Question

Ammonium cyanate, \(\mathrm{NH}_{4} \mathrm{NCO},\) in water rearranges to produce urea, a common fertilizer, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}:\) $$ \mathrm{NH}_{4} \mathrm{NCO}(a q) \longrightarrow\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}(a q) $$ The rearrangement is a seond-order reaction. It takes \(11.6 \mathrm{~h}\) for the concentration of \(\mathrm{NH}_{4} \mathrm{NCO}\) to go from \(0.250 \mathrm{M}\) to \(0.0841 \mathrm{M}\) (a) What is \(k\) for the reaction? (b) What is the half-life of the reaction when \(\mathrm{NH}_{4} \mathrm{NCO}\) is \(0.100 \mathrm{M} ?\)

Short answer

Question: Calculate (a) the rate constant (k) and (b) the half-life of the reaction when the initial concentration of NH4NCO is 0.100 M, given the initial concentration of NH4NCO is 0.250 M, its concentration after 11.6 hours is 0.0841 M, and it follows a second-order reaction. Answer: (a) The rate constant (k) for the reaction is \(4.68 \times 10^{-3} \, \mathrm{h}^{-1}\). (b) The half-life of the reaction when the initial concentration of \(\mathrm{NH}_{4} \mathrm{NCO}\) is 0.100 M is 21.4 hours.

Step-by-step solution

Find the rate constant (k)

Use the given values to calculate the rate constant (k) using the integrated rate law equation: $$ \frac{1}{[\mathrm{NH}_{4} \mathrm{NCO}]_{t }}-\frac{1}{[\mathrm{NH}_{4} \mathrm{NCO}]_{0}}=k t $$ We have: $$ [\mathrm{NH}_{4} \mathrm{NCO}]_{0} = 0.250 \, \mathrm{M} $$ $$ [\mathrm{NH}_{4} \mathrm{NCO}]_{t} = 0.0841 \, \mathrm{M} $$ $$ t = 11.6 \, \mathrm{h} $$ Rearrange the equation to solve for k: $$ k = \frac{\frac{1}{[\mathrm{NH}_{4} \mathrm{NCO}]_{t }}-\frac{1}{[\mathrm{NH}_{4} \mathrm{NCO}]_{0}}}{t} $$ Calculate k: $$ k = \frac{\frac{1}{0.0841 \, \mathrm{M}} - \frac{1}{0.250 \, \mathrm{M}}}{11.6 \, \mathrm{h}} = 4.68 \times 10^{-3} \, \mathrm{h}^{-1} $$

Find the half-life when NH4NCO is 0.100 M

The half-life formula for a second-order reaction is: $$ t_{1/2} = \frac{1}{k [\mathrm{NH}_{4} \mathrm{NCO}]_{0}} $$ Use the calculated k value and the given initial concentration to find the half-life: $$ t_{1/2} = \frac{1}{(4.68 \times 10^{-3} \, \mathrm{h}^{-1})(0.100 \, \mathrm{M})} = 21.4 \, \mathrm{h} $$ (a) The rate constant (k) for the reaction is \(4.68 \times 10^{-3} \, \mathrm{h}^{-1}\). (b) The half-life of the reaction when the initial concentration of \(\mathrm{NH}_{4} \mathrm{NCO}\) is 0.100 M is 21.4 hours.

Key concepts

Rate Law

The rate law is an equation that relates the rate of a chemical reaction to the concentration of the reactants. It is expressed by the equation:
\[\text{rate} = k[\text{A}]^{m}[\text{B}]^{n}\] where \[k\] is the reaction rate constant, \[\text{A}\] and \[\text{B}\] are the concentrations of the reactants and \[m\] and \[n\] are the reaction orders with respect to each reactant. In the context of our ammonium cyanate reaction, the rate law helps us to understand how the concentration of the substance affects the reaction rate. Since this is a second-order reaction, the rate law would be squared with respect to the concentration of ammonium cyanate, indicating a proportional increase in the rate if the concentration doubles.

Second-Order Reaction

A second-order reaction is one where the rate is proportional to the square of the concentration of one reactant, or to the product of the concentrations of two reactants. This means that if you were to double the concentration of the reactant, the reaction rate would increase by a factor of four. Second-order reactions have a distinctive integrated rate law format, which we use to solve for the reaction rate constant, \[k\], or to determine the half-life of a reactant under certain conditions. The second-order nature of the ammonium cyanate reaction greatly influences how we interpret the rate at which the reaction proceeds and how it depends on the concentration of the reactant.

Half-Life

The half-life of a chemical reaction is the time required for the concentration of a reactant to decrease to half of its initial value. It is a key concept in chemical kinetics as it provides a measure of the speed of a reaction. In a second-order reaction, the half-life is inversely proportional to the initial concentration of the reactant and the rate constant, \[k\]. Its equation is given by:
\[ t_{1/2} = \frac{1}{k [\text{Reactant}]_{0}} \] Thus, unlike in first-order reactions where the half-life is constant, in second-order reactions, the half-life increases as the concentration decreases. The exercise provided an excellent example where the calculation of the half-life was shown, given a certain concentration and rate constant.

Reaction Rate Constant

The reaction rate constant, \[k\], is a proportionality factor that relates the reaction rate to the concentrations of the reactants in the rate law equation. It is specific to a given reaction at a certain temperature. Its value is critical in determining the speed of the reaction, as it tells us how rapidly the reactants are converted into products. In our example, once the value of \[k\] was calculated from the known concentrations and time, it could be then used to find other important parameters such as the half-life of the reaction for different starting concentrations. Understanding the reaction rate constant is fundamental for predicting how a reaction will progress over time and for designing experiments to control the rate of a chemical process.