For the reaction A--->2B+C, the following data are obtained for [A] as a function of time:?

t=0 min, [A]=.80 M; 8 min, [A]=.60 M; 24 min, [A]=.35 M; 40 min, [A]=.20 M.





a) Establish the order of the reaction.


b) What is the value of the rate constant, k?


c) Calculate the rate of formation of B at 22 min.





Any help would be greatly appreciated, thank you.

For the reaction A---%26gt;2B+C, the following data are obtained for [A] as a function of time:?
(a)


Assume a simple power law kinetic:


r(A) = d[A]/dt = - k·[A]ⁿ





The corresponding integrated forms of the rate late law are:


ln( [A]₀/[A]) = k·t for n =1


[A]₀⁽¹ ⁻ⁿ⁾- [A]⁽¹ ⁻ⁿ⁾= (1-n)·k·t for n≠1





Standard procedure would be to plot [A]⁽¹ ⁻ⁿ⁾ (or ln[A] ) versus t for various n. The graph which yields a straight line is the one with correct reaction order n.





But plotting is a bit difficult here. So assume n and calculate calculate rate constant for each time. If k is (almost) constant assumed n was correct. If k tend to increase or decrease with time assumed n was incorrect





Assume n=1


k = ln( [A]₀/[A]) / t





t = 8min


k = ln( 0.8M / 0.6M) / 8min = 0.3596min⁻¹


t = 24min


k = ln( 0.8M / 0.35M) / 24min = 0.3444min⁻¹


t = 40min


k = ln( 0.8M / 0.2M) / 40min = 0.3466min⁻¹





That looks good. So reaction is first order:


r(A) = d[A]/dt = - k·[A]








(b)


You could simply take the average of the values calculated in (a). To minimize deviation from data points by linear regression.


Fitting


ln( [A]₀/[A]) = k·t


to the data points leads to


k = ( Σ ln([A]₀/[A])·t ) / ( Σ t² )


= (ln(0.8/0.6)·8min + (0.8/0.35)·24min + ln(0.8/0.2)·40min ) / ((8min)² + (24min)² + (40min)² )


= 0.03464min⁻¹








(c)


ln( [A]₀/[A]) = k·t


%26lt;=%26gt;


[A] = [A]₀· exp(-k·t)





Hence rate of reaction of A is


r(A) = -k·[A] = - k · [A]₀· exp(-k·t)





According to reaction equation 2moles of B are formed per mole A consumed.


Hence


d[B] = -2·d[A]


%26lt;=%26gt;


r(B) = d[B]/dt = -2·d[A]/dt = -2·r(A)


=%26gt;


r(B) = 2 · k · [A]₀· exp(-k·t)





at t=22min


r(B) = 2 · 0.03464min⁻¹ · 0.8M · exp(-0.03464min⁻¹


·22min)


= 0.02586 M/min = 0.02586 mol/Lmin