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For any first-order reaction, understanding its half-life is crucial. The half-life is the time it takes for half of the reactant to be used up, or depleted. In a first-order reaction, the half-life remains constant. This is one of the unique characteristics of first-order reactions. Even as the reaction progresses, the half-life does not change.
To calculate the half-life, the formula is:

• \( t_{1/2} = \frac{0.693}{k} \)

Where \( t_{1/2} \) is the half-life, and \( k \) is the rate constant. Given that \( k \approx 0.00307 \, s^{-1} \), the half-life is approximately 226 seconds for this reaction. This means every 226 seconds, half of the remaining reactant is consumed, maintaining a predictability crucial for many scientific calculations.Using the same formula, you can see why the second half-life remains identical to the first—it's due to the nature of first-order reactions where the concentration of the reactant reduces exponentially but the half-life remains constant.

The rate constant, often denoted as \( k \), plays a pivotal role in defining the speed of the reaction. This constant varies for different reactions and is heavily influenced by factors such as temperature and the nature of the reactants. In a first-order reaction, the rate is directly proportional to the concentration of a single reactant.To calculate the rate constant for our example, we used the formula from the integrated rate law:

• \( k = \frac{ln(1/0.25)}{320} \)

Here, 75% completion means that 25% of the reactant is still present, hence \( \frac{A_t}{A_0} = 0.25 \). Solving gives a \( k \) value of \( 0.00307 \, s^{-1} \). This number dictates how fast the reaction occurs and is fundamental to predicting reaction behavior over time.

The integrated rate law is an equation that links the concentration of reactant over time. For first-order reactions, it is expressed as:

• \( ln\frac{A_0}{A_t} = kt \)

This equation allows us to determine the concentration of reactants at any given time, as long as we know the initial concentration \( A_0 \) and the rate constant \( k \). For our problem, it was key in calculating how long it takes for the reaction to progress to various stages of completion, like 75% or 90%.For example, to find the completion time for 90% reaction:

• \( ln\frac{1}{0.10} = k \times t \)

With \( A_t = 0.10 \times A_0 \), the integrated rate law told us precisely how long it would take to reach this point—763 seconds. This formula is vital in field experiments and industrial applications to ensure reactions conclude as expected.

Completion time estimates how long it takes for a reaction to reach a specified level of completion. It's an essential concept, especially in processes where exact timing equates to efficiency and safety. For a reaction at 90% completion, only 10% reactant remains, thus the fraction \( \frac{A_t}{A_0} = 0.10 \).Using the integrated rate law, we solve for \( t \):

• \( ln\frac{1}{0.10} = 0.00307 \times t \)

After rearranging and solving this equation, the time for 90% completion emerges as approximately 763 seconds. Understanding completion times helps in planning and optimizing chemical processes, preventing overuse of resources or hazardous outcomes due to prolonged reactions.