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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 first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times\) \(10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})\) in a volume of \(2.0 \mathrm{~L}\). (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min}\) ? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70{ }^{\circ} \mathrm{C}\) ?

The addition of \(\mathrm{NO}\) accelerates the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{aligned} \mathrm{NO}(g)+& \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this reaction? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism?

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(28{ }^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.0 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2} \mathrm{~s} \mathrm{} 0.025 \mathrm{M}\), what is the rate of formation of \(\mathrm{Cl}^{-}\)?

The temperature dependence of the rate constant for a reaction is tabulated as follows: $$ \begin{array}{ll} \hline \text { Temperature (K) } & k\left(M^{-1} \mathrm{~s}^{-1}\right) \\ \hline 600 & 0.028 \\ 650 & 0.22 \\ 700 & 1.3 \\ 750 & 6.0 \\ 800 & 23 \\ \hline \end{array} $$ Calculate \(E_{a}\) and \(A\).

Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carbon- containing fragment such as \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2}\), and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a \(\mathrm{C}\) atom for " \(\mathrm{R}^{w}\) ). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

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