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Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:

  1. What is the order of the reaction with respect to that reactant?
  2. Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?

Short Answer

Expert verified
  1. Order of reaction with respect to the reactant is 2.
  2. Order of reaction with respect to the reactant is 1.

Step by step solution

01

Rate law

The rate law for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentration of the reactants participating in it.

\({\bf{rate = k(reactant}}{{\bf{)}}^{\bf{n}}}\)

Doubling the concentration of a reactant increases the rate of a reaction four times.

\({\bf{4(rate) = k(2reactant}}{{\bf{)}}^{\bf{n}}}\)

Hence, the order of the reaction can be calculated as

\(\begin{aligned}{c}\frac{{{\bf{4(Rate)}}}}{{{\bf{Rate}}}}{\bf{ = }}\frac{{\,{\bf{k(2(reactant)}}{{\bf{)}}^{\bf{n}}}}}{{{\bf{k(reactant}}{{\bf{)}}^{\bf{n}}}}}\\{\bf{n = 2}}\end{aligned}\)

The concentration of the reactant doubled and the rate quadrupled, hence the order with respect to the reactant is 2.

02

Rate of reaction

Tripling the concentration of a different reactant increases the rate of a reaction three times. It can be represented as

\({\bf{3(Rate) = }}\,{\bf{k(3(reactant)}}{{\bf{)}}^{\bf{n}}}\)

Hence, the order of the reaction can be calculated as

\(\begin{aligned}{}\frac{{{\bf{3(rate)}}}}{{{\bf{rate}}}}{\bf{ = }}\frac{{{\bf{k(3reactant}}{{\bf{)}}^{\bf{n}}}}}{{{\bf{k(reactant}}{{\bf{)}}^{\bf{n}}}}}\\{\bf{n = 1}}\end{aligned}\)

The concentration of the reactant and the rate both tripled, hence order with respect to this reactant is 1.

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Most popular questions from this chapter

What is the half-life for the first-order decay of phosphorus-32?\(_{{\bf{15}}}^{{\bf{32}}}{\bf{P}} \to _{{\bf{16}}}^{{\bf{32}}}{\bf{S + }}{{\bf{e}}^{\bf{ - }}}\)The rate constant for the decay is\({\bf{4}}{\bf{.85 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}{\bf{da}}{{\bf{y}}^{{\bf{ - 1}}}}\).

When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?

For the reaction\({\bf{A}} \to {\bf{B + C}}\), the following data were obtained at 30 ยฐC:

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Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as \({\bf{CC}}{{\bf{l}}_{\bf{2}}}{{\bf{F}}_{\bf{2}}}\), catalyse the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is:

\(\begin{aligned}{}{{\bf{O}}_{\bf{3}}}\overset{sunlight}{\rightarrow}{}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{O}}\\{{\bf{O}}_{\bf{3}}}{\rm{ }} + {\rm{ }}{\bf{Cl}}\to {{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{ClO}}\\{\bf{ClO}}{\rm{ }} + {\rm{ }}{\bf{O}}\to {\bf{Cl}}{\rm{ }} + {\rm{ }}{{\bf{O}}_{\bf{2}}}\end{aligned}\)

(a) Explain why chlorine atoms are catalysts in the gas-phase transformation:

\({\bf{2}}{{\bf{O}}_{\bf{3}}}\mathop {}\limits^{}\to {\bf{3}}{{\bf{O}}_{\bf{2}}}\)

(b) Nitric oxide is also involved in the decomposition of ozone by the mechanism: Is NO a catalyst for the decomposition? Explain your answer.

\(\begin{aligned}{}{{\bf{O}}_{\bf{3}}}\overset{sunlight}{\rightarrow}{\rm{ }}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{O}}\\{{\bf{O}}_{\bf{3}}}{\rm{ }} + {\rm{ }}{\bf{NO}}\rightarrow {\bf{N}}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{{\bf{O}}_{\bf{2}}}\\{\bf{N}}{{\bf{O}}_{\bf{2}}}{\rm{ }} + {\rm{ }}{\bf{O}}\rightarrow {\bf{NO}}{\rm{ }} + {\rm{ }}{{\bf{O}}_{\bf{2}}}\end{aligned}\)

Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the equation:\({{\bf{H}}_{\bf{2}}}{\bf{(g) + 2NO(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{O(g) + }}{{\bf{H}}_{\bf{2}}}{\bf{O}}\).Determine the rate law, the rate constant, and the orders with respect to each reactant from the following data:

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