Post-Laboratory Questions 1. spaces provided for the \"nswers and additional pap
ID: 484023 • Letter: P
Question
Post-Laboratory Questions 1. spaces provided for the "nswers and additional paper if necessary. solution, When you calculated k solution and Hrcos you in the rate reaction of KMno. 1, 2, and 3. (a) assumed k had the same value determinations Hzczo, solution in What assumption under the conditions of solution and those did you make about the reaction of KMno. eterminations allowed you to consider k to be a constant? sound practice to compare the results of determinations 4-6 when calculating k? (b) Briefly explain. your data substantiate the rule of thumb regarding the effect on the reaction rate of a 10-degree increase in reaction temperature? Briefly explain. 3. Consider the reaction that occurs when a Clo2 solution and a solution containing hydroxide ions (OH are mixed at 0 °C shown in Equation 18. 2ClOh (aq) 2 OH (aq) Clos (aq) Cloz (aq)+ H2O() (Eq. 18)Explanation / Answer
1(a)
The assumption is that the temperature of the systen remains same throughout the reaction.
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3. (a)
Rate = k [ClO2]^m[OH-]^n
(rate)1/(rate)2 = k [1.25 *10^-2]^m[1.30 *10^-3]^n/k[2.50 *10^-2]^m[1.30*10^-3]^n
or, 2.33*10^-4/9.34*10^-4 = k [1.25 *10^-2]^m[1.30 *10^-3]^n/k[2.50 *10^-2]^m[1.30*10^-3]^n
Cancel the similar terms.
(1/4) = (1/2) ^m
or, (1/2)^2 = (1/2)^m
m = 2
Rate is second order with respect to concentration of ClO2.
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(rate)2/(rate)3 = k [1.25 *10^-2]^m[1.30 *10^-3]^n/k[1.25 *10^-2]^m[2.6*10^-3]^n
or, 9.34*10^-4/1.87*10^-3 = (1/2) ^n
or, (1/2)= (1/2)^n
n = 1
The rate is first order with respect to concentration of OH-.
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(b) For determination 1
2.33 *10^-4 = k [1.25 *10^-2]^2[1.30*10^-3]
or, k = 14.34 L/mol
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(c)
rate = 14.34 L/mol *(8.25*10^-3 mol/L)^2(5.35*10^-2mol/L) = 5.22*10^-5 mol^2/L^2