Decomposition of Sucrose: Understanding the Rate and Order of Reaction

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SUMMARY

The decomposition of sucrose (C12H22O11) is identified as a first-order reaction based on the linearity of the ln[C12H22O11] versus time plot. The rate constant (k) is determined to be -0.45 hr-1, which is the negative slope of the plot. Confusion arises regarding the definition of "rate of reaction," which can be expressed as either d[sucrose]/dt or -d[sucrose]/dt. The discussion suggests that the exponent in the rate law may be a misprint, reinforcing the first-order classification.

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Homework Statement
In an acid solution, sucrose (C12H22O11) will decompose into fructose and glucose. A plot of ln[C12H22O11] versus time gives a straight line with a slope of -0.45 hr^(-1). What is the rate law for this decomposition reaction?
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I think this is a first order reaction because ln[C12H22O11] vs. time is linear. The k value is the negative of the slope. Therefore, my answer is rate = 0.45 hr-1 [C12H22O11].

The correct solution is rate = -0.45 hr-1 [C12H22O11]0. I don't understand why this is a zero order reaction, or why the k value is -0.45. Is the solution wrong? Thanks.
 
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It depends how you define "rate of reaction". The rate law is
d[sucrose]/dt = -k[sucrose]
Then it depends whether you define "rate" as d[sucrose]/dt or -d[sucrose]/dt
Note that the question as you quote it asks for the rate law, not the rate. Is it you that's defining "rate"?
You are right that a linear plot of lnC vs. t indicates a first order reaction. I suspect that the exponent 0 is a misprint.
 
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mjc123 said:
You are right that a linear plot of lnC vs. t indicates a first order reaction. I suspect that the exponent 0 is a misprint.
Or the lnC vs t should be just C vs t.
 
Possibly, but that wouldn't give a slope with units hr-1.
 

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