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Chapter 14: Problem 24

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

Short Answer

Expert verified
(a) Rate = -\(\frac{1}{2}\) \(\frac{d[\mathrm{H}_{2} \mathrm{O}]}{dt}\) = \(\frac{d[\mathrm{H}_{2}]}{dt}\) = \(\frac{1}{2}\) \(\frac{d[\mathrm{O}_{2}]}{dt}\) (b) Rate = -\(\frac{d[\mathrm{SO}_{2}]}{dt}\) = \(\frac{1}{2}\) \(\frac{-d[\mathrm{O}_{2}]}{dt}\) = \(\frac{d[\mathrm{SO}_{3}]}{dt}\) (c) Rate = -\(\frac{1}{2}\) \(\frac{d[\mathrm{NO}]}{dt}\) = -\(\frac{1}{2}\) \(\frac{d[\mathrm{H}_{2}]}{dt}\) = \(\frac{d[\mathrm{N}_{2}]}{dt}\) = \(\frac{1}{2}\) \(\frac{d[\mathrm{H}_{2} \mathrm{O}]}{dt}\) (d) Rate = -\(\frac{1}{2}\) \(\frac{d[\mathrm{H}_{2}]}{dt}\) = -\(\frac{d[\mathrm{N}_{2}]}{dt}\) = \(\frac{d[\mathrm{N}_{2} \mathrm{H}_{4}]}{dt}\)

Step by step solution

01

Identify the rate of the formation of products and the consumption of reactants

Using the coefficients of the balanced chemical equation, we can determine the relationship between the rate of formation of products and the rate of consumption of reactants: Rate of formation of \(\mathrm{H}_{2}\) = 2 times the rate of consumption of \(\mathrm{H}_{2} \mathrm{O}\) Rate of formation of \(\mathrm{O}_{2}\) = 1 times the rate of consumption of \(\mathrm{H}_{2} \mathrm{O}\)
02

Determine the rate expressions for reactants and products

Now that we know the relationship between the rates, we can write the rate expressions in terms of each reactant and product: Rate = -\(\frac{1}{2}\) \(\frac{d[\mathrm{H}_{2} \mathrm{O}]}{dt}\) = \(\frac{d[\mathrm{H}_{2}]}{dt}\) = \(\frac{1}{2}\) \(\frac{d[\mathrm{O}_{2}]}{dt}\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\)
03

Identify the rate of the formation of products and the consumption of reactants

Using the coefficients of the balanced chemical equation, we can determine the relationship between the rate of formation of products and the rate of consumption of reactants: Rate of formation of \(\mathrm{SO}_{3}\) = 1 times the rate of consumption of \(\mathrm{SO}_{2}\) and Rate of formation of \(\mathrm{SO}_{3}\) = \(\frac{1}{2}\) times the rate of consumption of \(\mathrm{O}_{2}\)
04

Determine the rate expressions for reactants and products

Now that we know the relationship between the rates, we can write the rate expressions in terms of each reactant and product: Rate = -\(\frac{d[\mathrm{SO}_{2}]}{dt}\) = \(\frac{1}{2}\) \(\frac{-d[\mathrm{O}_{2}]}{dt}\) = \(\frac{d[\mathrm{SO}_{3}]}{dt}\) (c) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2\mathrm{H}_{2} \mathrm{O}(g)\)
05

Identify the rate of the formation of products and the consumption of reactants

Using the coefficients of the balanced chemical equation, we can determine the relationship between the rate of formation of products and the rate of consumption of reactants: Rate of formation of \(\mathrm{N}_{2}\) = 1 times the rate of consumption of \(\mathrm{NO}\) Rate of formation of \(\mathrm{H}_{2} \mathrm{O}\) = \(\frac{1}{2}\) times the rate of consumption of \(\mathrm{H}_{2}\)
06

Determine the rate expressions for reactants and products

Now that we know the relationship between the rates, we can write the rate expressions in terms of each reactant and product: Rate = -\(\frac{1}{2}\) \(\frac{d[\mathrm{NO}]}{dt}\) = -\(\frac{1}{2}\) \(\frac{d[\mathrm{H}_{2}]}{dt}\) = \(\frac{d[\mathrm{N}_{2}]}{dt}\) = \(\frac{1}{2}\) \(\frac{d[\mathrm{H}_{2} \mathrm{O}]}{dt}\) (d) \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{H}_{4}(g)\)
07

Identify the rate of the formation of products and the consumption of reactants

Using the coefficients of the balanced chemical equation, we can determine the relationship between the rate of formation of products and the rate of consumption of reactants: Rate of formation of \(\mathrm{N}_{2} \mathrm{H}_{4}\) = 2 times the rate of consumption of \(\mathrm{H}_{2}\) Rate of formation of \(\mathrm{N}_{2} \mathrm{H}_{4}\) = 1 times the rate of consumption of \(\mathrm{N}_{2}\)
08

Determine the rate expressions for reactants and products

Now that we know the relationship between the rates, we can write the rate expressions in terms of each reactant and product: Rate = -\(\frac{1}{2}\) \(\frac{d[\mathrm{H}_{2}]}{dt}\) = -\(\frac{d[\mathrm{N}_{2}]}{dt}\) = \(\frac{d[\mathrm{N}_{2} \mathrm{H}_{4}]}{dt}\)

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Most popular questions from this chapter

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\). and a frequency factor of \(A=6.0 \times 10^{8} M^{-1} \mathrm{~s}^{-1}\). The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100{ }^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the shape for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\). The following two-step mechanism is proposed for the reaction: Step 1: \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}\) Step 2: A \(+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}\) \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

A colored dye compound decomposes to give a colorless product. The original dye absorbs at \(608 \mathrm{~nm}\) and has an extinction coefficient of \(4.7 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at that wavelength. You perform the decomposition reaction in a 1 -cm cuvette in a spectrometer and obtain the following data: From these data, determine the rate law for the reaction "dye \(\longrightarrow\) product" and determine the rate constant.

(a) In which of the following reactions would you expect the orientation factor to be least important in leading to reaction: \(\mathrm{NO}+\mathrm{O} \longrightarrow \mathrm{NO}_{2}\) or \(\mathrm{H}+\mathrm{Cl} \longrightarrow \mathrm{HCl}\) ? (b) How does the kinetic-molecular theory help us understand the temperature dependence of chemical reactions?

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M}\), the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k\) ? (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the halflife for this reaction at \(61.05^{\circ} \mathrm{C}\) ?
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