Give the individual reaction orders for all substances and the overall reaction order from this rate law: $$ \text { Rate }=k \frac{\left[\mathrm{HNO}_{2}\right]^{4}}{[\mathrm{NO}]^{2}} $$

Short Answer

Expert verified
The reaction order for \([\text{HNO}_{2}]\) is 4, for \([\text{NO}]\) is -2, and the overall reaction order is 2.

Step by step solution

01

Identify Reaction Order for [HNO2]

Look at the exponent of \([\text{HNO}_{2}]\) in the rate law \(\text { Rate }=k \frac{\big[\text{HNO}_{2}\big]^{4}}{\big[\text{NO}\big]^{2}}\). The exponent is 4, so the reaction order with respect to \([\text{HNO}_{2}]\) is 4.
02

Identify Reaction Order for [NO]

Look at the exponent of \([\text{NO}]\) in the rate law. The exponent in the denominator counts as negative, so it is -2, making the reaction order with respect to \([\text{NO}]\) equal to -2.
03

Calculate Overall Reaction Order

Sum the individual reaction orders for all species involved. For this rate law, the sum is 4 (for \([\text{HNO}_{2}]\)) + (-2) (for \([\text{NO}]\)) = 2. Therefore, the overall reaction order is 2.

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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 an equation that links the rate of a chemical reaction to the concentration of the reactants. It provides insight into how the speed of a reaction changes when you alter the concentrations of the substances involved. The general form of a rate law can be written as: \( \text{Rate} = k [A]^m [B]^n \).
In this equation:
  • Rate: The speed of the reaction.
  • k: The rate constant.
  • [A] and [B]: The concentrations of reactants A and B.
  • m and n: The reaction orders for A and B.
Understanding the rate law is crucial for comprehending how different factors such as concentration affect the reaction rate.
Individual Reaction Orders
Individual reaction orders indicate the power to which the concentration of each reactant is raised in the rate law. They show how the rate depends on the concentration of specific reactants. In our rate law, \( \text{Rate} = k \frac{[\text{HNO}_2]^4}{[\text{NO}]^2} \), we can determine the individual reaction orders as follows:
The reaction order for \( \text{HNO}_2 \) is 4 because the concentration of \( \text{HNO}_2 \) is raised to the fourth power in the numerator. This means that the rate of reaction increases significantly with an increase in the concentration of \( \text{HNO}_2 \).
Next, the reaction order for \( \text{NO} \) is -2, as the concentration of \( \text{NO} \) is in the denominator and therefore has a negative exponent. This implies that an increase in the concentration of \( \text{NO} \) will decrease the reaction rate. Hence, each reactant's contribution to the overall reaction rate depends on these individual orders.
Overall Reaction Order
The overall reaction order is the sum of the individual reaction orders of all the reactants in the rate law. It gives a broad sense of the reaction's dependence on the concentrations of the reactants.
For our given rate law, \( \text{Rate} = k \frac{[\text{HNO}_2]^4}{[\text{NO}]^2} \):
  • The reaction order for \( \text{HNO}_2 \) is 4.
  • The reaction order for \( \text{NO} \) is -2.
To find the overall reaction order, we simply add these individual orders:
\( 4 + (-2) = 2 \).
Therefore, the overall reaction order of this reaction is 2. This tells us how the rate of the reaction as a whole responds to changes in the concentrations of the reactants.

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Most popular questions from this chapter

In the lower troposphere, ozone is one of the components of photochemical smog. It is generated in air when nitrogen dioxide, formed by the oxidation of nitrogen monoxide from car exhaust, reacts by the following mechanism: Assuming the rate of formation of atomic oxygen in step 1 equals the rate of its consumption in step \(2,\) use the data below to calculate (a) the concentration of atomic oxygen [O] and (b) the rate of ozone formation. $$ \begin{array}{lr} k_{1}=6.0 \times 10^{-3} \mathrm{~s}^{-1} & {\left[\mathrm{NO}_{2}\right]=4.0 \times 10^{-9} \mathrm{M}} \\ k_{2}=1.0 \times 10^{6} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s} & {\left[\mathrm{O}_{2}\right]=1.0 \times 10^{-2} \mathrm{M}} \end{array} $$

The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both \(\mathrm{NOBr}\) and \(\mathrm{Br}_{2}\) are reddish brown: $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ Use the data below to answer the following: (a) Determine the average rate over the entire experiment. (b) Determine the average rate between 2.00 and \(4.00 \mathrm{~s}\). (c) Use graphical methods to estimate the initial reaction rate. (d) Use graphical methods to estimate the rate at \(7.00 \mathrm{~s}\). (e) At what time does the instantaneous rate equal the average rate over the entire experiment? $$ \begin{array}{cc} \text { Time (s) } & \text { [NOBr] (mol/L) } \\ \hline 0.00 & 0.0100 \\ 2.00 & 0.0071 \\ 4.00 & 0.0055 \\ 6.00 & 0.0045 \\ 8.00 & 0.0038 \\ 10.00 & 0.0033 \end{array} $$

The rate constant of a reaction is \(4.50 \times 10^{-5} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(195^{\circ} \mathrm{C}\) and \(3.20 \times 10^{-3} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(258^{\circ} \mathrm{C}\). What is the activation energy of the reaction?

Aqua regia, a mixture of \(\mathrm{HCl}\) and \(\mathrm{HNO}_{3},\) has been used since alchemical times to dissolve many metals, including gold. Its orange color is due to the presence of nitrosyl chloride. Consider this one-step reaction for the formation of this compound: \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \operatorname{NOCl}(g)+\mathrm{Cl}(g) \quad \Delta H^{\circ}=83 \mathrm{~kJ}\) (a) Draw a reaction energy diagram, given \(E_{\text {affwd }}=86 \mathrm{~kJ} / \mathrm{mol}\). (b) Calculate \(E_{\text {arrev }}\). (c) Sketch a possible transition state for the reaction. (Note: The atom sequence of nitrosyl chloride is \(\mathrm{Cl}-\mathrm{N}-\mathrm{O} .)\)

For the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \longrightarrow \mathrm{AB}(g),\) how many unique collisions between \(A\) and \(B\) are possible if 1.01 mol of \(\mathrm{A}(g)\) and \(2.12 \mathrm{~mol}\) of \(\mathrm{B}(g)\) are present in the vessel?

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