Compound A reacts with compound B according to the equation A + B C + D. In an e

Question

Compound A reacts with compound B according to the equation A + B C + D. In an experiment, 0.500 M compound A was combined with a large excess of compound B and allowed to react. The concentration of compound A was measured at 0.10-minute time intervals during the reaction and the data are summarized below. (Here, [At] is the concentration of A at a particular time.)

Using these data, determine the instantaneous rate of change (disappearance) of compound A at 0.23 minutes of reaction. M·min1

Time (min) [At] (mol/L) 0.00 0.500 0.10 0.396 0.20 0.314 0.30 0.249 0.40 0.198 0.50 0.157 0.60 0.124 0.70 0.099 0.80 0.078 0.90 0.062 1.00 0.049

Explanation / Answer

the rate of elementary reaction is

-rA= K[A] [B]

when large excess of B is taken, the reaction boils down to

-rA= K'[A], where K'= K[B]

-dCA/dt= K'[A]

when integrated noting that at t=0 [A]= [A]0 and t=t A=[A]

ln[A]= ln[A]0-K't

so a plot of ln A vs t gives a straight line whose slope is -K'

the plot of lnA vst is shown below

the equation of best fit is

lnA= 2.3202*t-0.6933

A= 0.5exp(-2.3202t)

dA/dt= 0.5*exp(-2.3202t)*(-2.3202)

at t= 0.23 minutes

dA/dt= 0.5*exp(-2.3202*0.23) (-2.3202)=-0.68 M/min

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