Question

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, C12H22O11+H2O⟶C6H12O6+C6H12O6 follows a first-order rate law for the disappearance of sucrose: rate = k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.) (a) In neutral solution, k = 2.1 × 10–11 s–1 at 27 °C and 8.5 × 10–11 s–1 at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature). (b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 × 10–7 M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible. (c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)?

Answer 1

To solve for the activation energy and frequency factor, we can use the Arrhenius equation with the given rate constants at 27 ℃ and 37 ℃. The rate constant at 47 ℃ can then be calculated. For estimating the time to reach equilibrium at 27 ℃, the first-order kinetics are applied, simplifying the calculation by assuming the reaction is irreversible due to the low concentration of sucrose at equilibrium.

Step-by-step explanation:

Calculating Activation Energy and Reaction Rates

The hydrolysis of sucrose into glucose and fructose follows a first-order rate law, meaning the rate of reaction is directly proportional to the concentration of sucrose. Using the Arrhenius equation and the rate constants provided at 27 ℃ and 37 ℃, we can calculate the activation energy (Ea) and the frequency factor (A).

To find the rate constant (k) at 47 ℃, we first need to determine Ea using the two rate constants given for 27 ℃ (k1 = 2.1 x 10-11 s-1) and 37 ℃ (k2 = 8.5 x 10-11 s-1). This can be done by rearranging the Arrhenius equation to solve for Ea using the natural logarithm of the rate constants and the respective temperatures.

For the time to reach equilibrium at 27 ℃, we apply the first-order rate law, starting from a concentration of sucrose (C0) of 0.150 M to an equilibrium concentration (Ce) of 1.65 x 10-7 M. Using the integrated rate law for first-order reactions, we can calculate the time required for the reaction to reach this concentration. Assuming an irreversible reaction simplifies the calculation as the concentration of sucrose at equilibrium is so low that it virtually has no impact on the reverse reaction's rate.

Answer 2

The activation energy, frequency factor, and rate constant can be determined using the Arrhenius equation. The time it takes for a sucrose solution to reach equilibrium can be calculated using the concept of half-life. Assuming the reaction is irreversible simplifies the calculation by considering only the one-way process.

Step-by-step explanation:

(a) The Arrhenius equation relates the rate constant of a reaction to the activation energy, temperature, and the frequency factor. The Arrhenius equation is given by k = Ae^(-Ea/RT), where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the ideal gas constant, and T is the temperature in Kelvin. To determine the activation energy and frequency factor at 47°C, we can use the given rate constant values at 27°C and 37°C. Taking the natural logarithm of both sides of the Arrhenius equation, we can write ln(k) = ln(A) - (Ea/RT), which is in the form of a linear equation. By using the given rate constants and temperatures, we can set up two equations for ln(k1) = ln(A) - (Ea/R)(1/T1) and ln(k2) = ln(A) - (Ea/R)(1/T2). Rearranging these equations, we get ln(k1) - ln(k2) = (Ea/R)(1/T2 - 1/T1). Substituting the values for k1, k2, T1, and T2, we can solve for Ea and ln(A), which will allow us to determine the frequency factor. Once we have Ea and ln(A), we can plug in the values for R, T, and solve for the rate constant at 47°C.

(b) To find the time it takes for a solution of sucrose to reach equilibrium at 27°C in the absence of a catalyst, we can use the concept of half-life. The half-life of a reaction is the time it takes for the concentration of the reactant to decrease by half. Since the reaction is assumed to be first-order, the half-life can be found using the equation t1/2 = (ln 2) / k, where t1/2 is the half-life and k is the rate constant. Given the initial concentration of sucrose and the concentration at equilibrium, we can use the half-life equation to find the time it takes for the concentration of sucrose to decrease from the initial concentration to the concentration at equilibrium.

(c) Assuming that the reaction is irreversible simplifies the calculation in part (b) because it means that the concentration of sucrose only decreases with time and reaches a final equilibrium concentration. In a reversible reaction, the reactants can react to form products, but the products can also react to reform the reactants. This leads to a dynamic equilibrium where the concentrations of reactants and products remain constant. However, since the concentration of sucrose at equilibrium is very low, it suggests that the reaction is mostly one-way and the reverse reaction is negligible. By assuming that the reaction is irreversible, we can simplify the calculations and treat the reaction as a one-way process.