Zero, First & Second Rate orders Chapter 14 Part III

Integrated Rate Law aA  Products Rate= -∆[A]/∆t= k[A]m For any reaction: Every reaction has a rate law of this form. We will develop the integrated rate law for n=1: first order n=2: second order n=0 : zero order aA  Products Rate= -∆[A]/∆t= k[A]m

First Order Rate Laws 2N2O5(soln) -> 4NO2(soln) + O2(g) Rate =(- ∆[N2O5]/∆t) = k[N2O5] As it is first order, this means that if the [N2O5] doubles the production of also NO2 & O2 doubles. By integrating the formula (calculus) ln [N2O5] = -kt + [N2O5]t=0 This is still the integrated rate law with [Reactant] as a function of time

1st order integrated rate law: Rate as function of time For any reaction aA-> products Rate =(- ∆[A]/∆t) = k[A] ln [A] = -kt + [A]t=0 The equation depends on time. If [A] initial is known and k is known, the [A] may be calculated at t.

ln [A] = -kt + [A]t=0 The equation is in the form y = mx +b. A plot of x (time) versus y (ln[A]) will yield a straight line. Therefore another way to determine rate order, if the plot of ln[A] versus time gives a straight line, then it is first order. If it is not a straight line, it is not first order. This rate law may also be expressed: ln ([A]t=0/[A])=kt

2N2O5(g) -> 4NO2(g) + O2(g) The decomposition of this gas was studied at a constant volume to collect these results. Verify the rate order and find the rate constant for rate =(- ∆[N2O5]/∆t)

First Order

Calculate k Slope = ∆y/∆x = ∆(ln[N2O5])/∆t Using first and last points Slope = -5.075-(-2.303)/(400s-0s) = -2.772/400s k = -6.93 x 10-3 s-1

Half life of a first order reaction The time required to reach one half the original concentration. Designated by t 1/2 We can see the half life from the data is 100 sec. Note is always takes 100 seconds.

Butadiene forms its dimer 2C4H6 (g)  C8H12 (g) Second Order Rate Laws Butadiene forms its dimer 2C4H6 (g)  C8H12 (g)

aA-> products If the reaction is first order: ln([A]0/[A])=kt By definition, when t= t 1/2 then, [A] = [A]0/2 ln([A]0/[A]0/2) = kt 1/2 ln(2)=kt 1/2 t 1/2= 0.693/k

Second Order Rate Laws Butadiene forms its dimer 2C4H6 (g) - > C8H12 (g)

2nd Order Data [C4H6] (mol/L) Time (sec) 0.01000 0.00625 1000 0.00476 0.00625 1000 0.00476 1800 0.00370 2800 0.00313 3600 0.00270 4400 0.00241 5200 0.00208 6200

Time (sec) [C4H6] (mol/L) ln[C4H6] 1/[C4H6] 0.01000 -4.60517 100.0 1000 0.00625 -5.07517 160.0 1800 0.00476 -5.34751 210.1 2800 0.00370 -5.59942 270.3 3600 0.00313 -5.76672 319.5 4400 0.00270 -5.9145 370.4 5200 0.00241 -6.02813 414.9 6200 0.00208 -6.17539 480.8

2nd order reaction using integrated law for first order

2nd order reaction and 2nd order integrated rate law