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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. (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 \(0.010 \mathrm{M}\) ?

Short Answer

Expert verified
(a) The rate law for this reaction is given by: \(rate = k [\mathrm{NO}]^2 [\mathrm{H}_{2}]^1\). (b) At concentrations of \([\mathrm{NO}] = 0.035\ \mathrm{M}\) and \([\mathrm{H}_{2}] = 0.015\ \mathrm{M}\), the reaction rate at 1000 K is \(5.355\ \mathrm{M}\,\mathrm{s}^{-1}\). (c) When the concentration of \(\mathrm{NO}\) is increased to 0.10 M and the concentration of \(\mathrm{H}_{2}\) is 0.010 M, the reaction rate at 1000 K is \(60.0\ \mathrm{M}\,\mathrm{s}^{-1}\).

Step by step solution

01

Write the rate law expression

The rate law for this reaction is first order in H2 and second order in NO. The general rate law expression is given by: $$rate = k [A]^m [B]^n$$In this case, the rate law will be: $$rate = k [\mathrm{NO}]^2 [\mathrm{H}_{2}]^1$$ #b) Calculate reaction rate with concentrations given#
02

Plug in given values for rate constant and concentrations

Now we have the rate law expression, and we are given the rate constant at 1000 K, which is \(6.0 × 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), along with the concentrations of NO (\(0.035\ M\)) and H2 (\(0.015\ M\)). Plug these values into the rate law expression: $$rate = (6.0 × 10^{4}\mathrm{M}^{-2} \mathrm{~s}^{-1})[(0.035\mathrm{M})]^2 [(0.015\mathrm{M})]^1$$
03

Simplify the expression to find the rate

Now, calculate the reaction rate by multiplying the values and their respective units: $$rate = (6.0 × 10^{4}\mathrm{M}^{-2} \mathrm{~s}^{-1})(0.035^2\mathrm{M}^{2})(0.015\mathrm{M})$$ $$rate = 6.0 × 10^{4}\ M^{-2}s^{-1} × 0.035^{2}\ M^{2} × 0.015\ M$$ $$rate = 5.355\ M\,s^{-1}$$ #c) Calculate new reaction rate with new concentrations given #
04

Plug in new values for concentrations

Now, we need to calculate the new reaction rate when the concentration of NO is increased to \(0.10\ M\), while the concentration of H2 is \(0.010\ M\) at 1000 K. Plug these values into the rate law expression: $$rate = (6.0 × 10^{4}\mathrm{M}^{-2} \mathrm{~s}^{-1})[(0.10\mathrm{M})]^2 [(0.010\mathrm{M})]^1$$
05

Simplify the expression to find the new reaction rate

Now, calculate the new reaction rate by multiplying the values and their respective units: $$rate = 6.0 × 10^{4}\ M^{-2}s^{-1} × 0.10^{2}\ M^{2} × 0.010\ M$$ $$rate = 60.0\ M\,s^{-1}$$

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Most popular questions from this chapter

The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). Catalysts that convert these species to \(\mathrm{N}_{2}\) are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of \(\mathrm{NO}, \mathrm{NO}_{2},\) and \(\mathrm{N}_{2} .(\mathbf{b})\) Using a resource such as Table 8.4 , look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2}\), describing what wavelengths of light need to be monitored as a function of time.

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{array}{l} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{array} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? How do you know? (c) What is the intermediate in the reaction? How do you distinguish it from the catalyst?

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{nm}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at 30.0 min. Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? Explain your answer. (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)

The reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Using the following kinetic data, determine the magnitude and units of the first order rate constant: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { Pressure } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { (atm) } \\ \hline 0 & 1.000 \\ 2,500 & 0.947 \\ 5,000 & 0.895 \\ 7,500 & 0.848 \\ 10,000 & 0.803 \\ \hline \end{array} $$
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