Experiment 21 Advance Study Assignment: Rates of Chemical Reactions, II. A Clock

Question

Experiment 21 Advance Study Assignment: Rates of Chemical Reactions, II. A Clock Reaction... Part 2

2. For Reaction Mixture 1 the student found that 85 seconds were required. On dividing Equation 5 for Reaction Mixture 1 by Equation 5 for Reaction Mixture 2, and after canceling out the common terms (k, terms in (BrO,) and (H*)), she got the following equation: 11.8 (0.0020 (1 Recognizing that 11.8/22 is about equal to h, she obtained an approximate value for m. What was that value? By taking logarithms of both sides of the equation, she got an exact value for m. What was that value? Since orders of reactions are often integers, she reported her approximate value as the order of the reac- tion with respect to I

Explanation / Answer

Ans. #I. Approximation method:

            (11.8 / 22) = (½)m

            Or, ½ = (½)m

            Hence, m = 1

#II. Logarithmic method-

            Log (11.8 / 22) = m log (1/2)

            Or, log 0.53636 = m log 0.5

            Or, (-0.27054) = m (-0.30103)

            Or, m = (-0.27054) / (-0.30103)

            Hence, m = 0.899

So, the exact value of m = 0.899

* The integer (whole number) nearest to 0.899 is 1.

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