Question

The reaction 2ClO2 + 2OH- -> ClO3- + ClO2 was studied with the following results:

[ClO2](M) [OH-](M) Rate (M/s)
0.060 0.030 0.0248
0.020 0.030 0.00276
0.020 0.090 0.00828

a) Determine the rate law for the reaction

b) Calculate the rate constant

c) Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M

Answer 1

Rate of the reaction is given by the following equation:

\[rate = k[A]^m[B]^n\]

where k is the rate constant, m and n are some exponent. and A and B are the initial molar concentrations of the reactants involved.

I see that temperature is not involved, so we can leave that out of our calculations and assume we are speaking of reactants at a constant temperature.

For trials 1 and 2 we have:

\[rate = k[0.060]^m[0.030]^n\] -1
\[rate = k[0.020]^m[0.030]^n\] -2

Divide the first rate by the second rate = 0.0248 /0.00276 = 8.98 = 9. therefore, the rate increased by a factor of 9 when ClO2 was increased.

In dividing this way, you cancel out k and OH-^n since OH remained constant in both trials.

= \[9 = 3^m\]

therefore m = 2

for the trials 2 and 3 we have:

\[rate = k[0.020]^m[0.030]^n\] -2
\[rate = k[0.020]^m[0.090]^n\] -3

Divide the second rate by the first rate = 0.00828/0.00276 = 3

Therefore the rate increased by a factor of 3 as OH- was increased.

therefore we get:

\[3=3^n \] hence n = 1.

therefore our rate law is: \[Rate = k[ClO _{2}]^2[OH-]^1\]

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Answers to b) and c) coming in a second.

Answer 2

In order to find the rate constant, we take our rate law:

\[rate = k[ClO _{2}]^2[OH-]^1\]

=\[\frac{ rate}{[ClO _{2}]^2[OH-]^1 }= k\]

Take any values from the above table, I'll use trial one values:

\[\frac{ 0.0248 }{ 0.060^2 *0.030^1 }\] \[= 229.62 = 2.3*10^2\]

Therefore, our rate constant \[ = 2.3*10^2\]

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Answer 3

In order to find the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M , simply plug in the values :

\[rate = 2.29*10^2[0.100]^2[0.050]\]

\[=0.1145\] = the rate.

I hope this was helpful! c: