Question

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 \(61.05^{\circ} \mathrm{C} ?\)

Short answer

The rate constant k for this reaction at 61.05 °C is \(4.28 \times 10^{-4} s^{-1}\). The concentration of urea after \(4.00 \times 10^{3}\) seconds is approximately 0.181 M, and the half-life of the reaction is approximately 1619 seconds.

Step-by-step solution

Finding the rate constant k

For a first-order reaction, the rate is given by: Rate = k * [urea] We are given: Rate = \(8.56 \times 10^{-5} M/s\) [Urea] = 0.200 M Substitute the given values into the rate equation and solve for k: \(8.56 \times 10^{-5} M/s = k \times 0.200 M\) k = \(\frac{8.56 \times 10^{-5} M/s}{0.200 M}\) k = \(4.28 \times 10^{-4} s^{-1}\) So, the rate constant k is \(4.28 \times 10^{-4} s^{-1}\).

Finding the concentration of urea after 4.00 x 10³ s

To find the concentration of urea at a given time for a first-order reaction, we can use the following equation: [urea]_t = [urea]_₀ * \({e}^{-kt}\), where [urea]_t is the concentration at time t, [urea]_₀ is the initial concentration, k is the rate constant, and t is the time. We are given: [urea]_₀ = 0.500 M k = \(4.28 \times 10^{-4} s^{-1}\) t = \(4.00 \times 10^{3}s\) Substitute the given values into the equation: [urea]_t = (0.500 M) * \({e}^{-(4.28 \times 10^{-4} s^{-1})(4.00 \times 10^{3}s)}\) [urea]_t = (0.500 M) * \({e}^{-1.71}\) [urea]_t ≈ 0.181 M So, the concentration of urea after \(4.00 \times 10^{3}\) s is approximately 0.181 M.

Finding the half-life of the reaction

For a first-order reaction, the half-life (t₁/₂) is given by the following equation: t₁/₂ = \(\frac{0.693}{k}\), where k is the rate constant. We know: k = \(4.28 \times 10^{-4} s^{-1}\) Substitute the value of k into the half-life equation: t₁/₂ = \(\frac{0.693}{4.28 \times 10^{-4} s^{-1}}\) t₁/₂ ≈ 1619 s So, the half-life for this reaction at 61.05 °C is approximately 1619 s.

Key concepts

Rate Constant

When studying the fascinating world of chemical reactions, one term that often emerges is the 'rate constant', typically denoted by the symbol 'k'. The rate constant is a crucial component in the equations of chemical kinetics that allows us to understand not only the speed but also the details of how a chemical reaction proceeds.

For a first-order reaction, as in the case of the urea decomposition example, the rate constant links the concentration of the reactants to the rate of the reaction. It encapsulates the effects of environmental factors such as temperature and pressure, and it’s unique for each reaction under specific conditions. The mathematical relationship is expressed as: Rate = k * [Reactant]. In the urea example, we calculated the rate constant 'k' by rearranging this equation to solve for 'k' given the rate and concentration of urea.

The value of 'k' doesn't merely offer a snapshot of the reaction's briskness; it helps predict future reaction rates and concentrations at any moment. However, it's important to bear in mind that 'k' can change with temperature, and this dependency can be quantified by the Arrhenius equation, adding another layer to its interpretation. But for one reaction at a constant temperature, the rate constant remains steadfast, a numerical sentinel of the reaction's relentless pace.

Half-life of Reaction

The 'half-life of a reaction' refers to the time required for the amount of a reactant to decrease by half its initial value. In the realm of chemical kinetics, this is a valuable measure as it gives a direct insight into the enduring nature of a reactant over time. For first-order reactions, the beauty lies in the constancy of the half-life; it remains unchanged regardless of the starting concentration.

This exceptional property is encapsulated in the simple equation: t₁/₂ = \(\frac{0.693}{k}\), with 't₁/₂' symbolizing the half-life and 'k' standing for our ever-important rate constant. In the case of urea's decomposition, once we know 'k', determining the half-life becomes straightforward. We found out that the urea half-life, at the given temperature, was approximately 1619 seconds.

Understanding half-life is not just academically valuable. It has practical implications, such as for pharmaceuticals within the body, where half-life informs dosage and frequency. Additionally, in environmental chemistry, the half-life tells us how long a pollutant might persist. It interlinks chemistry with time, giving us a yardstick for the unseen progress of reactions.

Chemical Kinetics

At the center of these discussions is the sweeping field of 'chemical kinetics', the study of rates of chemical processes and the factors influencing them. Chemical kinetics doesn't just provide a stopwatch to measure how quickly reactions occur; it unravels the molecular narrative behind the visible changes we observe.

It answers questions such as 'How speedy is this reaction?', 'What path does it take?', and 'What can affect this molecular hustle?'. Understanding chemical kinetics allows scientists and students alike to predict reaction behavior, design reactions for desired speeds, and even prevent unwanted reactions.

The kinetic investigation of the urea reaction reveals the first-order nature of its kinetics and provides a window into the understanding of more complex reactions. By studying the rate, rate constant, and half-life, we gain insights into not just the 'how fast' but also the 'how' of a reaction. As students delve into exercises like those of urea decomposition, they build a framework that will support their explorations into the intricate dance of atoms and molecules that is central to the science of chemistry.