Problem 57 - Chemistry and Chemical Reactivity

Question

Ammonium cyanate, NH_NCO, rearranges in water to give urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}.\) $$\mathrm{NH}_{4} \mathrm{NCO}(\mathrm{aq}) \rightarrow\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}(\mathrm{aq})$$ $$\begin{array}{cc}\text { Time (min) } & {\left[\mathrm{NH}_{4} \mathrm{NCO}\right](\mathrm{mol} / \mathrm{L})} \\\\\hline 0 & 0.458 \\\4.50 \times 10^{1} & 0.370 \\\1.07 \times 10^{2} & 0.292 \\\2.30 \times 10^{2} & 0.212 \\\6.00 \times 10^{2} & 0.114 \\\\\hline\end{array}$$ 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 hours.

Step-by-Step Solution

Answer

The reaction is first order with \(k = 0.00457 \text{ min}^{-1}\), a half-life of 151.6 min, and \([\mathrm{NH}_4 \mathrm{NCO}] = 0.0505 \text{ mol/L}\) after 12 hours.

Step 1: Determine Order Using Concentration Data

We can use the integrated rate laws for first and second-order reactions to compare the changes in concentration. The first-order integrated rate law is \( ext{ln}([A]_0/[A]) = kt\). For second order, it's \(1/[A] - 1/[A]_0 = kt\). Plot \(\text{ln}([\text{NH}_4\text{NCO}])\) vs. time and \(1/[\text{NH}_4\text{NCO}]\) vs. time. If the first plot is linear, it's first-order. If the second is linear, it's second-order.

Step 2: First Order Confirmation

Calculate \(\text{ln}([\text{NH}_4\text{NCO}]\)) at various times: \\(\text{Time = 0}\), \(\text{ln}(0.458) = -0.7803\)\\(\text{Time = 45.0}\), \(\text{ln}(0.370) = -0.9943\)\\(\text{Time = 107.0}\), \(\text{ln}(0.292) = -1.2335\)\\(\text{Time = 230.0}\), \(\text{ln}(0.212) = -1.5505\)\\(\text{Time = 600.0}\), \(\text{ln}(0.114) = -2.1691\)\The plot of \(\text{ln}([\text{NH}_4\text{NCO}])\) vs. time is linear, confirming first-order kinetics.

Step 3: Calculate Rate Constant \(k\)

Using the formula \(\text{ln}([A]_0/[A]) = kt\) for first-order reactions. Use \([A]_0 = 0.458\) and any other data point, e.g., \(t = 230.0\), \([A] = 0.212\): \\(\text{ln}(0.458/0.212) = k \times 230.0\) \Solve for \(k\): \\(k = (\text{ln}(2.1604))/230.0 = 0.00457\, \text{min}^{-1}\).

Step 4: Calculate Half-life

For a first-order reaction, the half-life \(t_{1/2}\) is \(t_{1/2} = \frac{0.693}{k}\). Substitute \(k = 0.00457 \text{ min}^{-1}\): \\(t_{1/2} = \frac{0.693}{0.00457} = 151.6\, \text{min}\).

Step 5: Calculate Concentration After 12 Hours

Convert 12 hours to minutes: \(12 \times 60 = 720\, \text{min}\). Use \([A] = [A]_0 e^{-kt}\) to calculate the concentration after 720 minutes: \\([A] = 0.458 \times e^{-0.00457 \times 720}\) \\([A] = 0.458 \times e^{-3.2904} = 0.0505\, \text{mol/L}\).

Key Concepts

Reaction Order

When it comes to understanding a chemical reaction's behavior, it's crucial to determine its reaction order. This essentially tells us how the rate of reaction depends on the concentration of the reactants.
In this context, we are analyzing the rearrangement of ammonium cyanate into urea, and we want to find if it's a first-order or second-order reaction.
To discern this, the integrated rate laws come into play.

• For a first-order reaction, the relationship is \( \ln([A]_0/[A]) = kt \).
• For a second-order reaction, it follows \( 1/[A] - 1/[A]_0 = kt \).

By plotting \( \ln([\text{NH}_4\text{NCO}]) \) against time and checking for a linear relationship, we can identify that this reaction follows first-order kinetics. The linearity confirms the order and helps in further calculations involving the rate constant and half-life.

Rate Constant

The rate constant, denoted as \( k \), is a vital part of understanding the speed of a chemical reaction.
Especially in first-order reactions, its value is unique to each reaction and independent of the concentrations of reactants.
In our ammonium cyanate to urea reaction example, we calculated it by harnessing the first-order rate law: \( \ln([A]_0/[A]) = kt \).After plugging in our values for

• \( [A]_0 = 0.458 \) mol/L,
• \( [A] = 0.212 \) mol/L at \( t = 230 \) minutes,

the equation becomes \( \ln(0.458 / 0.212) = k \times 230 \). Solving this, we find that \( k = 0.00457 \, \text{min}^{-1} \).This constant gives us a precise measure of how quickly the reaction progresses, providing a foundation for calculating other related kinetic properties.

Half-life

The half-life of a reaction is the time required for half of the initial reactant to transform into the product.
For first-order reactions, the half-life has a simple and interesting relationship with the rate constant: \( t_{1/2} = \frac{0.693}{k} \).
This explains why first-order reactions have a constant half-life that remains unaffected by the starting concentration.Using our calculated rate constant,

• \( k = 0.00457 \, \text{min}^{-1} \),

we plug this into the formula, yielding \( t_{1/2} = \frac{0.693}{0.00457} = 151.6 \, \text{min} \).Interesting point: As half-lives of first-order reactions do not change, you can predict how long it might take for any given percentage of the reactant to be consumed, simply by multiplying the half-life accordingly.

Concentration Calculation

Predicting the concentration of a reactant at a given time is crucial, particularly in industrial and laboratory settings.
In this example, our goal was to find the concentration of ammonium cyanate after 12 hours.As it is a first-order reaction, the formula used is

• \( [A] = [A]_0 e^{-kt} \).

Converting 12 hours to 720 minutes, we use\( [A] = 0.458 \times e^{-0.00457 \times 720} \).After solving this, we find the concentration is approximately\( [A] = 0.0505 \, \text{mol/L} \).This equation is incredibly useful as it allows chemists to anticipate the amount of reactant remaining, thus scheduling any necessary steps in the reaction or next experiments.