Problem 54 - Chemistry: Principles and Reactions

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 second-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 \(0.100 \mathrm{M}\) ? (c) How long will it take to rearrange \(39 \%\) of a \(0.450 \mathrm{M}\) solution? (d) How fast is a \(0.839 \mathrm{M}\) solution being changed to urea?

Step-by-Step Solution

Answer

The rate constant (k) for the reaction is 0.20028 h⁻¹M⁻¹. b) What is the half-life of the reaction when the initial concentration is 0.100 M? The half-life of the reaction when the initial concentration is 0.100 M is 49.876 hours. c) How long does it take to rearrange 39% of the 0.450 M solution? It takes 4.380 hours to rearrange 39% of the 0.450 M solution. d) What is the rate at which a 0.839 M solution is being changed to urea? The rate at which a 0.839 M solution is being changed to urea is 0.1406 M h⁻¹.

Step 1: (a) Find the rate constant (k) for the reaction

Given the initial and final concentrations and given time, we can use the second-order integrated rate law to find the value of k. $$\dfrac{1}{0.0841}-\dfrac{1}{0.250} = k \cdot 11.6$$ Solve for k: $$ k = \dfrac{\frac{1}{0.0841} - \frac{1}{0.250}}{11.6}$$ $$ k = 0.20028 \mathrm{h}^{-1}\mathrm{M}^{-1}$$

Step 2: (b) Find the half-life of the reaction when the initial concentration is \(0.100\,\text{M}\)

For a second-order reaction, the half-life is given by $$t_{1/2}=\dfrac{1}{k[\mathrm{A})_{0}]}$$ Plug in the values of k and \([\mathrm{A})_{0}]\) to find the half-life: $$t_{1/2}=\dfrac{1}{(0.20028\,\mathrm{h}^{-1}\mathrm{M}^{-1})(0.100\,\mathrm{M})}$$ $$t_{1/2}=49.876\,\text{hours}$$

Step 3: (c) Find the time it takes to rearrange \(39 \%\) of the \(0.450\,\text{M}\) solution

The initial concentration is given by \([\mathrm{A})_{0}]=0.450\,\mathrm{M}\). We want to find the time taken when \(39 \%\) has rearranged, hence, the final concentration is given by \([\mathrm{A})_{t}]=(1-0.39) \cdot 0.450\,\mathrm{M}\). Using the second-order integrated rate law: $$\dfrac{1}{[(1-0.39) \cdot 0.450]}-\dfrac{1}{0.450} = (0.20028\,\mathrm{h}^{-1}\mathrm{M}^{-1}) \cdot t$$ Solve for t: $$ t = \dfrac{\frac{1}{(1-0.39) \cdot 0.450} - \frac{1}{0.450}}{ 0.20028\,\mathrm{h}^{-1}\mathrm{M}^{-1}} $$ $$ t = 4.380\,\text{hours} $$

Step 4: (d) Find the rate at which a \(0.839\,\text{M}\) solution is being changed to urea

For a second-order reaction, the rate of the reaction is given by the rate law: $$\text{Rate} = k [\mathrm{A})]^2$$ Plug in the values of k and the given concentration: $$\text{Rate} = (0.20028\,\mathrm{h}^{-1}\mathrm{M}^{-1})(0.839\,\mathrm{M})^2$$ $$\text{Rate} = 0.1406\,\mathrm{M}\,\mathrm{h}^{-1}$$ The rate at which the \(0.839\,\text{M}\) solution is being changed to urea is \(0.1406\,\mathrm{M}\,\mathrm{h}^{-1}\).

Key Concepts

Rate Constant (k)

The rate constant, symbolized as 'k', is an essential parameter in chemical kinetics, representing the speed at which a chemical reaction proceeds. It's a proportionality constant that connects the reaction rate with the reactant concentrations.

In the context of second-order reactions, like the ammonium cyanate rearrangement to urea, the rate constant has units of \text{h}^{-1}\text{M}^{-1}, reflecting that the rate is dependent on the inverse of both concentration and time. When calculating 'k', we use the integrated second-order rate law which takes into account the changes in concentration of the reactants over time. The formula derived from this law for a second-order reaction is \[\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\] where \text{M} represents molarity. It's critical to note that the rate constant is determined experimentally and varies with temperature, so a different temperature would result in a different rate constant. In general, the higher the rate constant, the faster the reaction proceeds.

Reaction Half-Life

The half-life of a reaction, often denoted as \text{t}_{1/2}, is the time required for the concentration of a reactant to decrease to half of its initial value. It offers a straightforward way of understanding how fast a reaction moves towards its completion.

For second-order reactions, the half-life is inversely proportional to the initial concentration of the reactant and the rate constant, given by the formula \[\text{t}_{1/2} = \frac{1}{k[A]_0}\] Unlike first-order reactions, the half-life of a second-order reaction is not constant; it increases as the reactant concentration decreases. This is because, with fewer particles available to interact, the reaction slows down, so it takes longer to use up half of the reactants as the reaction progresses.

Integrated Rate Law

The integrated rate law for a second-order reaction describes how the concentration of reactants changes over time. Unlike the differential rate law, which gives the rate at a particular instant, the integrated rate law provides an equation that predicts reactant concentration at any given time after the reaction has started.

Applied to a real-life scenario like the ammonium cyanate rearrangement, we can calculate important features of the reaction, such as the time taken for a certain percentage of the reactant to be consumed. The formula for the integrated rate law for a second-order reaction is \[\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}\] where \text{t} is the time and \text{k} is the rate constant. This law is especially useful in solving problems related to reaction kinetics, as it allows us to directly relate reactant concentration to time.

Chemical Reaction Rate

Chemical reaction rate refers to the speed at which reactants are converted into products. For reaction rates concerning second-order kinetics, the rate is proportional to the square of the reactant concentration. The rate law for our ongoing example appears as \[\text{Rate} = k[A]^2\] where [A] is the concentration of the reactant. In this case, if you double the concentration of the reactant, the reaction rate increases by fourfold, not twofold, indicating the squared dependence on concentration.

Calculating the rate helps in understanding how quickly a given concentration of reactant is being used up. By knowing the rate constant and the concentration of reactants, it's possible to determine the rate at any point during the reaction. Understanding the chemical reaction rate is vital in many practical applications, such as controlling reaction conditions in industrial processes and determining the shelf-life of pharmaceutical products.