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 $$\mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)$$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 M,\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\) , (a) What is the rate constant, \(k ?\) units of \(s^{-1}\) . (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to 0.100\(?\)

Short answer

The rate constant, \(k = \frac{8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}}{0.200 \mathrm{M}} = 4.28 \times 10^{-4} \mathrm{s}^{-1}\). The half-life, \(t_{1/2} = \frac{0.693}{4.28 \times 10^{-4} \mathrm{s}^{-1}} = 1617.29\ \mathrm{s}\). After rearranging and solving for \(t\), we find that the time required for the absorbance to fall to 0.100 is \(t = 4806.70\ \mathrm{s}\).

Step-by-step solution

Calculate the rate constant, k

Since the given reaction is first-order overall and first-order with respect to urea, the rate expression can be written as: Rate = \(k[\mathrm{NH}_{2}\mathrm{CONH}_{2}]\) Given that the initial concentration of urea (\([\mathrm{NH}_{2}\mathrm{CONH}_{2}]\)) is 0.200 M and the rate at this concentration is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). We can now calculate the rate constant \(k\) using the above equation: \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s} = k(0.200 \mathrm{M})\) Rearrange to find the value of \(k\): \(k = \frac{8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}}{0.200 \mathrm{M}}\)

Calculate the half-life of the reaction

For a first-order reaction, the half-life (\(t_{1/2}\)) can be calculated using the following equation: \(t_{1/2} = \frac{0.693}{k}\) Now, we plug the value of rate constant \(k\) we calculated in Step 1 into the formula: \(t_{1/2} = \frac{0.693}{\frac{8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}}{0.200 \mathrm{M}}}\)

Calculate the time required for the absorbance to fall to 0.100

To find the time required for absorbance to fall to 0.100, we can use the following first-order integrated rate law equation: \(ln \frac{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_{initial}}{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_{final}} = kt\) We know the initial concentration and the absorbance we want to achieve, as well as the rate constant \(k\) from Step 1. So, let's plug it in to find the time, \(t\): \(ln \frac{0.200 \mathrm{M}}{0.100 \mathrm{M}} = \left(\frac{8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}}{0.200 \mathrm{M}}\right)t\) Rearrange and solve for \(t\) to find the time required for the absorbance to fall to 0.100.

Key concepts

First-Order Reaction

A first-order reaction is a type of chemical reaction where the reaction rate is directly proportional to the concentration of a single reactant. This means that if you double the concentration of that reactant, the rate of the reaction also doubles. First-order reactions have unique properties when it comes to their kinetic behavior. They are often expressed with a rate equation:

• Rate = k[A]

Here, \([A]\) represents the concentration of the reactant and \(k\) is the rate constant. In our context, when discussing the decomposition of urea, the reaction behaves according to first-order kinetics with respect to urea. The rate of decomposition depends only on the concentration of urea itself. As the urea concentration decreases over time, so does the reaction rate.

Rate Constant

The rate constant \(k\) is a crucial parameter in chemical kinetics as it provides a measure of the speed of a chemical reaction. It's unique to each reaction at a given temperature. For first-order reactions, the rate constant has units of s\(^{-1}\), reflecting that it relates the reaction rate to the concentration of the reactant.
To calculate \(k\) for the decomposition of urea, we use the equation from step 1:

• Given: \( 8.56 \times 10^{-5} \, \mathrm{M/s} = k(0.200 \, \mathrm{M}) \)
• Solving gives: \( k = \frac{8.56 \times 10^{-5} \, \mathrm{M/s}}{0.200 \, \mathrm{M}} = 4.28 \times 10^{-4} \, \mathrm{s^{-1}} \)

This value of \(k\) can be used to predict how quickly the decomposition reaction will proceed under the given conditions. It allows us to calculate other kinetic properties, such as the half-life of the reaction.

Half-Life

In chemical kinetics, the half-life \(t_{1/2}\) of a reaction is the time required for the concentration of a reactant to decrease to half its initial value. For first-order reactions, the half-life is independent of the initial concentration, which is a unique feature. The formula for half-life in a first-order reaction is:

• \(t_{1/2} = \frac{0.693}{k}\)

Using this formula, once we know the rate constant \(k\), we can determine the half-life of the urea decomposition reaction:

• \(t_{1/2} = \frac{0.693}{4.28 \times 10^{-4} \, \mathrm{s^{-1}}} \approx 1618 \mathrm{\, s} \)

This equation tells us that no matter how much urea we start with, it will always take about 1618 seconds for the concentration to halve, as long as the temperature remains constant.

Urea Decomposition

Urea decomposition is an important biological and chemical process, particularly in the context of nitrogen metabolism in living organisms. In acidic conditions, such as in 0.1 M HCl, urea decomposes into ammonium ions \(\mathrm{NH}_4^+\) and bicarbonate ions \(\mathrm{HCO}_3^-\). This process helps convert waste nitrogen into a form that can be excreted by organisms.
The reaction mechanism can be understood through kinetic studies, like those discussed in our exercise. We use its first-order nature to make predictions about how quickly urea will break down. By understanding the rate constant, half-life, and other dynamics of the reaction, researchers can better grasp how urea decomposition occurs both in natural environments and industrial applications. The information gleaned from studying the kinetics of urea decomposition helps in fields ranging from environmental science to medicine.