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Zero, First & Second Rate orders

Zero, First & Second Rate orders. Chapter 14 Part III. 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. Integrated Rate Law. aA  Products. Rate= -∆[A]/∆t= k[A] m.

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Zero, First & Second Rate orders

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  1. Zero, First & Second Rate orders Chapter 14 Part III

  2. 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 Integrated Rate Law aA  Products Rate= -∆[A]/∆t= k[A]m

  3. 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

  4. 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.

  5. 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

  6. 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)

  7. First Order

  8. 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

  9. 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.

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

  11. 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

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

  13. 2nd Order Data

  14. 2nd order reaction using integrated law for first order

  15. 2nd order reaction and 2nd order integrated rate law

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