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Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. (b) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M}\) ? (c) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M}\), while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M}\) ? (d) What is the reaction rate at \(1000 \mathrm{~K}\) if [NO] is decreased to \(0.010 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to \(0.030 \mathrm{M}\) ?

Short Answer

Expert verified
The rate law for this reaction is \(Rate = k[NO]^2[H_2]\). The reaction rates for the given concentrations are: (a) \(Rate = 2.65 M s^{-1}\) (b) \(Rate = 6.0 M s^{-1}\) (c) \(Rate = 0.018 M s^{-1}\)

Step by step solution

01

Write the rate law equation

The rate law for this reaction is given as first order in Hâ‚‚ and second order in NO. We can write this rate law as: \(Rate = k[NO]^2[H_2]\)
02

Use the given rate constant to calculate the reaction rate

The rate constant, k, is given as \(6.0 × 10^4 M^{-2}s^{-1}\). Now we will find the reaction rate for each scenario: (a) For [NO] = 0.035 M and [H₂] = 0.015 M: \(Rate = k[NO]^2[H_2] = (6.0 × 10^4 M^{-2}s^{-1})(0.035 M)^2(0.015 M)\)
03

Calculate the reaction rate for the given concentrations

Now, we will simply calculate the reaction rate for the given concentrations: (a) \(Rate = (6.0 × 10^4)(0.035^2)(0.015) = 2.65 M s^{-1}\) (b) For [NO] = 0.10 M and [H₂] = 0.010 M: \(Rate = (6.0 × 10^4)(0.10^2)(0.010) = 6.0 M s^{-1}\) (c) For [NO] = 0.010 M and [H₂] = 0.030 M: \(Rate = (6.0 × 10^4)(0.010^2)(0.030) = 0.018 M s^{-1}\) So the reaction rates for the given concentrations are: (a) \(Rate = 2.65 M s^{-1}\) (b) \(Rate = 6.0 M s^{-1}\) (c) \(Rate = 0.018 M s^{-1}\)

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Key Concepts

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

Reaction Kinetics
The study of reaction kinetics involves investigating the rates of chemical reactions and the mechanisms by which they occur. This field empowers students and chemists to understand how variables such as temperature, concentration, and catalysts affect the speed of reactions. In educational exercises, we typically focus on reactions progressing in a single step for simplicity; however, real-world scenarios can involve complex multi-step mechanisms. Reaction rates help predict how quickly reactants will be consumed or how soon products will form. By delving into reaction kinetics, students gain crucial insights into the controlled manipulation of industrial processes, environmental systems, and even biological pathways.
Rate Constant
The rate constant, denoted as 'k', is an essential parameter in the rate law equation, reflecting the intrinsic speed of a reaction. It is independent of the concentration of reactants but varies with temperature, as described by the Arrhenius equation. The rate constant is unique to each reaction and its specific conditions and is determined experimentally. In the given exercise, the rate constant is provided, allowing students to calculate the rate of reaction directly without needing to conduct an experiment themselves. The magnitude of 'k' can also hint at whether a reaction proceeds quickly or slowly, forming part of the foundation for understanding chemical kinetics.
Concentration Dependency
Concentration dependency is central to the concept of reaction rates, emphasizing that the rate often changes as the concentration of one or more reactants change. This association is quantified in the rate law for a reaction, which mathematically links the rate with the concentrations of reactants raised to their respective orders. The exercises provided illustrate how variations in reactant concentrations can lead to different reaction rates. Such exercises fortify a student's grasp on predicting the effects of concentration changes, a core skill in chemistry. It's also a practical aspect for real-life scenarios like diluting pollutants or optimizing the yield in industrial synthesis.
Reaction Order
Reaction order refers to the power to which the concentration of a reactant is raised in the rate law equation. It indicates how the reaction rate is affected by the concentration of that reactant. The orders are usually discovered through experiments and can be integers or fractions. In our exercise, the reaction is first order in \(\mathrm{H}_{2}\) and second order in NO, thus the reaction rate changes linearly with changes in the \(\mathrm{H}_{2}\) concentration but quadratically with changes in NO concentration. Understanding reaction order not only aids students in comprehending the specifics of a chemical reaction but also plays a crucial role in the design and analysis of many chemical processes.

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Most popular questions from this chapter

A first-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) has the rate constant \(k=3.2 \times 10^{-3} \mathrm{~s}^{-1}\). If the initial concentration of \(\mathrm{A}\) is \(2.5 \times 10^{-2} M\), what is the rate of the reaction at \(t=660 \mathrm{~s}\) ?

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) (b) \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)\) (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene $\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\(. A \)0.0400 \mathrm{M}\( solution of \)\mathrm{C}_{5} \mathrm{H}_{6}$ was monitored as a function of time as the reaction $2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}$ proceeded. The following data were collected: $$ \begin{array}{cc} \hline \text { Time (s) } & {\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)} \\\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \\ \hline \end{array} $$ Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, $\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\( versus time, and \)1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]$ versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \quad \text { (slow) } \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \text { (fast) } \end{aligned} $$ (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

The gas-phase decomposition of $\mathrm{NO}_{2}, 2 \mathrm{NO}_{2}(g) \longrightarrow\( \)2 \mathrm{NO}(g)+\mathrm{O}_{2}(g),$ is studied at \(383^{\circ} \mathrm{C}\), giving the following data: $$ \begin{array}{cc} \hline \text { Time (s) } & {\left[\mathrm{NO}_{2}\right](M)} \\ \hline 0.0 & 0.100 \\ 5.0 & 0.017 \\ 10.0 & 0.0090 \\ 15.0 & 0.0062 \\ 20.0 & 0.0047 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to the concentration of \(\mathrm{NO}_{2} ?(\mathbf{b})\) What is the rate constant? (c) Predict the reaction rates at the beginning of the reaction for initial concentrations of \(0.200 \mathrm{M}, 0.100 \mathrm{M},\) and $0.050 \mathrm{M} \mathrm{NO}_{2}$.

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