Nitro-glycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 °C, and their first-order decomposition was studied. Determine the average rate constants for each experiment using the following data:

Initial (\({{\bf{C}}_{\bf{3}}}{{\bf{H}}_{\bf{5}}}{{\bf{N}}_{\bf{3}}}{{\bf{O}}_{\bf{9}}}\)) (M)

4.88

3.52

2.29

1.81

5.33

4.05

2.95

1.72

t(s)

300

300

300

300

180

180

180

180

% Decomposed

52.0

52.9

53.2

53.9

34.6

35.9

36.0

35.4

Short Answer

Expert verified

The rate constant increases as the time period decrease because the rate constant is inversely proportional to the time period taken by the reaction. The average rate constant of the experimental data is \({\bf{0}}{\bf{.0065 se}}{{\bf{c}}^{{\bf{ - 1}}}}\).

Step by step solution

01

Reaction Rate

The reaction involved the effective collision of two reactants to produce the desired products. Reactions can be natural, which occur in the surrounding environment, whereas it can be artificially done in the laboratory to form the desired product.

The reaction rate can be defined as the reaction speed to produce the products. The reaction rate can be slow, fast or moderate. The reaction can take less than millisecond to produce products, or it can take years to produce the desired product.

The half-life period can be defined as the time period at which half the concentration of the reactants gets converted into a product.

02

Explanation

The half-life period of the first order is:

\({\bf{Half - life period = }}\frac{{{\bf{ln }}\left( {\bf{2}} \right)}}{{\bf{k}}}\)

The rate constant of first-order does not depend upon the concentration of the reactant, but it depends upon the time taken by the reaction.

Taking the first reading of the time = 300 second to calculate the rate constant.

\(\begin{align}k &= {\bf{ }}\frac{{2.303}}{t}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\k &= {\bf{ }}\frac{{2.303}}{{300s}}Log{\bf{ }}\frac{{4.88}}{{0.48}}\\k &= {\bf{ }}\frac{{2.303}}{{300s}}Log{\bf{ }}10.2\\k &= {\bf{ }}\frac{{2.303}}{{300s}} \times 1.0086\\k &= {\bf{ }}0.001\end{align}\)

The rate constant of the reaction is 0.001 sec-1.

Now, take the first reading of the time = 180 second to calculate the rate constant.

\(\begin{align}k &= {\bf{ }}\frac{{2.303}}{t}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\k &= {\bf{ }}\frac{{2.303}}{{180s}}Log{\bf{ }}\frac{{5.33}}{{0.65}}\\k &= {\bf{ }}\frac{{2.303}}{{180s}}Log{\bf{ }}8.2\\k &= {\bf{ }}\frac{{2.303}}{{180s}} \times 0.914\\k &= {\bf{ }}0.012\end{align}\)

The rate constant of the reaction is 0.012 sec-1.

The rate constant increases as the time period decrease because the rate constant is inversely proportional to the time period taken by the reaction.

Therefore, the average rate constant of the reaction is:

\(\begin{align}Average\,Rate\,Cons\tan t &= \frac{{0.001 + 0.012}}{2}\\Average\,Rate\,Cons\tan t &= 0.0065{\sec ^{ - 1}}\end{align}\)

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Most popular questions from this chapter

Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.

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

From the given data, use a graphical method to determine the order and rate constant of the following reaction: 2X⟶Y + Z

Time(s)

5.0

10.0

15.0

20.0

25.0

30.0

35.0.

40.0

(X)(M)

0.0990

0.0497

0.0332

0.0249

0.0200

0.0166

0.0143

0.0125

There are two molecules with the formula\({{\bf{C}}_{\bf{3}}}{{\bf{H}}_{\bf{6}}}\). Propene,\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CH = C}}{{\bf{H}}_{\bf{2}}}\), is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anaesthetic:

When heated to 499\({\bf{^\circ C}}\), cyclopropane rearranges (isomerizes) and forms propene with a rate constant of\({\bf{5}}{\bf{.95 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{{\bf{s}}^{{\bf{ - 1}}}}\). What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499.5\({\bf{^\circ C}}\)?

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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