Question

The following data were obtained in an experiment on inversion of cane sugar (a first order kinetics)

Time (min) 0 10

After a long time Total angle of rotation (degree) +40+15– 10

The rate constant (in second–1) is [ln 2 = 0.693]

• A. 0.0693
• B. 1.15 × 10–3
• C. 0.693
• D. 1.15× 10–2

Answer 1

Answer: (B)

1.15 × 10–3

Answer 2

$K = \frac{1}{10 \times 60}In\frac{r_{\infty} - r_{0}}{r_{\infty} - r_{t}} = \frac{1}{10 \times 60}In\frac{- 10 - 40}{10 - 15} = \frac{1}{10 \times 60} \times In2$

$= \frac{0.0693}{60} = 1.15 \times 10^{- 3}\sec^{- 1}{}$