Nitrogen monoxide reacts with chlorine according to the equation: \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) The following initial rates of reaction have been observed for certain reactant concentrations: $$\begin{array}{|c|c|c|}\hline \text { \([\mathrm{NO}]\left(\operatorname{mol} / L^{1}\right)\) } & \text { \(\left[\mathrm{Cl}_{2}\right](\mathrm{mol} / \mathrm{L})\) } & \text { Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{h}^{-1}\right)\) } \\\\\hline 0.50 & 0.50 & 1.14 \\\\\hline 1.00 & 0.50 & 4.56 \\ \hline 1.00 & 1.00 & 9.12 \\\\\hline\end{array}$$ What is the rate law that describes the rate's dependence on the concentrations of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) ? What is the rate constant? What are the orders with respect to each reactant?

Short Answer

Expert verified
Rate law: Rate = k[NO]^2[Cl2]. The reaction is second-order with respect to NO and first-order with respect to Cl2. The rate constant (k) is 4.56 L^2/mol^2h.

Step by step solution

01

Identify Changes in Concentration and Rates

Examine the given data to discern the influence of changing concentration on the rate of reaction. Focus on experiments where the concentration of only one reactant changes while the other remains constant to isolate the effect on the rate of reaction.
02

Determine the Order of Reaction with Respect to NO

Compare the first two experiments where the concentration of NO is doubled (from 0.50 to 1.00 mol/L) while the concentration of Cl2 remains constant at 0.50 mol/L. Observe the change in rate from 1.14 to 4.56 mol/L/h and calculate the rate order for NO using the rate equation.
03

Determine the Order of Reaction with Respect to Cl2

Compare the second and third experiments where the concentration of Cl2 is doubled (from 0.50 to 1.00 mol/L) while the concentration of NO remains constant at 1.00 mol/L. Observe the change in rate from 4.56 to 9.12 mol/L/h and calculate the order for Cl2 using the rate equation.
04

Write the Rate Law

Combine the orders of reaction calculated in Steps 2 and 3 to form the rate law, which describes the rate's dependence on the concentrations of NO and Cl2.
05

Calculate the Rate Constant

Use the rate law equation from Step 4 along with any of the sets of concentration and rate data to solve for the rate constant, k.

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Key Concepts

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

Rate Law
The rate law is a mathematical expression that links the rate of a chemical reaction to the concentration of its reactants. It has the general form \[\text{Rate} = k[\text{A}]^{m}[\text{B}]^{n}...\] where \( k \) is the rate constant, \( [\text{A}], [\text{B}], ... \) represents the concentration of reactants, and \( m \) and \( n \) are the reaction orders for each reactant respectively.

In the exercise provided, the rate law assists in understanding how the rate of production of \( \text{NOCl} \) is affected by the concentration of nitrogen monoxide (\( \text{NO} \)) and chlorine (\( \text{Cl}_2 \)). By examining the changes in concentration and observing the resultant changes in the reaction rate, one can deduce the powers to which each reactant concentration is raised in the rate law—this indicates the order of the reaction with respect to each reactant.

Understanding the rate law helps predict how a rate will change with varying reactant concentrations—an essential concept for controlling chemical reactions in industrial applications, and studying reaction mechanisms in research.
Reaction Order
Reaction order is a key term in chemical kinetics that indicates the dependency of the rate of reaction on the concentration of each reactant. It is often determined experimentally and is denoted by an exponent in the rate law equation.

The reaction order can be zero, first, second, or even fractional based on how the rate changes with concentration. A zero-order reaction means the rate is independent of the concentration of that reactant, while a first-order reaction shows that the rate is directly proportional to its concentration. A second-order dependency suggests the rate is proportional to the square of that reactant's concentration.
  • For \( \text{NO} \), if the rate increases by a factor of \(k\) when the concentration of \( \text{NO} \) alone is increased by a factor of \(n\), then \( \text{NO} \) has a reaction order of \(n\) in the reaction.
  • Similarly, for \( \text{Cl}_2 \), the reaction order is deduced by observing how the rate changes when only its concentration is varied.

By carefully studying the provided data and comparing rates with varying reactant concentrations, the reaction order with respect to each reactant is identified, which in turn assists in the determination of the rate law of the reaction.
Rate Constant
The rate constant, denoted as \( k \), is the proportionality factor in the rate law equation. Its value provides insight into the intrinsic speed of a reaction under given conditions and remains constant at a fixed temperature.

The units of the rate constant vary depending on the overall order of the reaction and can help you check if your rate law is consistent with the units of rate (usually \( \text{mol} \cdot \text{L}^{-1} \cdot \text{time}^{-1} \)). For instance, if the overall reaction order is one, the unit for \( k \) might be \( s^{-1} \), while for a second-order reaction, it might be \( M^{-1}s^{-1} \) or \( L\cdot mol^{-1}\cdot s^{-1} \).

In the exercise, once the reaction orders are known, any set of the provided concentration-rate data pairs can be used to calculate the rate constant by rearranging the rate law. The computed \( k \) allows for the prediction of the rate for any given concentrations of the reactants, making it a fundamental aspect of the rate law and indispensable for chemists in various practical applications.

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Most popular questions from this chapter

For the reaction \(Q \longrightarrow W+X,\) the following data were obtained at \(30^{\circ} \mathrm{C}:\) $$\begin{array}{|c|c|c|c|} \hline[Q]_{\text {initial }}(M) & 0.170 & 0.212 & 0.357 \\ \hline \text { Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{s}^{-1}\right)\) } & 6.68 \times 10^{-3} & 1.04 \times 10^{-2} & 2.94 \times 10^{-2} \\ \hline \end{array}$$ (a) What is the order of the reaction with respect to \([Q]\), and what is the rate law? (b) What is the rate constant?

The decomposition of acetaldehyde is a second order reaction with a rate constant of \(4.71 \times 10^{-8} \mathrm{L} \mathrm{mol}^{-1} \mathrm{s}^{-1}\). What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of \(5.55 \times 10^{-4}\) \(M ?\)

Nitrosyl chloride, NOCl, decomposes to NO and Cl\(_2\). $$2 \mathrm{NOCl}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ Determine the rate law, the rate constant, and the overall order for this reaction from the following data: $$\begin{array}{|l|c|c|c|}\hline \text {\([\mathrm{NOCl}](M)\)}) & 0.10 & 0.20 & 0.30 \\\\\hline \text { Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{h}^{-1}\right)\)} & 8.0 \times 10^{-10} & 3.2 \times 10^{-9} & 7.2 \times 10^{-9} \\\\\hline\end{array}$$

Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces: $$_{15}^{32} \mathrm{P} \longrightarrow_{16}^{32} \mathrm{S}+\mathrm{e}^{-}$$ rate \(=4.85 \times 10^{-2} \mathrm{day}^{-1}\left[^{32} \mathrm{P}\right]\) What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of \(0.0033 \mathrm{M}\) ?

Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction \(\mathrm{NO}+\mathrm{O}_{3} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}\) is first order with respect to both \(\mathrm{NO}\) and \(\mathrm{O}_{3}\) with a rate constant of \(2.20 \times 10^{7} \mathrm{L} / \mathrm{mol} / \mathrm{s}\). What is the instantaneous rate of disappearance of NO when \([\mathrm{NO}]=3.3 \times 10^{-6} \mathrm{M}\) and \(\left[\mathrm{O}_{3}\right]=5.9 \times 10^{-7} \mathrm{M} ?\)

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