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Ammonium cyanate, NH_NCO, rearranges in water to give urea, (NH\(_{2}\)) \(_{2}\) CO: $$\mathrm{NH}_{4} \mathrm{NCO}(\mathrm{aq}) \longrightarrow\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}(\mathrm{aq})$$ The rate equation for this process is "Rate \(=k\) \(\left[\mathrm{NH}_{4} \mathrm{NCO}\right]^{2}, "\) where \(k=0.0113 \mathrm{L} / \mathrm{mol} \cdot\) min. If the original concentration of \(\mathrm{NH}_{4} \mathrm{NCO}\) in solution is \(0.229 \mathrm{mol} / \mathrm{L}\) how long will it take for the concentration to decrease to \(0.180 \mathrm{mol} / \mathrm{L} ?\)

Short Answer

Expert verified
It takes approximately 105 minutes.

Step by step solution

01

Understand the Rate Equation

The reaction given has a rate equation: \( \text{Rate} = k \left[ \mathrm{NH}_{4} \mathrm{NCO} \right]^2 \). This means the reaction follows a second-order kinetics, specifically depending on the square of the concentration of \( \mathrm{NH}_{4} \mathrm{NCO} \).
02

Write the Integrated Rate Law for Second-Order Reactions

For second-order reactions, the integrated rate equation is: \[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \] where \( [A] \) is the concentration at time \( t \), \( [A]_0 \) is the initial concentration, and \( k \) is the rate constant.
03

Substitute Known Values into the Integrated Rate Law

Given: \( [A]_0 = 0.229 \, \mathrm{mol/L} \), \( [A] = 0.180 \, \mathrm{mol/L} \), and \( k = 0.0113 \, \mathrm{L} / \mathrm{mol} \cdot \mathrm{min} \). Substitute these into the integrated rate law: \[ \frac{1}{0.180} = \frac{1}{0.229} + 0.0113t \]
04

Solve for Time \( t \)

Rearrange the equation to solve for \( t \): \[ t = \frac{1}{0.0113} \left( \frac{1}{0.180} - \frac{1}{0.229} \right) \]. Calculate as follows:\[ \frac{1}{0.180} = 5.555 \]\[ \frac{1}{0.229} = 4.367 \]So, \[ t = \frac{1}{0.0113} (5.555 - 4.367) \]\[ t = \frac{1}{0.0113} (1.188) \]\[ t = 105.133 \] approximately.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ammonium cyanate
Ammonium cyanate, often represented as NH extsubscript{4}NCO, plays a fascinating role in chemistry, primarily because of its ability to undergo a rearrangement into urea when dissolved in water. It is an inorganic compound that illustrates the concept of chemical conversion through molecular rearrangement. This reaction is a classic example showcasing how a seemingly simple compound can transform into another with entirely different chemical properties and biological significance.

In this reaction, ammonium cyanate serves as a precursor to urea, a crucial organic compound identified by the formula \((NH_2)_2CO\). The process itself is important not just in educational settings but also in biological contexts, as it was historically significant in vitality theories during its discovery in the 19th century. This rearrangement emphasizes the unique abilities of chemical substances to change form while participating in fundamental chemical reactions.
Urea formation
Urea formation is an intriguing and significant chemical process with biological relevance. The reaction from ammonium cyanate (NH extsubscript{4}NCO) to urea ((NH extsubscript{2}) extsubscript{2}CO) is a splendid demonstration of the simple molecules achieving biological complexity.

Historically, this process was pivotal in organic chemistry, marking the discovery that organic compounds could be synthesized from inorganic materials. In 1828, Friedrich Wöhler discovered this transformation, which challenged the idea that organic compounds required a living origin, laying the foundation for organic synthesis.
  • Urea is a significant player in the metabolism of nitrogen-containing substances in animals.
  • It serves as a non-toxic compound used to excrete excess nitrogen primarily in the urine.
  • The ability to synthesize urea artificially has profound implications in agriculture, where it is used as a fertilizer, and in medical applications as a non-toxic byproduct.
Integrated rate law
The integrated rate law is a critical tool in understanding reaction kinetics, especially for second-order reactions such as the conversion of ammonium cyanate to urea. Kinetics allows chemists to describe how the concentration of reactants decreases over time by providing a mathematical relationship.

