Problem

Problem

The following data were obtained in an experiment on inversion of cane sugar (a first order kinetics)
Time (min)010after a long time
Total angle of rotation (degree)+40+15-10
The rate constant (in second-1) is? (a) 0.0693 (b) 1.15×10−31.15 \times 10^{-3} (c) 0.693 (d) 1.15×10−21.15 \times 10^{-2}
Intuitive explanation

This problem uses optical rotation as a proxy for concentration in a first-order reaction. The key insight is that the angle of rotation is proportional to the amount of unreacted sugar, so you can treat the rotation values like concentration data. Convert the rotation measurements into a form of "remaining reactant," then apply the first-order rate law ln⁡([A]t)−ln⁡([A]0)=−kt\ln([A]_t) - \ln([A]_0) = -kt to find kk—just remember to convert minutes to seconds for the final answer.

Intuitive explanation

This problem uses optical rotation as a proxy for concentration in a first-order reaction. The key insight is that the angle of rotation is proportional to the amount of unreacted sugar, so you can treat the rotation values like concentration data. Convert the rotation measurements into a form of "remaining reactant," then apply the first-order rate law ln⁡([A]t)−ln⁡([A]0)=−kt\ln([A]_t) - \ln([A]_0) = -kt to find kk—just remember to convert minutes to seconds for the final answer.

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