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Chapter 12: Q31 E (page 705)

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.

Short Answer

Expert verified

The rate equation is\({\bf{k = (}}{{\bf{I}}^{\bf{ - }}}{\bf{)(OC}}{{\bf{l}}^{\bf{ - }}}{\bf{)}}\), and the rate constant is \({\bf{6}}{\bf{.1}} \times {\bf{1}}{{\bf{0}}^{{\bf{ - 2}}}}{\bf{Lmo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\)

Step by step solution

01

Rate equation

Let us consider the rate equation as =\({\bf{k = (}}{{\bf{I}}^{\bf{ - }}}{{\bf{)}}^{\bf{m}}}{{\bf{(OC}}{{\bf{l}}^{\bf{ - }}}{\bf{)}}^{\bf{n}}}\)

Data points 1, and 2 show that the rate of reaction doubles as the concentration of (I-) doubles.It can be considered first order with respect to \({\bf{(}}{{\bf{I}}^{\bf{ - }}}{\bf{)}}\). So, m=1

From data points 1 and 3, we get two equations.

\(\begin{align}{l}3.05*{10^{ - 4}} = k{(0.10)^1}{(0.05)^n}\\1.83*{10^{ - 4}} = k{(0.30)^1}{(0.01)^n}\end{align}\)

Dividing the equations, we get,

\(\begin{align}{}3.05*{10^{ - 4}} = k{(0.10)^1}{(0.05)^n}\\1.83*{10^{ - 4}} = k{(0.30)^1}{(0.01)^n}\\\frac{{3.05*{{10}^{ - 4}}}}{{1.83*{{10}^{ - 4}}}} = {\left( {\frac{5}{3}} \right)^n}\\\frac{5}{3} = {\left( {\frac{5}{3}} \right)^n}\\n = 1\end{align}\)

Therefore, the reaction is also first order with respect to \((OC{l^ - })\)

Overall, the rate equation is \({\bf{k = (}}{{\bf{I}}^{\bf{ - }}}{\bf{)(OC}}{{\bf{l}}^{\bf{ - }}}{\bf{)}}\)

02

Calculation of Rate constant

From the rate equation,

\({\bf{k = (}}{{\bf{I}}^{\bf{ - }}}{\bf{)(OC}}{{\bf{l}}^{\bf{ - }}}{\bf{)}}\)

\(\begin{align}{l}k = \frac{{rate}}{{({I^ - })(OC{l^ - })}}\\k = \frac{{3.05*{{10}^{ - 4}}}}{{(0.10)(0.05)}}\\k = 6.1*{10^{ - 2}}Lmo{l^{ - 1}}{s^{ - 1}}\end{align}\)

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Most popular questions from this chapter

In the PhET Reactions & Rates (http://openstaxcollege.org/l/16PHETreaction) interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select 鈥淪how Bonds鈥� under Options.

  1. Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?
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The rate constant for the rate of decomposition of \({{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{5}}}\)to\({\bf{NO}}\) and \({{\bf{O}}_{\bf{2}}}\)in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K:

\({\bf{2}}{{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{5}}}{\bf{(g) - - - 4NO(g) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition.

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\({\bf{2NOCl(g)}} \to {\bf{2NO(g) + C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}}\)

Determine the rate law, the rate constant, and the overall order for this reaction from the following data:

The rate constant for the first-order decomposition at 45 掳C of dinitrogen pentoxide, N2O5, dissolved in chloroform, \({\bf{CHC}}{{\bf{l}}_3}\), is 6.2 脳 10鈭�4 min鈭�1.

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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): \(\begin{aligned}{}_{\bf{6}}^{{\bf{14}}}{\bf{C}} \to _{\bf{6}}^{{\bf{14}}}{\bf{N + }}{{\bf{e}}^{\bf{ - }}}\\{\bf{rate = k(}}_{\bf{6}}^{{\bf{14}}}{\bf{C)}}\end{aligned}\).

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