Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) hydrolyzes into glucose and fructose. The hydrolysis is a first-order reaction. The half-life for the hydrolysis of sucrose is \(64.2\) min at \(25^{\circ} \mathrm{C}\). How many grams of sucrose in \(1.25 \mathrm{~L}\) of a \(0.389 \mathrm{M}\) solution are hydrolyzed in \(1.73\) hours?

First, we have to calculate the initial amount of sucrose in grams. To do this, use the molarity and volume given: \(C_1 = 0.389\,\text{M}\) \(V_1 = 1.25\,\text{L}\) Next, we need to find the moles of sucrose: Moles of sucrose = \(C_1 \times V_1\) Moles of sucrose = \((0.389\,\text{M})(1.25\,\text{L}) = 0.48625\,\text{mol}\) Now, we can find the grams of sucrose using the molar mass of sucrose (C12H22O11), which is 342.3 g/mol. Initial grams of sucrose = Moles of sucrose × Molar mass: Initial grams of sucrose = \(0.48625\,\text{mol} \times 342.3\,\frac{\text{g}}{\text{mol}} \approx 166.39\,\text{g}\)

The half-life of sucrose (t1/2) is given as 64.2 minutes, and the rate constant (k) of a first-order reaction can be calculated using the formula: \(t_{1/2} = \frac{0.693}{k}\) To find k, rearrange the equation: \(k = \frac{0.693}{t_{1/2}}\) Before calculating k, convert the time given, 1.73 hours, to minutes: Time in minutes = \(1.73\,\text{hours} \times 60\,\frac{\text{min}}{\text{hour}} \approx 103.8\,\text{min}\) Now, calculate k: \(k = \frac{0.693}{64.2\,\text{min}} \approx 0.0108\,\text{min}^{-1}\)

With the rate constant (k) and time (t), we can find the final concentration of sucrose (C2) using the first-order reaction formula: \(C_2 = C_1 \times e^{-kt}\) \(C_2 = (0.389\,\text{M}) \times e^{-(0.0108\,\text{min}^{-1})(103.8\,\text{min})} \approx 0.213\,\text{M}\) Next, find the final moles of sucrose by multiplying the final concentration (C2) with the volume (V1): Final moles of sucrose = \(C_2 \times V_1\) Final moles of sucrose = \((0.213\,\text{M})(1.25\,\text{L}) \approx 0.26625\,\text{mol}\) Now, find the final grams of sucrose by multiplying the final moles of sucrose with the molar mass: Final grams of sucrose = \(0.26625\,\text{mol} \times 342.3\,\frac{\text{g}}{\text{mol}} \approx 91.13\,\text{g}\)

Now that we have both the initial and final amounts of sucrose in grams, we can calculate the hydrolyzed amount: Hydrolyzed amount of sucrose = Initial grams of sucrose - Final grams of sucrose Hydrolyzed amount of sucrose = \(166.39\,\text{g} - 91.13\,\text{g} \approx 75.26\,\text{g}\) Therefore, 75.26 grams of sucrose in a 1.25 L solution of 0.389 M concentrate are hydrolyzed in 1.73 hours at 25°C.