Problem 100 - Chemistry The Central Science In Si Units

Question

Urea (NH_2CONH \(_{2}\) ) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{MHCl}\) occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ 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 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 \(0.500 \mathrm{M} ?\) (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) The rate constant, \(k\), is \(4.28 \times 10^{-4} \text{ s}^{-1}\). (b) The concentration of urea after 4000 s is \(0.090 \text{ M}\). (c) The half-life at 61.05°C is approximately 1619 s.

Step 1: Determine the Rate Constant, k

Given the reaction is first order in urea, we can use the rate equation for a first-order reaction:\[ \text{Rate} = k \times [\text{NH}_2\text{CONH}_2] \]Substituting the values given:\[ 8.56 \times 10^{-5} \text{ M/s} = k \times 0.200 \text{ M} \]Solving for \(k\):\[ k = \frac{8.56 \times 10^{-5} \text{ M/s}}{0.200 \text{ M}} = 4.28 \times 10^{-4} \text{ s}^{-1} \]

Step 2: Calculate Concentration After Time t

For a first-order reaction, the concentration as a function of time is given by:\[ [A]_t = [A]_0 e^{-kt} \]Where \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is time.Substitute the known values:\([A]_t = 0.500 \text{ M} \times e^{-4.28 \times 10^{-4} \text{ s}^{-1} \times 4000 \text{ s}} \)Calculate:\[ = 0.500 \text{ M} \times e^{-1.712} \]\[ = 0.500 \text{ M} \times 0.180 \]\[ \approx 0.090 \text{ M} \]

Step 3: Calculate the Half-Life

The half-life \(t_{1/2}\) for a first-order reaction is calculated using the formula:\[ t_{1/2} = \frac{0.693}{k} \]Substitute the rate constant \(k\):\[ t_{1/2} = \frac{0.693}{4.28 \times 10^{-4} \text{ s}^{-1}} \]Calculate:\[ \approx 1619 \text{ s} \]

Key Concepts

First-Order Reaction

A first-order reaction is a type of chemical reaction where the rate depends linearly on the concentration of one reactant. In simple terms, if you double the concentration of that reactant, the rate of the reaction will also double. This implies a direct proportionality between the concentration of the reactant and the reaction rate. It is especially important because it simplifies the analysis of the reaction kinetics.
In a first-order reaction, the rate equation is:\[ \text{Rate} = k \times [\text{Reactant}] \]
Here, \( k \) is the rate constant specific to the reaction and conditions, and \([\text{Reactant}]\) represents the concentration of the reactant. Understanding this concept is crucial because it lays the foundation for calculating other parameters such as the rate constant and concentration over time.
Studying first-order reactions helps in various scientific fields. They occur frequently in not only chemistry labs but also in biological processes, like radioactive decay or certain metabolic reactions.

Rate Constant

The rate constant, denoted as \( k \), is a crucial factor in determining the rate at which a reaction occurs. It is a proportionality constant in the rate equation of a reaction, linking the rate of reaction to the concentration of the reactants.
For a first-order reaction, the rate constant has units of \( \text{s}^{-1} \), which indicates the probability of a reaction occurring per second. The rate constant gives us insight into how fast or slow a reaction takes place under certain conditions.
To find \( k \), you can rearrange the rate equation:\[ k = \frac{\text{Rate}}{[\text{Reactant}]} \]
This was demonstrated using the given information, where calculating the rate constant required dividing the observed rate by the concentration of urea. By understanding \( k \), chemists can predict the behavior of the reaction under various conditions.
Moreover, the rate constant can change if temperature changes or if a catalyst is introduced, highlighting the dynamic nature of chemical reactions.

Half-Life

Half-life is a term commonly associated with the rate of reaction decay, particularly in first-order reactions. It refers to the time required for the concentration of a reactant to decrease to half its initial value.
For first-order reactions, the half-life is calculated using the formula:\[ t_{1/2} = \frac{0.693}{k} \]
This expression shows that the half-life of a first-order reaction is independent of its initial concentration, setting it apart from other reaction types. The half-life concept is important because it provides a tangible measure of how fast a reaction proceeds.

• It allows predictions of how much of a reactant remains after a certain period.
• It can simplify the monitoring of reaction progress over time.

Understanding half-life is essential in many real-world applications, such as pharmacokinetics in medicine or nuclear chemistry, where it helps in determining the stability and required dosage of pharmaceuticals.

Reaction Mechanism

The reaction mechanism outlines the step-by-step sequence of elementary reactions by which a chemical change occurs. It details the pathway from reactants to products, encompassing all intermediate stages and structures.
Understanding the reaction mechanism is vital because:

• It helps predict the intermediate species and transition states during a reaction.
• It provides insight into how the reaction conditions affect the rate and outcome.
• It assists in proposing modifications to improve reaction efficiency or selectivity.

For the given decomposition of urea, understanding that it occurs via a first-order mechanism simplifies predictions about its behavior over time. Such knowledge can aid in both practical applications and theoretical understanding, making it a cornerstone in chemical kinetics studies.
The reaction mechanism can often be complex, involving several steps, but by breaking it down, chemists can better control and utilize these reactions in various scientific fields.