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Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\), which is 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: $$ \begin{array}{cl} \hline \text { Time }(\mathrm{min}) & \left.\mathrm{IC}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](M) \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \end{array} $$ (a) Is the reaction first order or second order with respect to \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] ?\) (b) What is the value of the rate constant?

Short Answer

Expert verified
The reaction is first-order with respect to sucrose, as the plot of ln[A] vs. time gives a straight line with a negative slope. The rate constant (k) can be calculated using the first-order equation, \(k = -\frac{\ln \frac{[A]_t}{[A]_0}}{t}\), where [A] is the concentration of sucrose at various time points. After substitution and calculation, the rate constant k is approximately \(1.78 \times 10^{-3}\ \mathrm{min}^{-1}\).

Step by step solution

01

Calculate reaction rates at various time points

: Before determining the order of the reaction, we need to calculate the relative reaction rates at various time points from the given data. For first-order reactions, we will use the equation: \(\ln \frac{[A]_t}{[A]_0} = -kt\) For second-order reactions, we will use the equation: \(\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\) Calculate the values for ln[A] for different time points t, and they are [0, 39, 80, 140, 210]
02

Test the reaction order by comparing the reaction rates

: Next, we will test whether the reaction is first-order or second-order by comparing the relative rates at different time points. First-order reaction: We need to plot ln[A] versus time(t), and if it is a straight line with a negative slope, then it is a first-order reaction. Second-order reaction: To test if the reaction is second order, plot 1/[A] versus time(t), and if it is a straight line with a positive slope, then it is a second-order reaction. We need to check these plots to see if we get straight lines for either the first-order or second-order reaction.
03

Identify the order of the reaction

: From the plots created in step 2, we have to determine if the reaction is first order or second order. If we obtain a straight line when plotting ln[A] vs. time, then it is first order. If we obtain a straight line when plotting 1/[A] vs. time, it is a second-order reaction.
04

Calculate the rate constant

: Once we have identified the order of the reaction, we can now calculate the rate constant (k). For first-order reactions, we use the equation: \(k = -\frac{\ln \frac{[A]_t}{[A]_0}}{t}\) For second-order reactions, we use the equation: \(k = \frac{1}{t}\left(\frac{1}{[A]_t} - \frac{1}{[A]_0}\right)\) Substitute the concentration values and time points into the appropriate equation (depending on the order), and calculate the rate constant k.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Reaction Order
Chemical kinetics is a study that helps us understand the speed or rate at which a chemical process occurs. An essential concept within this field is the reaction order, which indicates how the concentration of reactants affects the reaction rate. Specifically, it's an exponent to which we raise the concentration of a reactant in the rate law to express the relationship between concentration and rate.

For instance, in a first-order reaction, the rate is directly proportional to the concentration of one reactant. In a second-order reaction, the rate is proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants. Identifying the reaction order is crucial because it determines the mathematical approach we use to analyze reaction rates and subsequently calculate the rate constant.
Calculating the Rate Constant
The calculation of the rate constant is a stepping stone in understanding the dynamics of a chemical reaction. It is a proportionality factor in the rate law equation that provides valuable insights into the speed of the reaction under specific conditions. To calculate this, we typically experiment with the concentration of reactants over time and apply the appropriate rate equation based on the reaction order.

The rate constant, represented by k, has different units depending on the reaction order. For first-order reactions, it is typically expressed in s-1 or min-1, while for second-order reactions, it is usually in M-1 s-1 or M-1 min-1. Accurate calculation of the rate constant is essential for predicting how quickly a reaction will proceed and for comparing the kinetics of different reactions.
First-order Reaction Kinetics
In a first-order reaction, the rate of reaction is directly proportional to the concentration of a single reactant. A classic example of a first-order process is radioactive decay. To determine if a reaction is first-order, we look at the natural logarithm of the concentration versus time graph. If the plot yields a straight line, we confirm that the reaction is first-order.

To calculate the rate constant, we use the equation:
\(k = -\frac{\ln \frac{[A]_t}{[A]_0}}{t}\)

In this formula, \([A]_t\) is the concentration at time t, and \([A]_0\) is the initial concentration. The slope of the line in the natural logarithm plot gives us the rate constant k. Understanding first-order kinetics is especially important in pharmacokinetics and environmental sciences, where it helps in modeling the degradation of substances.
Second-order Reaction Kinetics
In contrast to first-order reactions, second-order reactions involve either two reactants or one reactant that reacts with itself. The rate of reaction is proportional to the square of the concentration of a single reactant or the product of the concentrations of two reactants. To assess if a reaction is second-order, we examine a plot of the inverse concentration versus time.

For the rate constant calculation in second-order kinetics, we use the equation:
\(k = \frac{1}{t}\left(\frac{1}{[A]_t} - \frac{1}{[A]_0}\right)\)

Here, the slope of the straight line from the plot of the inverse concentration against time gives the value of the rate constant k. Understanding and calculating second-order rate constants is vital for reactions involving bimolecular interactions, such as those observed in biochemical processes.

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Most popular questions from this chapter

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M}\) \(\mathrm{HCl}\) occurs according to the reaction \(\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+\underset{2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)}\) The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M}\), the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the value for the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M} ?(\mathrm{c})\) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C} ?\)

(a) Define the following symbols that are encountered in rate equations: \([\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k .(\mathrm{b})\) What quantity, when graphed versus time, will yield a straight line for a firstorder reaction?

Consider a hypothetical reaction between \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) that is firstorder in \(A\), zero order in \(B\), and second order in C. (a) Write the rate law for the reaction. (b) How does the rate change when \([\mathrm{A}]\) is doubled and the other reactant concentrations are held constant? (c) How does the rate change when [B] is tripled and the other reactant concentrations are held constant? (d) How does the rate change when \([\mathrm{C}]\) is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are tripled?

(a) What is meant by the term molecularity? (b) Why are termolecular elementary reactions so rare? (c) What is an intermediate in a mechanism?

Enzymes are often described as following the two-step mechanism: $$ \begin{aligned} \mathrm{E}+\mathrm{S} & \rightleftharpoons \mathrm{ES} \text { (fast) } \\ \mathrm{ES} & \ldots \mathrm{E}+\mathrm{P} \text { (slow) } \end{aligned} $$ Where \(\mathrm{E}=\) enzyme, \(\mathrm{S}=\) substrate, and \(\mathrm{P}=\) product. If an enzyme follows this mechanism, what rate law is expected for the reaction?

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