Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} .\) At \(23{ }^{\circ} \mathrm{C}\) and in \(0.5 \mathrm{M} \mathrm{HCl}\), the following data were obtained for the disappearance of sucrose: $$ \begin{array}{rl} \hline \text { Time }(\mathrm{min}) & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](M)} \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] ?(\mathbf{b})\) What is the rate constant? (c) Using this rate constant, calculate the concentration of sucrose at 39,80,140 , and 210 min if the initial sucrose concentration was \(0.316 \mathrm{M}\) and the reaction was zero order in sucrose.

Step by step solution

01

Identify the rate law for the reaction

The rate law indicates the relationship between the rate of a chemical reaction and the concentration of the reactants. We need to determine the order of the reaction with respect to the sucrose concentration. Two potential rate laws can be considered: 1) First-order reaction: Rate = k [Sucrose] 2) Second-order reaction: Rate = k [Sucrose]^2 In both cases, k is the rate constant. We will analyze the data and determine which rate law best matches the given data.

02

Calculate the change in concentration over time

We need to calculate the change in sucrose concentration over time. Using the given data, we can create a table that displays the time, sucrose concentration, change in sucrose concentration, and ratio of change in concentration to initial concentration. $$ \begin{array}{|c|c|c|c|} \hline \text{Time(min)} & [\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}](\mathrm{M}) & \Delta [\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}] & \frac{\Delta [\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}]}{[C_{12}H_{22}O_{11}]_0} \\ \hline 0 & 0.316 & 0 & \\ 39 & 0.274 & -0.042 & 0.133 \\ 80 & 0.238 & -0.078 & 0.247 \\ 140 & 0.190 & -0.126 & 0.398 \\ 210 & 0.146 & -0.170 & 0.538 \\ \hline \end{array} $$

03

Test for first-order reaction

First, we will test if the reaction is of first order with respect to sucrose concentration. For first-order reactions, we have the equation: \[ t = \frac{1}{k} \ln \frac{[C_{12}H_{22}O_{11}]_0}{[C_{12}H_{22}O_{11}]_t} \] We can determine the average value of k for different time periods and see if it remains consistent: 1) From 0 to 39 minutes: \[ k_1 = \frac{\ln(\frac{0.316}{0.274})}{39} = 3.139 \times 10^{-3}\ \text{min}^{-1} \] 2) From 39 to 80 minutes: \[ k_2 = \frac{\ln(\frac{0.274}{0.238})}{80 - 39} = 3.144 \times 10^{-3}\ \text{min}^{-1} \] 3) From 80 to 140 minutes: \[ k_3 = \frac{\ln(\frac{0.238}{0.190})}{140 - 80} = 3.111 \times 10^{-3}\ \text{min}^{-1} \] It can be observed that the rate constant, k, remains reasonably consistent over different time periods. Thus, we can affirm that the reaction is first-order with respect to sucrose concentration.

04

Calculate the average rate constant

We can now calculate the overall average rate constant, k, for the first-order reaction: \[ k_{\text{avg}} = \frac{k_1 + k_2 + k_3}{3} = \frac{3.139 \times 10^{-3} + 3.144 \times 10^{-3} + 3.111 \times 10^{-3}}{3} = 3.131 \times 10^{-3}\ \text{min}^{-1} \] The reaction thus has the rate law: Rate = \(k_{\text{avg}} \times\) [Sucrose] Answer: (a) The reaction is first-order with respect to sucrose concentration. (b) The rate constant is \(3.131 \times 10^{-3}\) min^{-1}.

05

Calculate the concentration of sucrose at given times

Since we know that reaction's rate constant and the initial sucrose concentration, we can use the first-order reaction equation to calculate the concentration of sucrose at given times: (a) t = 39 min: \[ [C_{12}H_{22}O_{11}]_{39} = [C_{12}H_{22}O_{11}]_0 \times e^{-k \times t} = 0.316 \times e^{-3.131 \times 10^{-3} \times 39} = 0.274\ \mathrm{M} \] (b) t = 80 min: \[ [C_{12}H_{22}O_{11}]_{80} = 0.316 \times e^{-3.131 \times 10^{-3}\times 80} = 0.238\ \mathrm{M} \] (c) t = 140 min: \[ [C_{12}H_{22}O_{11}]_{140} = 0.316 \times e^{-3.131 \times 10^{-3}\times 140} = 0.190\ \mathrm{M} \] (d) t = 210 min: \[ [C_{12}H_{22}O_{11}]_{210} = 0.316 \times e^{-3.131 \times 10^{-3}\times 210} = 0.146\ \mathrm{M} \] Answer: (c) The concentration of sucrose at the given times can be calculated as follows: 39 min: 0.274 M, 80 min: 0.238 M, 140 min: 0.190 M, 210 min: 0.146 M.

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