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Determine the rate law for the reaction described by $$ \mathrm{NO}(g)+\mathrm{H}_{2}(g) \rightarrow \text { products } $$ from the initial-rate data tabulated below: \begin{tabular}{ccc} \hline Initial pressure of \(P_{\mathrm{H}_{2}} /\) Torr & Initial pressure of \(P_{\mathrm{No}} /\) Torr & Initial rate of reaction/Torr \(\cdot \mathrm{s}^{-1}\) \\ \hline 400 & 159 & 34 \\ 400 & 300 & 125 \\ 289 & 400 & 160 \\ 205 & 400 & 110 \\ 147 & 400 & 79 \\ \hline \end{tabular} Calculate the value of the rate constant for this reaction.

Short Answer

Expert verified
The rate law is \( \text{Rate} = k [\text{NO}]^2 [\text{H}_2] \) with \( k \approx 0.000034 \text{ Torr}^{-2} \text{s}^{-1} \).

Step by step solution

01

Understand the Rate Law Equation

The rate law for a chemical reaction is generally given by the expression \( \text{Rate} = k [A]^m [B]^n \), where - \( k \) is the rate constant,- \( [A] \) and \( [B] \) are the concentrations of the reactants,- \( m \) and \( n \) are the orders of the reaction with respect to each reactant.
02

Set Up Proportionality Equations

From the given data, we can set up equations based on the proposed rate law.For example: \( 34 = k (159)^m (400)^n \) and \( 125 = k (300)^m (400)^n \). We will do this for each dataset available.
03

Determine Reaction Order with Respect to NO

Compare the rates when \( P_{\mathrm{H}_2} \) is constant (400 Torr):\[ \frac{125}{34} = \frac{k (300)^m (400)^n}{k (159)^m (400)^n} \rightarrow 3.68 \approx \left(\frac{300}{159}\right)^m \] Solving \( m \approx 2 \).
04

Determine Reaction Order with Respect to H2

Compare the rates when \( P_{\mathrm{NO}} \) is constant (400 Torr):\[ \frac{160}{110} = \frac{k (400)^m (289)^n}{k (400)^m (205)^n} \rightarrow 1.45 \approx \left(\frac{289}{205}\right)^n \] Solving \( n \approx 1 \).
05

Write the Rate Law for the Reaction

Based on the calculated reaction orders \( m \approx 2 \) and \( n \approx 1 \), the rate law is:\[ \text{Rate} = k [\text{NO}]^2 [\text{H}_2]^1 \]
06

Calculate the Rate Constant k

We can use any of the experimental data sets to calculate \( k \). Using the first dataset:\[ 34 = k (159)^2 (400)^1 \]Solving for \( k \):\[ k \approx \frac{34}{(159)^2 (400)} \approx 0.000034 \text{ Torr}^{-2} \text{s}^{-1} \]

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Key Concepts

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

Rate Law
Chemical reactions have a speed at which they proceed, termed as the reaction rate. To predict this speed under varying conditions, scientists use the rate law. The rate law is an algebraic equation that describes how the rate of a reaction is dependent on the concentrations of the reactants. In general, it is written as - \( ext{Rate} = k [A]^m [B]^n\) where - \(k\) is the rate constant, - \([A]\) and \([B]\) are the molar concentrations of reactants, - \(m\) and \(n\) are the reaction orders concerning each reactant.
For instance, if a chemical reaction involves two gases, NO and Hâ‚‚, like in our exercise, its rate law will be constructed by examining variations in reaction rates when one reactant's concentration changes while the other remains constant.
  • The rate law provides scientists with a way to predict the effect of changes in reactant concentrations on the reaction rate.

