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Chapter 14: Problem 74

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\)

Short Answer

Expert verified
The molecularity of the elementary reaction \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) is 2. The rate law for this reaction is \(Rate = k[\mathrm{NO}]^{2}\).

Step by step solution

01

Determine the molecularity of the elementary reaction

We can determine the molecularity of the elementary reaction by looking at the reactants. The given reaction is: \[2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\] There are 2 molecules of NO(g) required for the reaction, we can see it from the coefficient "2" in front of NO(g). Thus, the molecularity of the reaction is 2.
02

Write the rate law for the elementary reaction

To write the rate law for the elementary reaction, we can use the formula: \[Rate = k[\mathrm{Reactant}]^{m}\] Where Rate is the rate of the reaction, k is the rate constant, [Reactant] is the concentration of the reactant, and m is the order of the reaction with respect to the reactant. Since this is an elementary reaction with molecularity 2, the order of the reaction with respect to NO(g) is the same as its stoichiometric coefficient, which is 2. Therefore, the rate law for this reaction is: \[Rate = k[\mathrm{NO}]^{2}\]

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Most popular questions from this chapter

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\) \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) .\) In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{NH}_{3}(g)\)

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 \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c}) \mathrm{What}\) 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 \(0.010 \mathrm{M}\) ?

The following is a quote from an article in the August 18,1998 , issue of The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from \(77^{\circ} \mathrm{F}\) to \(\left.59^{\circ} \mathrm{F}\right]\) lowers the reaction rate six times; a 36 -degree drop [from \(77^{\circ} \mathrm{F}\) to \(\left.41^{\circ} \mathrm{F}\right]\) produces a fortyfold decrease in the rate." (a) Calculate activation energies for the breakdown process based on the two estimates of the effect of temperature on rate. Are the values consistent? (b) Assuming the value of \(E_{a}\) calculated from the 36 -degree drop and that the rate of breakdown is first order with a half-life at \(25^{\circ} \mathrm{C}\) of 2.7 years, calculate the half-life for breakdown at a temperature of \(-15^{\circ} \mathrm{C}\).

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