Determine the overall orders of the reactions to which the following rate laws apply: (a) rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\), (b) rate \(=k,(\mathrm{c})\) rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]^{\frac{1}{2}}\) (d) rate \(=\) \(k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\)

Short Answer

Expert verified
The overall order of reactions for the given rate laws are respectively: (a) 2, (b) 0, (c) 1.5, (d) 3

Step by step solution

01

for rate law (a)

The given rate law is \(rate = k[\mathrm{NO}_{2}]^{2}\). The exponent of the concentration term, which is 2 in this case, gives the order of the reaction with respect to \(\mathrm{NO}_{2}\). Thus, the order of this reaction is 2.
02

for rate law (b)

The given rate law is \(rate = k\). Since there are no reactant concentrations in the rate law, this means the rate of the reaction is zero order, unaffected by the concentrations of the reactants and thus is 0.
03

for rate law (c)

The given rate law is \(rate = k[\mathrm{H}_{2}][\mathrm{Br}_{2}]^{1/2}\). The order with respect to \(\mathrm{H}_2\) is 1 and the order with respect to \(\mathrm{Br}_2\) is 1/2. So the overall order of the reaction is the sum of these, or 1 + 1/2 = 1.5.
04

for rate law (d)

The given rate law is \(rate = k[\mathrm{NO}]^{2}[\mathrm{O}_{2}]\). The order with respect to \(\mathrm{NO}\) is 2 and the order with respect to \(\mathrm{O}_2\) is 1. Therefore, the overall order of the reaction is 2 + 1 = 3.

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Most popular questions from this chapter

The thermal decomposition of phosphine \(\left(\mathrm{PH}_{3}\right)\) into phosphorus and molecular hydrogen is a first-order reaction: $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ The half-life of the reaction is \(35.0 \mathrm{~s}\) at \(680^{\circ} \mathrm{C}\). Calculate (a) the first-order rate constant for the reaction and (b) the time required for 95 percent of the phosphine to decompose.

A certain reaction is known to proceed slowly at room temperature. Is it possible to make the reaction proceed at a faster rate without changing the temperature?

A protein molecule, \(\mathrm{P}\), of molar mass \(\mathscr{A}\) dimerizes when it is allowed to stand in solution at room temperature. A plausible mechanism is that the protein molecule is first denatured (that is, loses its activity due to a change in overall structure) before it dimerizes: $$ \begin{array}{rlr} \mathrm{P} & \stackrel{k}{\longrightarrow} \mathrm{P}^{*}(\text { denatured }) & \text { slow } \\ 2 \mathrm{P}^{*} \longrightarrow \mathrm{P}_{2} & \text { fast } \end{array} $$ where the asterisk denotes a denatured protein molecule. Derive an expression for the average molar mass (of \(\mathrm{P}\) and \(\mathrm{P}_{2}\) ), \(, \overline{\mathscr{M}}\), in terms of the initial protein concentration \([\mathrm{P}]_{0}\) and the concentration at time \(t,\) \([\mathrm{P}]_{,}\) and \(\mathscr{A} .\) Describe how you would determine \(k\) from molar mass measurements.

The rate law for the reaction \(2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) is rate \(=k\left[\mathrm{H}_{2}\right][\mathrm{NO}]^{2} .\) Which of the following mechanisms can be ruled out on the basis of the observed rate expression? Mechanism I $$ \begin{array}{c} \mathrm{H}_{2}+\mathrm{NO} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N} \\ \mathrm{N}+\mathrm{NO} \longrightarrow \mathrm{N}_{2}+\mathrm{O} \\ \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \end{array} $$ $$ \begin{aligned} &\text { Mechanism II }\\\ &\begin{array}{l} \mathrm{H}_{2}+2 \mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{array} \end{aligned} $$ Mechanism III $$ \begin{aligned} 2 \mathrm{NO} & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} \\ \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} & \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow & \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{aligned} $$

The thermal decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) obeys firstorder kinetics. At \(45^{\circ} \mathrm{C},\) a plot of \(\ln \left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\) versus \(t\) gives a slope of \(-6.18 \times 10^{-4} \mathrm{~min}^{-1} .\) What is the half-life of the reaction?

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