Nitrogen monoxide reacts with chlorine according to the equation:

2NO(g) + Cl\({}_2\)(g)⟶ 2NOCl(g) The following initial rates of reaction have been observed for certain reactant concentrations:

What is the rate law that describes the rate’s dependence on the concentrations of NO and Cl2? What is the rate constant? What are the orders with respect to each reactant?

Short Answer

Expert verified

Rate law that describes the rate’s dependence on the concentration of NO and Cl is

represented as

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

The value of the rate constant is\({\bf{9}}{\bf{.1}}\,{{\bf{L}}^{\bf{2}}}\,{\bf{mo}}{{\bf{l}}^{{\bf{ - 2}}\,}}{{\bf{h}}^{{\bf{ - 1}}}}\)

The order with respect to NO is 2, and the order with respect to Cl is 1

Step by step solution

01

Rate law

General rate law

\({\bf{Rate = k(NO}}{{\bf{)}}^m}{{\bf{(Cl}}{}_{\bf{2}}{\bf{)}}^n}\)

From the table, we can make the following equation

\(\begin{align}1.14mol{L^{ - 1}}{h^{ - 1}} &= k{(0.50)^m}{(0.50)^n}\,\,\,\,\,......(1)\\4.56mol{L^{ - 1}}{h^{ - 1}} &= k{(1.00)^m}{(0.50)^n}\,\,\,\,\,......(2)\\9.12mol{L^{ - 1}}{h^{ - 1}} &= k{(1.00)^m}{(1.00)^n}\,\,\,\,\,......(3)\end{align}\)

02

Order of NO

Cl2remains constant, but NO is double, and the rate becomes 4 times large. so, m=2

Second-order in NO.

It is calculated by dividing equation (2) by (1)

\(\begin{align}\frac{{1.14mol{L^{ - 1}}{h^{ - 1}} = k{{(0.50)}^m}{{(0.50)}^n}}}{{4.56mol{L^{ - 1}}{h^{ - 1}} = k{{(1.00)}^m}{{(0.50)}^n}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.25 &= {(0.5)^m}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m &= 2\end{align}\)

03

Order of Cl

NO remains constant, but Cl\({}_2\) is double, and the rate becomes times large. so, \({\bf{n = 1}}\)

First-order in Cl2

The order of Cl can be calculated by dividing equation (3) from (2)

\(\begin{align}\frac{{4.56mol{L^{ - 1}}{h^{ - 1}} = k{{(1.00)}^m}{{(0.50)}^n}}}{{9.12mol{L^{ - 1}}{h^{ - 1}} = k{{(1.00)}^m}{{(1.00)}^n}}}\\0.5 &= {(0.5)^m}\\m &= 1\end{align}\)

04

Rate constant

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

It can be calculated as

\(\begin{align}k &= \frac{{rate}}{{{{(NO)}^m}{{(Cl{}_2)}^n}}}\\ &= \frac{{1.14mol\,{L^{ - 1}}\,{h^{ - 1}}}}{{{{(0.50\,mol\,{L^{ - 1}})}^2}(0.50\,mol\,{L^{ - 1}})}}\\ &= 9.1\,{L^2}\,mo{l^{ - 2\,}}{h^{ - 1}}\end{align}\)

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Most popular questions from this chapter

Atomic chlorine in the atmosphere reacts with ozone in the following pair of elementary reactions:

\({\bf{Cl + \;O3}}\left( {\bf{g}} \right){\bf{ - - - ClO}}\left( {\bf{g}} \right){\bf{ + \;O2}}\left( {\bf{g}} \right)\)(rate constant k1 )

\({\bf{ClO}}\left( {\bf{g}} \right){\bf{ + O - - - Cl}}\left( {\bf{g}} \right){\bf{ + \;O2}}\left( {\bf{g}} \right)\)(rate constant k2 )

Determine the overall reaction, write the rate law expression for each elementary reaction, identify any intermediates, and determine the overall rate law expression.

Account for the relationship between the rate of a reaction and its activation energy.

The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:

Temperature(K)

k(s-1)

293

0.054

298

0.100

For each of the following reaction diagrams, estimate the activation energy \(\left( {{E_a}} \right)\)of the reaction:

Thefollowingdatahave been determined for the reaction: \({{\bf{I}}^{\bf{ - }}}{\bf{ + OC}}{{\bf{l}}^{\bf{ - }}} \to {\bf{I}}{{\bf{O}}^{\bf{ - }}} + {\bf{C}}{{\bf{l}}^{\bf{ - }}}\)

1

2

3

\({{\bf{(}}{{\bf{I}}^{\bf{ - }}}{\bf{)}}_{{\bf{initial}}}}\)(M)

0.10

0.20

0.30

\({{\bf{(OC}}{{\bf{l}}^{\bf{ - }}}{\bf{)}}_{{\bf{initial}}}}\)(M)

0.050

0.050

0.010

Rate(mol/l/s)

\({\bf{3}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

\({\bf{6}}.{\bf{2*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

\({\bf{1}}.{\bf{83*1}}{{\bf{0}}^{{\bf{ - 4}}}}\)

Determine the rate equation and the rate constant for this reaction.

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