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 \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) $$ \text { is } 8.56 \times 10^{-5} \mathrm{M} / \mathrm{s} $$ (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 \(4.28 \times 10^{-4}\, s^{-1}\). (b) The concentration of urea after \(4.00 \times 10^3\, s\) is \(0.213\, M\). (c) The half-life for this reaction at \(61.05^{\circ}C\) is \(1.62 \times 10^3\, s\).

Step-by-step solution

Write the rate law for the reaction

The decomposition of urea follows first-order kinetics. Since the reaction is first order in urea and first order overall, the rate law for this reaction is: $$ \text{rate} = k[\mathrm{NH}_{2}\mathrm{CONH}_{2}] $$

Calculate the rate constant k using the given rate and initial concentration

We are given that the reaction rate at an initial concentration of \(0.200 M\) is \(8.56 \times 10^{-5} M/s\). We can use this information to find the rate constant k. $$ \text{rate} = k[\mathrm{NH}_{2}\mathrm{CONH}_{2}] $$ Plug in the given values: $$ 8.56 \times 10^{-5} M/s = k (0.200 M) $$ Solve for k: $$ k = \frac{8.56 \times 10^{-5} M/s}{0.200 M} = 4.28 \times 10^{-4} s^{-1} $$

Calculate the concentration of urea after a certain time

To find the concentration of urea after \(4.00 \times 10^3 s\), we can use the first-order kinetics equation: $$ \ln\frac{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t}{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_0} = -kt $$ We know \([\mathrm{NH}_{2}\mathrm{CONH}_{2}]_0 = 0.500 M\), \(k = 4.28 \times 10^{-4} s^{-1}\), and \(t = 4.00 \times 10^3 s\). Now, plug in these values and solve for \([\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t\): $$ \ln\frac{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t}{0.500 M} = -(4.28 \times 10^{-4} s^{-1})(4.00 \times 10^3 s) $$ $$ \ln\frac{[\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t}{0.500 M} = -1.712 $$ Now, find \([\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t\): $$ [\mathrm{NH}_{2}\mathrm{CONH}_{2}]_t = 0.500 M\, e^{-1.712} = 0.213 M $$

Calculate the half-life of the reaction

The half-life (\(t_{1/2}\)) of a first-order reaction can be calculated using the formula: $$ t_{1/2} = \frac{0.693}{k} $$ Using the calculated rate constant \(k = 4.28 \times 10^{-4} s^{-1}\), we can find the half-life: $$ t_{1/2} = \frac{0.693}{4.28 \times 10^{-4} s^{-1}} = 1.62 \times 10^3 s $$ In summary: (a) The rate constant k is \(4.28 \times 10^{-4} s^{-1}\). (b) The concentration of urea after \(4.00 \times 10^3 s\) is \(0.213 M\). (c) The half-life for this reaction at \(61.05^{\circ} C\) is \(1.62 \times 10^3 s\).

Key concepts

First-order reaction

Understanding the kinetics of a first-order reaction is crucial in chemical kinetics. In such reactions, the rate of the reaction is directly proportional to the concentration of one reactant. This means that if you double the concentration of the reactant, the rate also doubles. First-order reactions are characterized by a simple exponential decay of the reactant concentration over time.

The rate law for a first-order reaction is given by the equation:

• \( ext{rate} = k[A] \)

where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant. Notice that the units of the rate constant \( k \) are important, as they depend on the overall order of the reaction. For first-order reactions, \( k \) typically has units of \( ext{s}^{-1} \).

First-order kinetics are often seen in processes like radioactive decay and certain chemical reactions, including the decomposition of compounds. The fact that the rate depends only on the concentration makes it mathematically straightforward to describe using exponential functions.

Rate constant

The rate constant \( k \) is a central figure in the equation governing first-order reactions. Despite its name, the rate constant is not truly constant, as its value can change with varying temperature. It determines how fast a reaction proceeds. The larger the \( k \), the faster the reaction. It encapsulates all the factors influencing the reaction rate, except the reactant concentration which is addressed separately in the rate law.

To calculate the rate constant for a first-order reaction, one can rearrange the rate law equation:

• \( k = \frac{\text{rate}}{[A]} \)

Here, by knowing the reaction rate and the concentration of the reactant, the rate constant can be derived. It is essential for predicting how long it will take for reactants to be depleted and products to be formed in a reaction process. Understanding \( k \) provides great insight into the dynamics of a reaction at a molecular level.

Half-life

Half-life is a concept that describes the time required for a reactant's concentration to decrease by half. For first-order reactions, half-life is a constant, meaning it is independent of the initial concentration of the reactant. This unique characteristic simplifies calculations and predictions for first-order kinetics, making them very predictable and easy to analyze.

The formula for calculating half-life \( t_{1/2} \) for a first-order reaction is:

• \( t_{1/2} = \frac{0.693}{k} \)

In this equation, \( 0.693 \) is the natural logarithm of 2, which is inherent to the exponential nature of decay (or growth) in first-order reactions. Knowing the half-life allows scientists to determine how quickly concentrations diminish over time, which is crucial in fields ranging from pharmaceuticals to environmental science.

Urea decomposition

The decomposition of urea in aqueous solutions is a practical example of a first-order chemical reaction. Urea, a major end product of protein metabolism in the body, decomposes into ammonium ion \( ext{NH}_4^+ \) and bicarbonate ion \( ext{HCO}_3^- \) when mixed with water and some acid, such as hydrochloric acid (HCl).

This process can be represented by the chemical equation:

• \( \text{NH}_2 \text{CONH}_2 (aq) + \text{H}^+ (aq) + 2 \text{H}_2\text{O} (l) \rightarrow 2 \text{NH}_4^+ (aq) + \text{HCO}_3^- (aq) \)

The reaction being first-order means its rate is solely dependent on the concentration of urea. As urea breaks down, it provides a clear demonstration of how first-order kinetics work. This makes it a valuable exercise not only in theoretical chemistry but also in real-world applications where understanding decay or decomposition rates is necessary.