This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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To explain this solution to your lecturer, you would emphasize the fundamental principles of chemical kinetics for first-order reactions. The problem asks us to determine the fraction of diethyl ether remaining after a specific time, given its half-life for a first-order decomposition reaction. Step 1: Determine the rate constant (k) for the first-order reaction. For a first-order reaction, the rate of reaction is directly proportional to the concentration of a single reactant. A key characteristic of a first-order reaction is its half-life (t_1/2), which is the time required for the concentration of the reactant to decrease to half of its initial value. For a first-order reaction, the half-life is constant and is related to the rate constant (k) by the following equation: t_1/2 = ((2))/(k) Here, (2) is the natural logarithm of 2, which is a constant approximately equal to 0.6931. We are given the half-life t_1/2 = 1670 s. To determine the rate constant k, we rearrange the formula: k = ((2))/(t_1/2) Substitute the given value for t_1/2: k = (0.6931)/(1670 s) k ≈ 0.0004150 s^-1 This calculated rate constant k quantifies the intrinsic speed of the decomposition reaction. Step 2: Calculate the fraction of diethyl ether remaining after 4710 s. To determine the amount of reactant remaining after a specific time t, we utilize the integrated rate law for a first-order reaction. This equation relates the concentration of the reactant at time t ([A]_t) to its initial concentration ([A]_0), the rate constant (k), and the elapsed time (t): (([A]_t)/([A]_0)) = -kt The ratio ([A]_t)/([A]_0) represents the fraction of the reactant remaining at time t. To solve for this fraction, we need to eliminate the natural logarithm (). This is achieved by taking the exponential (base e) of both sides of the equation: e^(([A]_t)/([A]_0)) = e^-kt This simplifies to: ([A]_t)/([A]_0) = e^-kt Now, we substitute the calculated rate constant k = 0.0004150 s^-1 and the given time t = 4710 s: ([A]_t)/([A]_0) = e^-(0.0004150 s^-1 × 4710 s) First, we compute the product in the exponent: 0.0004150 × 4710 = 1.9545 Thus, the equation becomes: ([A]_t)/([A]_0) = e^-1.9545 Finally, we evaluate the exponential term: ([A]_t)/([A]_0) ≈ 0.1416 This result indicates that after 4710 s, approximately 0.1416 (or 14.16\%) of the initial diethyl ether remains. Note: The value 0.750 was not used in these calculations. The constant derived from the half-life formula for a first-order reaction is (2) ≈ 0.6931. The fraction of diethyl ether remaining after 4710 s is approximately 0.1416. What's next?