Chemistry rate law & rate constant question? Help?!?
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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
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Answer:
a) First compare reaction 2 with reaction 1. The [ClO2] is tripled, and the rate goes up by 0.0248/0.00276 = 9-fold. Therefore, the reaction is second order with respect to [ClO2]. Next compare reaction 2 with reaction 3. The [OH-] is tripled, and the rate goes up by 0.00828 / 0.00276 = 3, so the reaction is first order with respect to [OH-]. So, the rate law is: rate = k[ClO2]^2[OH-] b) To calculate the value of the rate constant, just pick any reaction, plug the rate and the concentrations into the rate law and solve for k. c) Once you have the value of k, you can plug that and the given concentrations into the rate law and calculate the rate.
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Other answers
Tripling the ClO2 concentration from 0.020 to 0.060 causes the rate to increase by 0.0248/0.00276 = 9.0 times, while holding [OH-] constant. R is proportional to [ClO2]^n 9R is propotional to [3ClO2]^n, so n = 2 because 3^2 = 9. The rate is second order with respect to [ClO2] Tripling OH- from 0.030 to 0.090 causes the rate to increase by 0.00828/0.00276 = 3, while holding [ClO2] constant. R is proportional to [OH-]^m 3R is proportional to [3OH-]^m, so m = 1 because 3^1 = 3. The rate is first order with respect to [OH-] . The rate law is R = k[ClO2]^2[OH-]^1 Using trial 1 data: k = R / {[ClO2]^2[OH-]} = 0.0248 / {(0.060)^2(0.030)^1} = 230 1/M^2 R = 230 1/s M^2 x (0.100)^2 x (0.030)^1 = 0.069 M/s
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