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First-order reactions exhibit a specific behavior in terms of their reaction rate. The reaction rate law for a first-order reaction provides a mathematical way to express how the concentration of a reactant decreases over time. This rate is expressed using the formula:\[ c = c_0 e^{-kt} \]- \(c\) is the concentration of the reactant at time \(t\).- \(c_0\) is the initial concentration of the reactant.- \(k\) is the rate constant which remains constant throughout the reaction duration.- \(e\) is the base of natural logarithms, approximately equal to 2.718.This formula indicates that the rate at which the reactant decreases is directly proportional to its concentration at any given time, which is a signature of first-order reactions. It highlights how even as the reaction progresses and the concentration of the reactants decreases, the exponential term governs the decrease to ensure a constant proportionality.

The rate constant \(k\) is a crucial component for understanding the speed of a reaction and is uniquely determined for each reaction under specific conditions. For a first-order reaction, calculating \(k\) involves using known concentrations and times from experimental data.Given the reaction scenario where 80% of 10 mol/L reactant is transformed into product in 10 seconds, the rate constant is calculated by setting up the formula:\[ c = c_0 e^{-kt} \]Here, after 10 seconds only 20% remains, thus:- Remaining concentration \( c = 2 \text{ mol/L} \)- Initial concentration \( c_0 = 10 \text{ mol/L} \)Plugging these values into the equation:\[ 2 = 10 e^{-10k} \]Solving for \(k\), take the natural logarithm:\[ e^{-10k} = 0.2 \] \[ -10k = ext{ln} (0.2) \]\[ k = -\frac{\text{ln} (0.2)}{10} \]This method provides the rate constant that can be applied under the same conditions, emphasizing its role in influencing reactions over time.

Concentration is a measure of the number of moles of a reactant in a certain volume and plays an integral role in understanding chemical reactions. For first-order reactions, the change in concentration over time is key to determining reaction progress. When starting with different concentrations, one may wonder how it affects the amount left after a common period. - In the provided exercise, the difficulties arise from different initial concentrations, yet time remains constant. Initially with 10 mol/L, we observed that after 10 seconds only 2 mol/L was left, demonstrating an 80% conversion. Using the same rate constant, applying it to a second initial concentration (5 mol/L) results in 1 mol/L remaining after 10 seconds, again - This percentage conversion seems unaffected by the initial volume due to the nature of first-order kinetics, showcasing the uncertainty with changing initial conditions yet consistent formula application.

Percentage conversion refers to the fraction of reactant transformed into product after a given time and provides a quick insight into how effective a reaction is under given conditions.For a consistent timeframe, irrespective of starting concentrations:- Knowing initial concentration is 5 mol/L and final is 1 mol/L,Calculate percentage conversion using:\[ \text{Converted percentage} = \left(\frac{c_0 - c}{c_0}\right) \times 100 \]From our exercise, the fraction converted from reactant to product is:- \((5 - 1)/5 = 0.8\)Meaning:\\[80\%\]Even with a lower starting concentration compared to the initial 10 mol/L, the percentage conversion remains the same at 80%. This consistency, irrespective of concentration variation, underscores a fundamental characteristic of first-order reactions: - They have no dependency on the initial concentration when calculating percentage conversion for a given period.