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The hydration of the ethylene oxide in aqueous solution proceeds with first order kinetics such that ethylene glycol is formed. The kinetics of the equation is determined by measuring the change in volume of the liquid system. The change in volume i noted by measurement of height at various instants in a thin capillary tube. From the data, calculate rate of disappearance of ethylene oxide at a concentration of 2.2 gm/lit. Express answer in terms of M/day. (write answer after multiplying by 10)
Time (Sec.) 0 3600 $\infty$
Height (mm) 10 $8 + \frac{10}{e}$ 8

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answer is 12.

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Detailed Solution

$\frac{k \times 3600}{24 \times 3600} = \ln (\frac{18 - 8}{8 + \frac{10}{e} - 8})$

$\Rightarrow k = 24 {day}^{- 1}$

r = k [ethylene oxide]

$= 24 \times \frac{2 .2}{44} = 1 .2 M {day}^{- 1}$

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