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Chapter 14: Problem 30

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. \((\mathbf{b})\) 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} ?(\mathbf{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 \([\mathrm{NO}]\) 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 is \( Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] \). (b) The reaction rate is \( 8.82 \times 10^{-1} M~s^{-1} \). (c) The reaction rate is \( 6.0 M~s^{-1} \). (d) The reaction rate is \( 1.8 \times 10^{-2} M~s^{-1} \).

Step by step solution

01

Write the Rate Law for the Reaction

Given that the reaction is first order in H₂ and second order in NO, we can express the rate law as follows: \( Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] \), where k is the rate constant.
02

Find the reaction rate at given concentrations (part b)

The rate constant at 1000 K is given as \(6.0 \times 10^4 M^{-2}s^{-1}\). We are given the concentrations of NO and H₂ as 0.035 M and 0.015 M, respectively. Using the rate law derived in Step 1, the reaction rate can be calculated: \[ Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] = (6.0 \times 10^4 M^{-2}s^{-1})(0.035 M)^2 (0.015 M) \] Calculate the rate to find the answer.
03

Find the reaction rate with new NO concentration (part c)

We are given a new concentration for NO (0.10 M) with the H₂ concentration remaining at 0.010 M. Use the same rate law equation as before: \[ Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] = (6.0 \times 10^4 M^{-2}s^{-1}) (0.10 M)^2 (0.010 M) \] Calculate the rate for this new set of concentrations.
04

Find the reaction rate with new concentrations for both reactants (part d)

The concentrations of NO and H₂ are now given as 0.010 M and 0.030 M, respectively. Use the rate law equation once again: \[ Rate = k[\mathrm{NO}]^2[\mathrm{H}_{2}] = (6.0 \times 10^4 M^{-2}s^{-1}) (0.010 M)^2 (0.030 M) \] Calculate the rate for this last set of reactant concentrations.

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Most popular questions from this chapter

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) $\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q) \longrightarrow \mathrm{HCN}(a q)$ (b) \(\mathrm{CH}_{3} \mathrm{Cl}(\) solv \()+\mathrm{OH}^{-}(\) solv $) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\( solv \))+\mathrm{Cl}^{-}($ solv \()\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}\)

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is $2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$. The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(55^{\circ} \mathrm{C}\) the rate constant is \(4.12 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when $\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.050 \mathrm{M} ?(\mathbf{c})$ What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is tripled to $0.150 \mathrm{M}$ ? (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is reduced by \(10 \%\) to \(0.045 \mathrm{M}\) ?

Consider two reactions. Reaction (1) has a half-life that gets longer as the reaction proceeds. Reaction (2) has a half-life that gets shorter as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

The enzyme urease catalyzes the reaction of urea, $\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$, with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at $100^{\circ} \mathrm{C}$. In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at $21^{\circ} \mathrm{C}$. (a) Write out the balanced equation for the reaction catalyzed by urease. \((\mathbf{b})\) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C}\), what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

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 $30{ }^{\circ} \mathrm{C}\( is \)4.0 \times 10^{-2} M^{-1} \mathrm{~s}^{-1}$. If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is $2.5 \times 10^{-4} \mathrm{M}\( and that of \)\mathrm{Cl}_{2}\( is \)2.0 \times 10^{-2} \mathrm{M},$ what is the rate of formation of \(\mathrm{H}^{+}\) ?
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