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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 \(\mathrm{NO}\) . Write the rate law. (b) If the rate constant for this reaction at 1000 \(\mathrm{K}\) is \(6.0 \times 10^{4} M^{-2} \mathrm{s}^{-1}\) what is the reaction rate when \([\mathrm{NO}]=0.035 M\) and \(\left[\mathrm{H}_{2}\right]=0.015 M ?\) (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\(M ?\) (d) What is the reaction rate at 1000 \(\mathrm{K}\) if \([\mathrm{NO}]\) is decreased to 0.010 \(\mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to 0.030 \(\mathrm{M}\) ?

Short Answer

Expert verified
(a) The rate law for this reaction is: \[Rate = k[\mathrm{NO}]^2[\mathrm{H}_2]\] (b) The reaction rate when [NO] = 0.035 M and [H₂] = 0.015 M is 0.0525 M.s⁻¹. (c) The reaction rate when [NO] = 0.10 M and [H₂] = 0.010 M is 0.060 M.s⁻¹. (d) The reaction rate when [NO] = 0.010 M and [H₂] = 0.030 M is 0.0018 M.s⁻¹.

Step by step solution

01

Part A: Writing the rate law

The rate law is a mathematical equation that expresses the reaction rate as a function of the reactant concentrations. The given reaction has a first order dependence on H₂ concentration and a second order dependence on NO concentration. Therefore, the rate law can be written as: \[Rate = k[\mathrm{NO}]^2[\mathrm{H}_2]\]
02

Part B: Determining the reaction rate

We are given the values of the rate constant(k), NO, and H₂ concentrations. Now, we can substitute these values into the rate law equation and calculate the reaction rate: \[Rate = (6.0 \times 10^4 M^{-2}s^{-1})(0.035 M)^2(0.015 M)\] \[Rate = 0.0525 M.s^{-1}\]So, the reaction rate when [NO] = 0.035 M and [H₂] = 0.015 M is 0.0525 M.s⁻¹.
03

Part C: Reaction rate with increased NO concentration

We are given new concentrations for NO and H₂. We can use the same rate law equation and substitute these new values to calculate the reaction rate: \[Rate = (6.0 \times 10^4 M^{-2}s^{-1})(0.10 M)^2(0.010 M)\] \[Rate = 0.060 M.s^{-1}\]Therefore, the reaction rate when [NO] = 0.10 M and [H₂] = 0.010 M is 0.060 M.s⁻¹.
04

Part D: Reaction rate with decreased NO concentration and increased H₂ concentration

We are given new concentrations for NO and H₂. Again, we can use the same rate law equation and substitute these new values to calculate the reaction rate: \[Rate = (6.0 \times 10^4 M^{-2}s^{-1})(0.010 M)^2(0.030 M)\] \[Rate = 0.0018 M.s^{-1}\]Hence, the reaction rate when [NO] = 0.010 M and [H₂] = 0.030 M is 0.0018 M.s⁻¹.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the branch of physical chemistry that concerns itself with understanding the rates of chemical reactions. It's a fascinating area of study because it helps scientists understand how quickly a reaction occurs and what can affect this speed.

At the heart of chemical kinetics is the reaction rate, which is the speed at which reactants are transformed into products. This rate can vary greatly, from the near-instantaneous nature of some ionic reactions to the painstakingly slow rusting of iron. Factors that affect reaction rates include the concentration of reactants, temperature, presence of a catalyst, and surface area of solid reactants or catalysts.

An integral part of studying chemical kinetics is to conduct experiments to measure reaction rates and then use the data to deduce the rate law, which provides a mathematical relationship between the reaction rate and the concentration of reactants. Understanding chemical kinetics is essential not just for academic research, but also has practical applications in designing industrial processes, pharmaceuticals, and even in environmental science for pollution control.
Rate Constant
The rate constant, symbolized by the letter 'k' in kinetic equations, is a proportionality factor that provides the relationship between the reaction rate and the concentrations of reactants as dictated by the rate law. Its value is determined experimentally and is a quantification of how rapidly a reaction occurs.

Importantly, the rate constant isn't constant in all scenarios—it changes with temperature, and its units vary depending on the overall order of the reaction. For instance, a reaction with a second-order rate law would have a rate constant with units of M-1s-1. Understanding the rate constant is essential when comparing different reactions or the same reaction under different temperatures, as it is an intrinsic value that indicates the inherent speed at which a process happens under specific conditions.
Reactant Concentration
Reactant concentration often plays a starring role in the dynamics of a chemical reaction. Reflected in the reaction's rate law, it tells us how the rate of a reaction will change when we alter the amount of one or more of the reactants.

For a given reaction, if the rate law indicates a first-order dependence on a reactant, it means that the rate is directly proportional to that reactant's concentration. Double the amount of reactant, and you double the reaction rate. Conversely, a second-order dependence implies that the rate changes with the square of the reactant's concentration. An increase in the reactant concentration results in a reaction rate that speeds up quadratically, making these reactions highly sensitive to changes in concentration. This concept is fundamental in controlling reaction rates in various applications, from industrial synthesis to pharmacokinetics in medicine.

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Most popular questions from this chapter

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) .\) If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.036 \mathrm{M} / \mathrm{s},\) what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?(\mathbf{b})\) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is 74 torr per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\) -fold increase in the reaction rate?

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbondioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 A. (a) Estimate how many platinum atoms would fit into a 2.0 -nm sphere; the volume of a sphere is \((4 / 3) \pi r^{3} .\) Recall that \(1 \hat{\mathrm{A}}=1 \times 10^{-10} \mathrm{m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{m} .\) (b) Estimate how many platinum atoms are on the surface of a \(2.0-\mathrm{nm}\) Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of 2.8 A. (c) Using your results from (a) and (b), calculate the percentage of Pt atoms that are on the surface of a 2.0 -nm nanoparticle. (d) Repeat these calculations for a 5.0 -nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. \(\begin{array}{ll}{\text { Step } 1 :} & {\mathrm{O}_{3}(g) \Longrightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g) \text { (fast) }} \\ {\text { Step } 2 :} & {\mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(g) \quad(\text { slow })}\end{array}\) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is O a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Consider the reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} .\) Is each of the following statements true or false? (a) The rate law for the reaction must be Rate \(=k[\mathrm{A}][\mathrm{B}] .\) (b) If the reaction is an elementary reaction, the rate law is second order. (c) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be greater than that for the forward reaction.

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