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Chapter 12: Q22E (page 704)

Under certain conditions, the decomposition of ammonia on a metal surface gives the following data:

Determine the rate law, the rate constant, and the overall order for this reaction.

Short Answer

Expert verified

The rate law for decomposition of ammonia is equal to the rate constant. The value of rate law and rate constant is \({\bf{1}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{mol/L/h}}\)

The overall order of the reaction is zero.

Step by step solution

01

Decomposition of ammonia on a metal surface

From the given table, it is clear that the rate of reaction is independent of the concentration of reactant which is the condition of zero order reaction.

Hence, decomposition of ammonia on a metal surface is a zero-order reaction. The gas is absorbed on the metal surface due to pressure.

02

Rate law and rate constant

The rate law for a zero-order reaction is as follows:

\({\bf{Rate = k}}\)

Where k is a rate constant.

So, for decomposition of ammonia rate law will be equal the rate constant and does not depend on the concentration of reactant.

Value of rate constant\({\bf{k = 1}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{mol/L/h}}\)

Rate law\({\bf{ = k = 1}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{mol/L/h}}\)

Hence, the overall rate of reaction will be zero.

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Most popular questions from this chapter

Does the following data fit a second-order rate law?

Trial

Time(s)

(A) (M)

1

5

0.952

2

10

0.625

3

15

0.465

4

20

0.370

5

25

0.308

6

35

0.230

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, \({{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}{\bf{ + }}{{\bf{H}}_{\bf{2}}}{\bf{O}} \to {{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}{\bf{ + }}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}\) follows a first-order rate equation for the disappearance of sucrose: \({\bf{Rate = k}}\left( {{{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}} \right)\) (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)

  1. In neutral solution, \({\bf{k = 2}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 27 掳C and \({\bf{8}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 37 掳C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 掳C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
  2. When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is\({\bf{1}}{\bf{.65 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{ M}}\). How long will it take the solution to reach equilibrium at 27 掳C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
  3. Why does assuming that the reaction is irreversible simplify the calculation in part (b)?

Ingeneral, can we predict the effect of doubling the concentration of A on the rate of the overall reaction A + B鉄禖? Can we predict the effect if the reaction is known to be an elementary reaction?

The annual production of \({\bf{HN}}{{\bf{O}}_{\bf{3}}}\) in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.

\(\begin{align}\left( a \right){\bf{ }}4N{H_3}{\bf{ }}\left( g \right){\bf{ }} + {\bf{ }}5{O_2}{\bf{ }}(g) \to 4NO\left( g \right){\bf{ }} + {\bf{ }}6{H_2}O\left( g \right)\\\left( b \right){\bf{ }}2NO\left( g \right){\bf{ }} + {\bf{ }}{O_{2{\bf{ }}}}(g) \to 2N{O_{2{\bf{ }}}}\left( g \right)\\\left( c \right){\bf{ }}3N{O_2}{\bf{ }}\left( g \right){\bf{ }} + {\bf{ }}{H_2}O(l) \to 2HN{O_3}(aq) + NO(g)\end{align}\)

The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in \({{\bf{O}}_{\bf{2}}}\), what is the rate of formation of \({\bf{N}}{{\bf{O}}_{\bf{2}}}\) when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is \({\bf{5}}{\bf{.8 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ L}}{{\bf{ }}^{\bf{2}}}{\bf{ mo}}{{\bf{l}}^{{\bf{ - 2}}}}{\bf{ s}}{{\bf{ }}^{{\bf{ - 1}}}}\).

The reaction of \({\bf{CO}}\) with \({\bf{C}}{{\bf{l}}_{\bf{2}}}\) gives phosgene \(\left( {{\bf{COC}}{{\bf{l}}_{\bf{2}}}} \right)\), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:(fast, \({{\bf{k}}_{\bf{1}}}\) represents the forward rate constant, \({k_{ - {\bf{1}}}}\)the reverse rate constant)\({\bf{CO}}\left( g \right){\rm{ }} + {\rm{ }}{\bf{Cl}}\left( g \right) \to {\bf{COCl}}\left( g \right)\)(slow, \({k_{\bf{2}}}\) the rate constant)\({\bf{COCl}}\left( g \right){\rm{ }} + {\rm{ }}{\bf{Cl}}\left( g \right) \to {\bf{COC}}{{\bf{l}}_{\bf{2}}}\left( g \right)\)(fast,\({k_{\bf{3}}}\)the rate constant)(a) Write the overall reaction.(b) Identify all intermediates.(c) Write the rate law for each elementary reaction.(d) Write the overall rate law expression.
See all solutions

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