  • It helps in understanding the relationship between reactant concentrations and the speed of the reaction.
Reaction Order
The reaction order in a chemical reaction tells us how the rate is affected by the concentration of reactants. It is a critical component of the rate law. Each reactant involved in a reaction has its own order, indicated as a power to which its concentration is raised in the rate law.
To determine the reaction order, you compare rates from experiments where the concentration of one reactant is varied while others are held constant. For our reaction, let's break it down:
- **Order with respect to NO:** By comparing experiments where only the concentration of NO varies, it can be determined that the reaction is of second order in NO (m = 2).
- **Order with respect to Hâ‚‚:** Conversely, by examining cases where only the concentration of Hâ‚‚ changes, we find it is first-order in Hâ‚‚ (n = 1).
Hence, for a simple reaction \( ext{Rate} = k [ ext{NO}]^2 [ ext{H}_2]^1\).
  • The sum of these orders gives the overall reaction order, which tells us the overall dependence of the reaction rate on reactant concentrations.

  • The reaction order is usually determined experimentally and can differ from the stoichiometric coefficients in the balanced reaction equation.
Rate Constant
The rate constant \(k\) is a proportionality factor in the rate law equation that links the reaction rate to the concentrations of reactants. Its value gives insights into how fast a reaction can occur.
The rate constant is distinct for each reaction and depends on factors like temperature and the presence of a catalyst. In our chemical reaction, once the rate equation \( ext{Rate} = k [ ext{NO}]^2 [ ext{H}_2]^1\) was established, the constant could be calculated from experimental data.
Let's highlight key points:
  • The rate constant \(k\) is determined from experimental data by rearranging the rate law equation to solve for k.

  • Its units depend on the overall order of the reaction. For a reaction with an overall order of 3 (second order in NO and first-order in Hâ‚‚), the units are \( ext{Torr}^{-2} ext{s}^{-1}\).

This constant remains unchanged unless the temperature is altered or a catalyst is introduced.

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Most popular questions from this chapter

The rate of the reaction described by the equation $$ 2 \mathrm{CO}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{C}(s) $$ was studied by injecting some \(\mathrm{CO}(g)\) into a reaction vessel and measuring the total pressure while maintaining a constant reaction volume: \begin{tabular}{cc} \hline \(\boldsymbol{P}_{\text {total }} /\) Torr & \(t / \mathrm{s}\) \\ \hline 250 & 0 \\ 288 & 398 \\ 224 & 1002 \\ 210 & 1801 \\ 196 & 3000 \\ \hline \end{tabular} Assuming that only \(\mathrm{CO}(g)\) is present initially, determine the value of the reaction rate constant.

Strontium-90 is a radioactive isotope that is produced in nuclear explosions. It decays by \(\beta\) -emission with a half-life of \(29.1\) years. Suppose that an infant ingests strontium-90 in mother's milk. Calculate the fraction of the ingested strontium-90 that remains in the body when the infant reaches 74 years of age, assuming no loss of strontium-90 except by radioactive decay.

The reaction described by the chemical equation $$ 3 \mathrm{BrO}^{-}(a q) \rightarrow \operatorname{BrO}_{3}^{-}(a q)+2 \mathrm{Br}^{-}(a q) $$ is second order in \(\mathrm{BrO}^{-}(a q)\) in basic solution with a rate constant equal to \(0.056 \mathrm{M}^{-1} \cdot \mathrm{s}^{-1}\) at \(80^{\circ} \mathrm{C}\). If \(\left[\mathrm{BrO}^{-}\right]_{0}\) \(=0.212 \mathrm{M}\), what will \(\left[\mathrm{BrO}^{-}\right]\) be \(1.00\) minute later?

One of the waste products of a uraniumpowered nuclear reactor is plutonium-239, which has a half-life of \(2.41 \times 10^{9}\) years. How long will it take for \(10 \%\) of the plutonium-239 waste from a nuclear power plant to decay? How long will it take for \(99 \%\) of the same waste to decay? Why is the storage of nuclear waste such a difficult problem to solve?

Azomethane, \(\mathrm{CH}_{3} \mathrm{~N}_{2} \mathrm{CH}_{5}(g)\), decomposes according to the equation $$ \mathrm{CH}_{3} \mathrm{~N}_{2} \mathrm{CH}_{3}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{3}(g)+\mathrm{N}_{2}(g) $$ Given that the decomposition is a first-order process with \(k=4.0 \times 10^{-4} \mathrm{~s}^{-1}\) at \(300^{\circ} \mathrm{C}\), calculate the fraction of azomethane that remains after \(1.0\) hour.

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