The following data were obtained in an experimental study for hydrocarbon biodeg

Question

The following data were obtained in an experimental study for hydrocarbon biodegradation in oily wastewater. Time (min Oil (mg/L) 100 200 30 40 50 2980.96 1510.20 765.09 387.61 196.37 99.48 ) What order is this reaction? ) Estimate the reaction constant. Why might your estimation be inaccurate? uestion, 4 points] Sometimes oil degradation may be interpreted as second order. Given that the rate of a second- order reaction is proportional to the square of the concentration of a given reactant. How can we determine whether the biodegradation in this study is a second order or not? [You need to derive the equation for the concentration of a reactant for a second-order reaction]

Explanation / Answer

the method adopted is based on integral method of analysis where the relation between rate of degradation (-dO/dt) and order (n) are established as -d[O]/dt= K[O]n, K is rate constant.

In the integral method of analysis, order is assumed and the integrated equation is verified for its validity,

to begin with let n=0, for zero order, -d[O]/dt= K when integrated [O]= [O]0-Kt, where [O]o is concentration of oil at zero time, [O] is concentration of oil at any time

so a plot of [O] vs t if gives straight line suggests that the order is 1st order or else different order willl have to be assumed. this is done 1st.

since the plot is not straight line, zeor order is discarded.

nezxt assume n=1, hence -dO/dt= K[O] when integrated ln[O]= ln[O]0-Kt, so a plot of ln[O] vs t gives straight line whose slope is -K and intercept is ln[O]0. so this is vefified now by creating the plot of ln[O] vs time

the plot is straight line suggesting the oil degradation is 1st order. The slope from the equation of best fit is

K=0.068/min

the rate of oil degradation for second order can be written as

-dO/dt= K[O]2, where [O] is oil concentration and K is rate constant

-d[O]/[O2]2= Kdt

when the above equation is integrated noting that at t=0 [O]= [O]0 and t=t [O]= [O]

1/[O] = 1/[O]O + Kt

so a plot of 1/[O] vs t is straight line whose slope is K and intercept is 1/[O]o if the degradation is second order.

So a plot of 1/[O] vs time is prepared and is shown below. if this is straight line, the reaction order can be considered as second order.

the plot is not straight line suggesting that the degradation is not second order.

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