For second-order reactions, the integrated rate law is:\[ \frac{1}{[A]} = \frac{1}{[A]_0} + kt \]where:
  • \([A]\) is the concentration of the reactant at time \(t\).
  • \([A]_0\) is the initial concentration.
  • \(k\) is the rate constant, which reflects how quickly reactants convert to products.
  • \(t\) is the time period over which the reaction occurs.
This formula reveals how the concentration of ammonium cyanate decreases quadratically over time, being directly proportional to the rate constant and time. In practical applications, calculations involving the integrated rate law help determine how long it will take for a reactant's concentration to reach a desired level, as demonstrated in solving problems with known initial and final concentrations.

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Most popular questions from this chapter

Data in the table were collected at \(540 \mathrm{K}\) for the following reaction: $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g})$$ (a) Derive the rate equation. (b) Determine the reaction order with respect to each reactant. (c) Calculate the rate constant, giving the correct units for \(k.\) $$\begin{array}{lll}\hline \text { Initial Concentration }(\mathrm{mol} / \mathrm{L}) & {\text { Initial Rate }} \\\\\hline[\mathrm{C} 0] & {\left[\mathrm{NO}_{2}\right]} & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\\\\hline 5.1 \times 10^{-4} & 0.35 \times 10^{-4} & 3.4 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.70 \times 10^{-4} & 6.8 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.18 \times 10^{-4} & 1.7 \times 10^{-8} \\\1.0 \times 10^{-3} & 0.35 \times 10^{-4} & 6.8 \times 10^{-8} \\\1.5 \times 10^{-3} & 0.35 \times 10^{-4} & 10.2 \times 10^{-8} \\\\\hline\end{array}$$

The reaction $$2 \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ was studied at \(904^{\circ} \mathrm{C},\) and the data in the table were collected. $$\begin{array}{lll}\hline \begin{array}{l}\text { Reactant Concentration } \\\\(\mathrm{mol} / \mathrm{L})\end{array} & & \\\\\hline[\mathrm{N} 0] & {\left[\mathrm{H}_{2}\right]} & \begin{array}{l}\text { Rate of Appearance of } \mathrm{N}_{2} \\\\(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s})\end{array} \\\\\hline 0.420 & 0.122 & 0.136 \\\0.210 & 0.122 & 0.0339 \\\0.210 & 0.244 & 0.0678 \\\0.105 & 0.488 & 0.0339 \\\\\hline\end{array}$$ (a) Determine the order of the reaction for each reactant. (b) Write the rate equation for the reaction. (c) Calculate the rate constant for the reaction. (d) Find the rate of appearance of \(\mathrm{N}_{2}\) at the instant when \([\mathrm{NO}]=0.350 \mathrm{mol} / \mathrm{L}\) and \(\left[\mathrm{H}_{2}\right]=0.205 \mathrm{mol} / \mathrm{L}.\)

The following statements relate to the reaction with the following rate law: Rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right].\) $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})$$ Determine which of the following statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

Hypofluorous acid, HOF, is very unstable, decomposing in a first-order reaction to give HF and \(\mathrm{O}_{2}\), with a halflife of only 30 min at room temperature: $$\mathrm{HOF}(\mathrm{g}) \longrightarrow \mathrm{HF}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ If the partial pressure of HOF in a \(1.00-\mathrm{L}\). flask is initially \(1.00 \times 10^{2} \mathrm{mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C},\) what is the total pressure in the flask and the partial pressure of HOF after exactly 30 min? After 45 min?

In the synthesis of ammonia, if \(-\Delta\left[\mathrm{H}_{2}\right] / \Delta t=4.5 \times 10^{-4}\) \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{min},\) what is \(\Delta\left[\mathrm{NH}_{3}\right] / \Delta t ?\) \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\)

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