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Chapter 1: Problem 10

A proposed mechanism for a reaction is as follows: $\begin{array}{cc}{\mathrm{NO}_{2}+\mathrm{F}_{2} \rightarrow \mathrm{NO}_{2} \mathrm{F}+\mathrm{F}} & {\text { Slow step }} \\ {\mathrm{F}+\mathrm{NO}_{2} \rightarrow \mathrm{NO}_{2} \mathrm{F}} & {\text { Fast step }}\end{array}$ What is the order of the overall reaction? (A) Zero order (B) First order (C) Second order (D) Third order

Short Answer

Expert verified
The order of the overall reaction is \(C\) Second order

Step by step solution

01

Identify the slowest step

Look at the given exercise and identify the slowest step of the reaction, because it determines the overall rate of reaction. Here, the slowest step is \(\mathrm{NO}_{2}+\mathrm{F}_{2} \rightarrow \mathrm{NO}_{2}\mathrm{F}+\mathrm{F}\)
02

Determine the reactants involved in the slowest step

The reactions involved in the slowest step are \(\mathrm{NO}_{2}\) and \(\mathrm{F}_{2}\).
03

Calculate the reaction order

The overall order of reaction is the sum of the orders with respect to each reactant in the rate-determining step. As there are two different reactants \(\mathrm{NO}_{2}\) and \(\mathrm{F}_{2}\) each with a stoichiometric coefficient of 1 in the rate-determining step, the overall order of the reaction will be the sum of these, that is, 2. This means the overall reaction is second order.

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Most popular questions from this chapter

$2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \leftrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) \Delta H=167 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}$ During the reaction above, the product yield can be increased by increasing the temperature of the reaction. Why is this effective? (A) The reaction is endothermic; therefore adding heat will shift it to the right. (B) Increasing the temperature increases the speed of the molecules, meaning there will be more collisions between them. (C) The reactants are less massive than the products, and an increase in temperature will cause their kinetic energy to increase more than that of the products. (D) The increase in temperature allows for a higher percentage of molecular collisions to occur with the proper orientation to create the product.

A strong acid/strong base titration is completed using an indicator which changes color at the exact equivalence point of the titration. The protonated form of the indicator is HIn, and the deprotonated form is In. At the equivalence point of the reaction: (A) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]\) (B) \([\mathrm{HIn}]=1 /\left[\mathrm{In}^{-}\right]\) (C) \([\mathrm{HIn}]=2\left[\mathrm{In}^{-}\right]\) (D) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]^{2}\)

\(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu} \quad E^{\circ}=+0.3 \mathrm{V}\) \(\mathrm{Fe}^{2+}+2 e^{-} \rightarrow \mathrm{Fe} \quad E^{\circ}=-0.4 \mathrm{V}\) Based on the reduction potentials given above, what is the reaction potential for the following reaction? \(\mathrm{Fe}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Fe}+\mathrm{Cu}^{2+}\) (A) \(-0.7 \mathrm{V}\) (B) \(-0.1 \mathrm{V}\) (C) \(+0.1 \mathrm{V}\) (D) \(+0.7 \mathrm{V}\)

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) What is the enthalpy change for reaction 1\(?\) (A) \(-148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (B) \(-56 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (C) \(-18 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (D) \(+148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\)

A laboratory technician wishes to create a buffered solution with a pH of 5. Which of the following acids would be the best choice for the buffer? (A) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \quad K_{a}=5.9 \times 10^{-2}\) (B) \(\mathrm{H}_{3} \mathrm{AsO}_{4} \quad K_{a}=5.6 \times 10^{-3}\) (C) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \quad K_{a}=1.8 \times 10^{-5}\) (D) \(\mathrm{HOCl}\) \(\quad K_{a}=3.0 \times 10^{-8}\)
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