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Chapter 12: Q26 E (page 704)

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:

Short Answer

Expert verified

The rate law represented as\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{1}}}\)

Reaction order with respect to NO is 2 and with respect to Cl is 1

Value of rate constant \({\bf{9}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}{{\bf{L}}^{\bf{2}}}{\bf{Mo}}{{\bf{l}}^{{\bf{ - 2}}}}{\bf{s}}\)

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.

General rate law for the given reaction is represented as

\({\bf{rate = (NO}}{{\bf{)}}^{\bf{m}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{n}}}\)

From the given table we can make the following equations

\(\begin{align}2.835 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}} &= k{(0.30)^m}{(0.60)^n}\,\,\,\,\,\,......(1)\left( {0,1} \right)\\1.134 \times {10^{ - 2}}mol{L^{ - 1}}{s^{ - 1}} &= k{(0.35)^m}{(0.35)^n}\,\,\,\,\,\,......(2)\\2.268 \times {10^{ - 2}}mol{L^{ - 1}}{s^{ - 1}} &= k{(0.60)^m}{(0.70)^n}\,\,\,\,\,\,......(3)\end{align}\)

02

Order of each reactant

The order of reaction refers to the power dependence of the rate on the concentration of each reactant.

On dividing equation (2) by (1), we get

\(\frac{{Rate(R{}_2)}}{{Rate(R{}_1)}} = \frac{{0.01134}}{{0.002835}} = \frac{{k{{(0.60)}^m}{{(0.35)}^n}}}{{k{{(0.30)}^m}{{(0.35)}^n}}}\)

\(\begin{align}{(2)^m} = 4\\m = 2\end{align}\)

On dividing equation (3) by (2), we get

\(\frac{{Rate(R{}_3)}}{{Rate(R{}_2)}} = \frac{{0.02268}}{{0.01134}} = \frac{{k{{(0.60)}^m}{{(0.70)}^n}}}{{k{{(0.60)}^m}{{(0.35)}^n}}}\)

\(\begin{align}{(2)^n} = 2\\n = 1\end{align}\)

Hence the rate of given reaction is

\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{1}}}\)

The overall order of reaction \({\bf{ = m + n = 2 + 1 = 3}}\)

03

Rate constant

The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and concentration of reacting substances.

\({\bf{Rate = k}}\left( {{\bf{NO}}} \right){{\bf{\;}}^{\bf{2}}}{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}\)

\(\begin{align}2.835 \times {10^{ - 3}}mol{L^{ - 1}}{s^{ - 1}} &= K{(0.30M)^2}(0.35M)\\k &= \frac{{2.835 \times {{10}^{ - 3}}mol{L^{ - 1}}{s^{ - 1}}}}{{{{(0.30M)}^2}(0.35M)}}\\ &= 9.0 \times {10^{ - 2}}{L^2}Mo{l^{ - 2}}{s^{ - 1}}\end{align}\)

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Most popular questions from this chapter

Experiments were conducted to study the rate of the reaction represented by this equation.(2)\({\rm{2NO(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\)Initial concentrations and rates of reaction are given here.

Experiment Initial Concentration

\(\left( {{\bf{NO}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)

Initial Concentration, \(\left( {{{\bf{H}}_{\bf{2}}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)Initial Rate of Formation of \({{\bf{N}}_{\bf{2}}}{\rm{ }}\left( {{\bf{mol}}/{\bf{L}}{\rm{ }}{\bf{min}}} \right)\)
1\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{1}}.{\bf{8}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
2\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{3}}.{\bf{6}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
3\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{30}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
4\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)

Consider the following questions:(a) Determine the order for each of the reactants, \({\bf{NO}}\) and \({{\bf{H}}_{\bf{2}}}\), from the data given and show your reasoning.(b) Write the overall rate law for the reaction.(c) Calculate the value of the rate constant, k, for the reaction. Include units.(d) For experiment 2, calculate the concentration of \({\bf{NO}}\)remaining when exactly one-half of the original amount of \({{\bf{H}}_{\bf{2}}}\) had been consumed.(e) The following sequence of elementary steps is a proposed mechanism for the reaction.Step 1:Step 2:Step 3:Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.

A study of the rate of the reaction represented as 2A鉄� B gave the following data:

  1. Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s.
  2. Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus (A). What are the units of this rate?
  3. Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and 10.0 s, and the instantaneous rate of formation of B at 15.0 s.

If the rate of decomposition of ammonia, \({\bf{N}}{{\bf{H}}_{\bf{3}}}\), at 1150 K is \(2.10 \times 1{0^{ - 6}}mol/L/s\), what is the rate of production of nitrogen and hydrogen?

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

What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is \({\bf{8}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}\)L/mol/s.
See all solutions

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