Rate data were collected for the following reaction at a...

Rate data were collected for the following reaction at a particular temperature. $2 \mathrm{ClO}_{2}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{ClO}_{3}^{-}(\mathrm{aq})+\mathrm{ClO}_{2}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$ (a) What is the rate-law expression for this reaction? (b) Describe the order of the reaction with respect to each reactant and to the overall order. (c) What is the value, with units, for the specific rate constant?

Rate data were collected for the following reaction at a particular temperature.
$2 \mathrm{ClO}_{2}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{ClO}_{3}^{-}(\mathrm{aq})+\mathrm{ClO}_{2}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$
(a) What is the rate-law expression for this reaction?
(b) Describe the order of the reaction with respect to each reactant and to the overall order. (c) What is the value, with units, for the specific rate constant?

Key Concepts

Rate Law Expression

The rate law expresses the reaction rate as a function of the concentrations of the reactants raised to a power determined by the reaction’s order with respect to each reactant. This formulation usually includes a rate constant (k) and terms that represent each reactant's concentration, which allows one to examine how different concentration changes influence the overall reaction rate.

Reaction Order

Reaction order refers to the exponents applied to the concentration terms in the rate law, indicating how sensitive the rate is to changes in each reactant's concentration. The order with respect to an individual species shows the influence of that species on the rate, while the overall order is the sum of the individual orders. This concept plays a crucial role in understanding the kinetics of a reaction and interpreting experimental data.

Specific Rate Constant

The specific rate constant (k) is a proportionality factor in the rate law that quantifies the speed of the reaction under given conditions. Its value, along with the units, depends on the overall reaction order and is determined experimentally. The units of k must be consistent with the rate expression, ensuring that when reactant concentrations are plugged into the rate law, the computed rate ends up with appropriate units such as mol·L?¹·s?¹.

Transcript

00:02 In this problem, rate data were collected for the following reaction.

00:08 2 moles of clo2 react with 2 moles of oh minus ion to give 1 mole cle3 minus iron and cl2 minus ion and 1 1 -mole h2o liquid.

00:37 Suppose rate expression for this reaction can be expressed as rate equals to rate constant k times concentration of clo2 raised to the power x into concentration of oh minus raised to the power y.

01:18 Consider this as equation 1.

01:25 Because we are describing the same reaction in each experiment, all the experiments are governed by the same rate law expression.

01:39 So, with the given data of experiment 1, rate is equal to 2 .07 into 10.

02:01 The power minus 4 molar per second which is equal to k times.

02:08 Here concentration of clo2 is 0 .012 molar raised to the power x and concentration of oh minus ion is 0 .012 molar raised to the power y.

02:26 Consider this as equation 2.

02:32 Now again substitute the value of rate, concentration of rate.

02:35 Of clo2 and concentration of os minus from experiment 2.

02:45 We get 4 .14 into 10 raised to the power minus 4 molar per second is equal to k into 0 .01 molar raised to the power x into 0 .024 molar raised to the power y.

03:06 Consider this as equation from experiment 4, we have rate equals to 1 .66 into 10 raised to the power minus 3 molar per second, which is equal to k into concentration of co2 is 0 .04 molar raised to the power x into concentration of o is minus 0 .0 to 4 molar raise to the power x into concentration of o is minus 0 .0 to 4 molar raised to the power y.

03:46 Consider this as equation 4.

03:50 Now we have three variables and three equations.

03:56 So using these three equations we can evaluate the value of k, x and y.

04:03 From experiment 1 and 2, we observe that concentration of clo2 is remained constant and concentration of oh minus is doubled and rate of this reaction is also.

04:26 So doubled.

04:33 Divide equation 3 by equation 2.

04:59 We get 2 is equal to 2 to the power y.

05:13 Here the value of k always cancels from such a ratio because it is constant at a particular temperature...

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