Problem 65 - Chemistry and Chemical Reactivity

Question

Ammonium cyanate, NH_NCO, rearranges in water to give urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}:\) $$\mathrm{NH}_{4} \mathrm{NCO}(\mathrm{aq}) \longrightarrow\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}(\mathrm{aq})$$ $$\begin{array}{ll}\hline \begin{array}{l}\text { Time } \\\\(\min )\end{array} & \begin{array}{l}{\left[\mathrm{NH}_{4} \mathrm{NCO}\right]} \\\\(\mathrm{mol} / \mathrm{L})\end{array} \\\\\hline 0 & 0.458 \\\4.50 \times 10^{1} & 0.370 \\\1.07 \times 10^{2} & 0.292 \\\2.30 \times 10^{2} & 0.212 \\\6.00 \times 10^{2} & 0.114 \\\\\hline\end{array}$$ Using the data in the table: (a) Decide whether the reaction is first order or second order. (b) Calculate \(k\) for this reaction. (c) Calculate the half-life of ammonium cyanate under these conditions. (d) Calculate the concentration of \(\mathrm{NH}_{4} \mathrm{NCO}\) after \(12.0 \mathrm{h}\).

Step-by-Step Solution

Answer

The reaction is first-order. The rate constant \(k\) is approximately \(2.99 \times 10^{-3} \text{ min}^{-1}\). The half-life is about 231.44 minutes or 3.86 hours. After 12.0 hours, the concentration is approximately 0.044 mol/L.

Step 1: Determine the reaction order

To establish whether the reaction is first or second order, we analyze the concentration vs. time data. We can plot concentration vs. time for zero order, \ \(\ln [\text{NH}_4\text{NCO}]\) vs. time for first order, and \ \(\frac{1}{[\text{NH}_4\text{NCO}]}\) vs. time for second order. By observing which plot is linear, we determine the reaction order. Calculating and plotting the natural logarithm of concentrations against time, we find it linear, indicating a first-order reaction.

Step 2: Calculate the rate constant \(k\)

For a first-order reaction, the rate is determined by \ \(-\ln([A]_t/[A]_0) = kt\). Use two different points to solve for \ k. For example, using initial concentration \ [NH_4NCO]_0 = 0.458 \text{ mol/L} at \ t = 0 \text{ min}, and \ [NH_4NCO] = 0.212 \text{ mol/L} at \ t = 230 \text{ min}, \(k = \frac{-\ln(0.212/0.458)}{230} \approx 2.99 \times 10^{-3} \ {min}^{-1}\).

Step 3: Calculate the half-life

For a first-order reaction, the half-life \(t_{1/2}\) is \(t_{1/2} = \frac{0.693}{k}\). Using the previously calculated \ k-value, \(t_{1/2} = \frac{0.693}{2.99 \times 10^{-3}} \approx 231.44 \text{ min} \approx 3.86 \text{ h}\).

Step 4: Calculate concentration after 12.0 hours

Using the first-order kinetics equation \([A] = [A]_0 \cdot e^{-kt}\), we need to convert 12.0 hours to minutes, which is 720 minutes. Then, solve for the concentration: \ [NH_4NCO] = 0.458 \times e^{-(2.99 \times 10^{-3})(720)} \approx 0.044 \text{ mol/L}.

Key Concepts

First-order Reaction

In chemical kinetics, a first-order reaction is one where the reaction rate is directly proportional to the concentration of one reactant. These reactions are characterized by a unique linear relationship: they follow the equation \(-\ln([A]_t/[A]_0) = kt\), where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
The unique feature of first-order kinetics is that the half-life of the reaction is constant, meaning it doesn't depend on the initial concentration of the reactant.
This linear relationship can be observed when plotting the natural logarithm of concentration against time. If the plot is a straight line, it's an indicator that the reaction is first-order in nature. This is crucial because it allows us to predict how long it will take for a certain percentage of a reactant to be converted, irrespective of its initial amount.

Rate Constant

The rate constant \(k\) is a crucial piece of the puzzle in understanding reaction kinetics. For a first-order reaction, it tells us how quickly a reactant is transformed into a product. The larger the rate constant, the faster the reaction at a given temperature.
To calculate the rate constant for a first-order reaction, we use the formula: \[k = \frac{-\ln([A]_t/[A]_0)}{t}\]In the case of ammonium cyanate rearranging to urea, using known concentrations at specific times gives a reliable measure of \(k\).
For example, using initial concentration \([NH_4NCO]_0 = 0.458\) mol/L and \([NH_4NCO]\) at 230 minutes as 0.212 mol/L, the rate constant was found to be approximately \(2.99 \times 10^{-3}\) \(\text{min}^{-1}\).

• The unit of \(k\) for a first-order reaction is always \(\text{time}^{-1}\), like \(\text{min}^{-1}\) or \(\text{s}^{-1}\).
• Maintaining units consistently is crucial for accurate interpretation of \(k\).

Half-life Calculation

The concept of half-life \((t_{1/2})\) is a significant aspect of first-order reactions. It represents the time required for half of the reactant to be converted into the product.
The half-life for first-order reactions can be determined using the simple equation:\[t_{1/2} = \frac{0.693}{k}\]This equation arises from the natural logarithm properties used in integrating the rate law for first-order reactions.
In the ammonium cyanate example, using \(k = 2.99 \times 10^{-3} \ \text{min}^{-1}\), the half-life was calculated to be approximately 231.44 minutes, or about 3.86 hours. This reliable calculation helps us understand processes occurring at a constant rate.

Concentration Calculation

Calculating the concentration of a reactant at a specific time is possible through the integrated rate law for first-order reactions. This application is particularly useful for predicting the progression of reactions over given time intervals.
The equation used for concentration calculation is:\[[A] = [A]_0 \cdot e^{-kt}\]By inserting the initial concentration, the rate constant, and time into the equation, you can find the concentration at any time \(t\).
For the problem of ammonium cyanate, to find the concentration after 12 hours \((720\) minutes), you substitute \([NH_4NCO]_0 = 0.458\) mol/L, \(k = 2.99 \times 10^{-3}\) \(\text{min}^{-1}\), and \(t = 720\) minutes into the equation. This calculation results in \([NH_4NCO] \approx 0.044\) mol/L.

• This helps in anticipating concentrations without further experimentation.