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Chapter 12: Q33 E (page 706)

Usethe data provided to graphically determine the order and rate constant of the following reaction: \({\bf{S}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}} \to {\bf{S}}{{\bf{O}}_{\bf{2}}}{\bf{ + C}}{{\bf{l}}_{\bf{2}}}\)

Time(hr)

0

5.00*\({\bf{1}}{{\bf{0}}^{\bf{3}}}\)

1.00*\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)

1.50*\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)

2.50*\({\bf{1}}{{\bf{0}}^{\bf{4}}}\)

3.00*104

4.00*104

\({\bf{(S}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{)}}\)(M)

0.100

0.0896

0.0802

0.0719

0.0577

0.0517

0.0415

Short Answer

Expert verified

The reaction is first order.

The rate constant of the reaction is \({\bf{2}}{\bf{.20 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)

Step by step solution

01

Definition

In first-order kinetics, the rate of reaction is directly proportional to the concentration of reactant.

\({\bf{Rate of reaction}} \propto \left( {{\bf{ concentration of reactant}}} \right)\)

\({\bf{Rate of reaction = K }}\left( {{\bf{concentration of reactant}}} \right)\)

Where K is the rate constant.

The rate constant of a first-order reaction is given as

\(k{\bf{ }} = \frac{{\ln \frac{{{{(P)}_0}}}{{(P)}}}}{t}\)

02

Calculation of rate constant

Time

\({\bf{(S}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{)}}\)

ln\({\bf{(S}}{{\bf{O}}_{\bf{2}}}{\bf{C}}{{\bf{l}}_{\bf{2}}}{\bf{)}}\)

0

0.100

-2.302

\({\bf{5 \times 1}}{{\bf{0}}^{\bf{3}}}\)

0.0896

-2.412

\({\bf{1 \times 1}}{{\bf{0}}^{\bf{4}}}\)

0.0802

-2.523

\({\bf{1}}{\bf{.5 \times 1}}{{\bf{0}}^{\bf{4}}}\)

0.0719

-2.632

\({\bf{2}}{\bf{.5 \times 1}}{{\bf{0}}^{\bf{4}}}\)

0.0577

-2.852

\({\bf{3 \times 1}}{{\bf{0}}^{\bf{4}}}\)

0.05174

-2.962

\({\bf{4 \times 1}}{{\bf{0}}^{\bf{4}}}\)

0.0415

-3.18

Now, the slope of the plot is given as

\(\begin{align}Slope &= \frac{{( - 2.412) - ( - 2.302)}}{{(5 \times {{10}^3}) - 0}}\\ &= - 2.20 \times {10^{ - 5}}{s^{ - 1}}\end{align}\)

03

Calculation

The slope of the curve for the variation in the concentration vs time plot is equal to the negative of the rate constant or the reaction.

So,

\(\begin{align}Rate &= - \left( { - 2.20 \times {{10}^{ - 5}}{s^{ - 1}}} \right)\\ &= 2.20 \times {10^{ - 5}}{s^{ - 1}}\end{align}\)

04

Plotting of graph

The plot of ln(P) vs time is a linear plot with a negative slope. This indicates first-order reaction kinetics.

Thus, the reaction is first order, and the rate constant is \({\bf{2}}{\bf{.2 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)

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Most popular questions from this chapter

Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?

\(\begin{array}{c}{\rm{(a) C}}{{\rm{l}}_2}{\rm{ + CO }} \to {\rm{ C}}{{\rm{l}}_2}{\rm{CO}}\\{\rm{rate = }}k{{\rm{(C}}{{\rm{l}}_2}{\rm{)}}^{\frac{3}{2}}}{\rm{(CO)}}\\{\rm{(b) PC}}{{\rm{l}}_3}{\rm{ + C}}{{\rm{l}}_{\rm{2}}}{\rm{ }} \to {\rm{ PC}}{{\rm{l}}_{\rm{5}}}\\{\rm{rate = }}k{\rm{(PC}}{{\rm{l}}_{\rm{3}}}{\rm{) (C}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\\{\rm{(c) 2NO + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\\{\rm{rate = }}k{\rm{(NO)(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\\{\rm{(d) 2NO + }}{{\rm{O}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2N}}{{\rm{O}}_{\rm{2}}}\\{\rm{rate = }}k{{\rm{(NO)}}^{\rm{2}}}{\rm{(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\\{\rm{(e) NO + }}{{\rm{O}}_{\rm{3}}}{\rm{ }} \to {\rm{ N}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{O}}_{\rm{2}}}\\{\rm{rate = }}k{\rm{(NO)(}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\end{array}\)

The rate constant for the radioactive decay of 14C is \({\bf{1}}{\bf{.21 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{ yea}}{{\bf{r}}^{{\bf{ - 1}}}}\). The products of the decay are nitrogen atoms and electrons (beta particles): \(\begin{aligned}{}_{\bf{6}}^{{\bf{14}}}{\bf{C}} \to _{\bf{6}}^{{\bf{14}}}{\bf{N + }}{{\bf{e}}^{\bf{ - }}}\\{\bf{rate = k(}}_{\bf{6}}^{{\bf{14}}}{\bf{C)}}\end{aligned}\).

What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of \({\bf{ 6}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 9 }}}}{\bf{M}}\)?

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.

The rate law for the reaction: \({{\bf{H}}_{\bf{2}}}\) (g) + 2\({\bf{NO}}\) (g) 鉄禱({\bf{N}}2{\bf{O}}\) (g) + \({{\bf{H}}_{\bf{2}}}\)O(g) has been determined to be rate = k(\({\bf{NO}}\))2 (\({{\bf{H}}_{\bf{2}}}\)). What are the orders with respect to each reactant, and what is the overall order of the reaction?

Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:

\(\begin{aligned}{l}_{{\bf{15}}}^{{\bf{32}}}{\bf{P}} \to _{{\bf{16}}}^{{\bf{32}}}{\bf{S + }}{{\bf{e}}^{\bf{ - }}}\\{\bf{rate = 4}}{\bf{.85 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}\,{\bf{da}}{{\bf{y}}^{{\bf{ - 1}}}}{{\bf{(}}^{{\bf{32}}}}{\bf{p)}}\end{aligned}\)

What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of \({\bf{0}}{\bf{.0033 M}}\)?

See all solutions

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