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The rate law for the reaction: $$ \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is given by rate \(=k\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{NO}_{2}^{-}\right]\). At \(25^{\circ} \mathrm{C},\) the rate constant is \(3.0 \times 10^{-4} / M \cdot \mathrm{s} .\) Calculate the rate of the reaction at this temperature if \(\left[\mathrm{NH}_{4}^{+}\right]=0.36 M\) and \(\left[\mathrm{NO}_{2}^{-}\right]=0.075 \mathrm{M}\).

Short Answer

Expert verified
The rate of the reaction is \(8.1 \times 10^{-6} \ M/s\).

Step by step solution

01

Identify the Given Values

We are given the rate constant \(k = 3.0 \times 10^{-4} \ M^{-1} \cdot s^{-1}\), and the concentrations \([\text{NH}_4^+] = 0.36 \ M\) and \([\text{NO}_2^-] = 0.075 \ M\). We need to find the rate of the reaction using these values.
02

Write the Rate Law Expression

The rate law for the reaction is given by \( \text{Rate} = k[\text{NH}_4^+][\text{NO}_2^-]\). This expression shows that the rate depends on the concentration of \(\text{NH}_4^+\) and \(\text{NO}_2^-\), and the rate constant \(k\).
03

Substitute the Values into the Rate Law

Substitute the given values into the rate law: \( \text{Rate} = (3.0 \times 10^{-4} \ M^{-1} \cdot s^{-1})(0.36 \ M)(0.075 \ M)\).
04

Calculate the Reaction Rate

Perform the multiplication: \( \text{Rate} = 3.0 \times 10^{-4} \times 0.36 \times 0.075\). First, calculate \(0.36 \times 0.075 = 0.027\). Then, multiply this result by \(3.0 \times 10^{-4}\): \(3.0 \times 10^{-4} \times 0.027 = 8.1 \times 10^{-6} \ M/s\).

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Key Concepts

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

Reaction Rate
The reaction rate provides us with valuable insight into how fast a chemical reaction proceeds. It is defined as the change in concentration of a reactant or product per unit time. In the context of this reaction involving ammonium ions
  • \(\text{NH}_4^+\) and nitrite ions \(\text{NO}_2^-\), the rate tells us how quickly these reactants are transformed into nitrogen gas \(\text{N}_2\) and water \(2\text{H}_2\text{O}\).
This rate is often expressed using the rate law, which is an equation that relates the reaction rate to the concentration of the reactants, incorporating a specific rate constant.
Rate Constant
The rate constant, denoted as \(k\), is a crucial factor in the rate law equation. It serves as a proportionality constant, connecting the reaction rate to the concentration of the reactants. The value of \(k\) is determined experimentally and varies with different reactions and conditions such as temperature.
  • For the given reaction, the rate constant at \(25^\circ\)C is \(3.0 \times 10^{-4}\, M^{-1} \cdot s^{-1}\), reflecting the specific conditions under which the reaction occurs quickly enough to observe.
  • As a characteristic of every unique reaction, the rate constant indicates how different factors like temperature and catalysts can influence the speed of that reaction.
Chemical Kinetics
Chemical kinetics, the area of chemistry concerned with the rates of chemical reactions, helps scientists understand how and why reactions occur. This field involves studying the variables affecting the reaction rate, including:
  • Concentration of reactants, which can alter how often molecules collide.
  • Temperature changes, as higher temperatures typically increase reaction rates.
  • The presence of catalysts, which can speed up reactions by lowering activation energy.
By analyzing experimental data, chemists can propose rate laws and determine the mechanisms through which reactions proceed. Understanding kinetics is crucial for fields ranging from pharmaceuticals to environmental science.
Concentration
The concentration of reactants plays a key role in determining the rate of a chemical reaction. In simple terms, concentration refers to how much of a given substance is present in a mixture. The higher the concentration of reactants, the greater the frequency of collisions between molecules, which often leads to faster reaction rates. For our specific reaction, the concentrations of \(\text{NH}_4^+\) at \(0.36\, M\) and \(\text{NO}_2^-\) at \(0.075\, M\) are essential in calculating how quickly the reaction will proceed at \(25^\circ\)C. Understanding concentration and its effect on reaction rates helps chemists manipulate reaction conditions to achieve desired outcomes in both industrial and laboratory settings.

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Most popular questions from this chapter

Determine the overall orders of the reactions to which the following rate laws apply: (a) rate \(=k\left[\mathrm{NO}_{2}\right]^{2},(\mathrm{~b})\) rate \(=k\), (c) rate \(=k\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{Br}_{2}\right]^{1 / 2}\) (d) rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\)

The rate constant of a first-order reaction is \(66 \mathrm{~s}^{-1}\). What is the rate constant in units of minutes?

Consider the reaction: $$ \mathrm{A}+\mathrm{B} \longrightarrow \text { products } $$ From the following data obtained a a certain temperature, determine the order of the reaction and calculate the rate constant. $$ \begin{array}{ccc} {[\mathbf{A}](\boldsymbol{M})} & {[\mathbf{B}](\boldsymbol{M})} & \text { Rate }(\boldsymbol{M} / \mathbf{s}) \\ \hline 1.50 & 1.50 & 3.20 \times 10^{-1} \\ 1.50 & 2.50 & 3.20 \times 10^{-1} \\ 3.00 & 1.50 & 6.40 \times 10^{-1} \end{array} $$

Classify the following elementary reactions as unimolecular, bimolecular, or termolecular: (a) \(2 \mathrm{NO}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{NOBr}\) (b) \(\mathrm{CH}_{3} \mathrm{NC} \longrightarrow \mathrm{CH}_{3} \mathrm{CN}\) (c) \(\mathrm{SO}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{2}+\mathrm{O}\)

The thermal decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) into phosphorus and molecular hydrogen is a first-order reaction: $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ The half-life of the reaction is \(35.0 \mathrm{~s}\) at \(680^{\circ} \mathrm{C}\). Calculate (a) the first-order rate constant for the reaction and (b) the time required for 95 percent of the phosphine to decompose.

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