Why is sucrose decomposition zero order if ln[C12H22O11] vs time is linear?

[i_love_science]

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?

Relevant Equations

rate law

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.

 

[mjc123]

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.

 

[chemisttree]

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.

 

[mjc123]

Possibly, but that wouldn't give a slope with units hr^-1.