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 $$ \begin{aligned} \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) \end{aligned} $$ 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 \(61.05^{\circ} \mathrm{C}\) ?

Short answer

(a) The rate constant, \(k\), is \(4.28 \times 10^{-4} \mathrm{s^{-1}}\). (b) The concentration of urea after \(4.00 \times 10^{3}\mathrm{s}\) is approximately \(0.181\text{ M}\). (c) The half-life for this reaction at \(61.05^{\circ}\mathrm{C}\) is approximately \(1.62 \times 10^3\mathrm{s}\).

Step-by-step solution

Find the rate constant, k

From the given information, the rate of the reaction is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\) when the concentration of urea is \(0.200 \mathrm{M}\). Using the rate law for a first order reaction, we can write: $$\text{rate} = k [\text{urea}]$$ Now, we can solve for k: $$k = \frac{\text{rate}}{[\text{urea}]} = \frac{8.56 \times 10^{-5}\mathrm{M/s}}{0.200\mathrm{M}} = 4.28 \times 10^{-4} \mathrm{s^{-1}}$$

Find the concentration of urea after 4000s

To find the concentration of urea after \(4.00 \times 10^{3}\mathrm{~s}\), we can use the integrated rate law for a first order reaction: $$\ln\frac{[\text{urea}]_{0}}{[\text{urea}]_{t}} = kt$$ Plugging in the given values and the calculated value of k: $$\ln\frac{0.500\text{ M}}{[\text{urea}]} = (4.28 \times 10^{-4} \mathrm{s^{-1}})(4.00 \times 10^3\mathrm{s})$$ Solving for the concentration of urea: $$\ln\frac{0.500\text{ M}}{[\text{urea}]} = 1.712$$ $$[\text{urea}] = 0.500\text{ M} e^{-1.712} \approx 0.181\text{ M}$$

Find the half-life for the reaction

To find the half-life for a first order reaction, we can use the formula: $$t_{1/2} = \frac{0.693}{k}$$ Plugging in the calculated value for k: $$t_{1/2} = \frac{0.693}{4.28 \times 10^{-4} \mathrm{s^{-1}}} \approx 1.62 \times 10^3\mathrm{~s}$$ So the answers are: (a) The rate constant, k, is \(4.28 \times 10^{-4} \mathrm{s^{-1}}\). (b) The concentration of urea after \(4.00 \times 10^{3}\mathrm{~s}\) is approximately \(0.181\text{ M}\). (c) The half-life for this reaction at \(61.05^{\circ}\mathrm{C}\) is approximately \(1.62 \times 10^3\mathrm{~s}\).

Key concepts

First Order Reaction

Understanding chemical kinetics is crucial for students and scholars of chemistry, and the concept of a first order reaction is a fundamental component of this area. A first order reaction is characterized by its rate depending solely on the concentration of one reactant. Mathematically, the rate of a first order reaction is directly proportional to the reactant's concentration.

This can be expressed by the rate law, which in the simplest form is: \[\text{rate} = k [\text{A}]\]Where \(k\) is the rate constant and \([\text{A}]\) is the concentration of the reactant A. What makes first order reactions particularly manageable in calculations, is the linear relationship between the logarithm of the concentration and time, which can be described using the integrated rate law. In the case of our exercise involving urea decomposition, since the reaction is first order in urea, the rate at which urea decomposes is directly proportional to its concentration at any given time.

Rate Constant

The rate constant, often represented by the symbol \(k\), is a crucial parameter in the rate law equation, serving as a proportionality factor that links the rate of reaction to the concentration(s) of the reactant(s). The value of the rate constant can vary widely and is influenced by various factors such as temperature, the presence of a catalyst, and the nature of the reactants.

For a first order reaction, the rate constant has a unit of \(\text{s}^{-1}\), indicating the reaction rate per second. Having a higher value for the rate constant means a faster reaction. By calculating the rate constant, as shown in the solution for the urea reaction exercise, students can predict how fast the reaction will proceed under certain conditions and this provides valuable insights when comparing different reactions or altering reaction conditions.

Half-life of Reaction

The half-life of a reaction, denoted as \(t_{1/2}\), is the time required for half of the reactant to be consumed or transformed. It's a concept that provides insight into the duration a reaction takes to reach a certain state. Importantly, for first order reactions, the half-life is constant and does not depend on the initial concentration. This unique feature of first order reactions makes them particularly interesting in fields like radiochemistry and pharmacokinetics.

Using the equation\[t_{1/2} = \frac{0.693}{k}\], students can calculate the half-life once they know the rate constant, as seen in the urea decomposition example. This constant relationship also helps validate if a reaction follows first order kinetics—any variation in half-life with concentration would imply the reaction is not first order.

Integrated Rate Law

The integrated rate law is a powerful tool, allowing chemists to relate the concentrations of reactants over time. For first order reactions, the law takes the form:\[\ln\frac{[\text{A}]_{0}}{[\text{A}]_{t}} = kt\]

Here, \([\text{A}]_{0}\) is the initial concentration, \([\text{A}]_{t}\) is the concentration at time \(t\), and \(k\) is the rate constant. This equation is derived by integrating the rate law over time and provides a direct link between kinetics and time, which can be used to determine how much reactant remains after a certain period. It also allows calculation of reaction rates at different time points without constant monitoring. When students are faced with a task such as determining the concentration of urea after 4000 seconds, the integrated rate law is their formula of choice, as outlined in step two of our problem's solution.