for t:
\(t = \frac{\ln([A]_t/[A]_0)}{-k}\)
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, \(\frac{[A]_t}{[A]_0} = 0.5:\)
\(0.5 = e^{-k \times 45 \text{ min}}\)
Taking the natural logarithm of both sides:
\(\ln(0.5) = -k \times 45 \text{ min}\)
Solving for k:
\(k = \frac{\ln(0.5)}{-45 \, \text{min}}\)
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):
\(t = \frac{\ln\left(\frac{[A]_t}{[A]_0}\right)}{-k}\)
\(t = \frac{\ln(0.001)}{-k}\)
\(t = \frac{\ln(0.001)}{\frac{\ln(0.5)}{-45 \, \text{min}}}\)
Using a calculator:
\(t ≈ 7.5\ hours\)
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.
For a first-order reaction, the time required to complete a certain percentage is given by:
\( t = \frac{2.303}{k} \log_{10} \left( \frac{[R]_0}{[R]} \right) \)
Half-life of a first-order reaction is:
\( t_{1/2} = \frac{0.693}{k} \)
Given: \( t_{1/2} = 45 \, \text{min} \)
So, \( k = \frac{0.693}{45} \)
For 99.9% completion, \( \frac{[R]_0}{[R]} = \frac{100}{0.1} = 1000 \)
Now plug into the first equation:
\( t = \frac{2.303}{k} \log_{10}(1000) = \frac{2.303}{k} \times 3 \)
Substitute \( k = \frac{0.693}{45} \):
\( t = \frac{2.303 \times 3 \times 45}{0.693} \approx \frac{310.905}{0.693} \approx 448.5 \, \text{min} \)
Convert to hours:
\( \frac{448.5}{60} \approx 7.5 \, \text{hours} \)
Correct Answer: 7.5 Hours
The following data were obtained during the first order thermal decomposition of \( \text{N}_2\text{O}_5(g) \) at constant volume: