Question on Half-Life PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon. It is thermally unstable and rearranges to propene at 1000 0C via the following first-order reaction: CH2H2CCH2(g)!H3CCHCH2(g)The rate constant is found to be 9.2 s-1(a)What is the half-life of the reaction? (b)How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? An Overview of Zero-Order, First-Order, and Simple Second-Order ReactionsZero Order First Order Second Order Half-life (t1/2) = Rate law rate = krate = k [A] rate = kUnits for kmol/L*s 1/s L/mol*s Reactions with More than one ReactantUse the half-life equation, t1/2= 0.693 k to solve One-quarter of the initial value means two half-lives have passed. 2 t1/2= 2(0.075s) = 0.150s (b) t1/2= 0.693/9.2 s-1= 0.075 s (a) Plot for straight line Slope, y-intercept Integrated rate law in straight-line form [A]t= kt + [A]0ln[A]t= -kt + ln[A]01/[A]t= kt + 1/[A]0[A]tvs. t ln[A]tvs. t 1/[A]t= t k, [A]0-k, ln[A]0k, 1/[A]0[A]0/2k(ln 2)/k1/(k[A]0) Key assumption so far: forward reactions only This assumption is generally good at short timesHow to solve:

[A]2

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2/12/12 15 Reaction Kinetics – Why Study ? Reaction Mechanism

2/12/12 16 NO2 (g)+ CO(g)àฏNO(g)+ CO2(g)If this reaction proceeds in one step as written, the rate law should be, Rate = k[NO2] [CO]However kinetics expts show: Rate = k[NO2]2The rate of this reaction is independent of [CO] and 2nd-order with respect to [NO2] The reaction therefore does not proceed in one Elementary Step as written above An Elementary Step is a reaction whose rate law can be written from its molecularity

2/12/12 17 Reaction Mechanisms: Series of Elementary Steps Stop Feb. 14

2/12/12 Reaction Mechanisms A reaction may actually consist of several consecutive steps A Rate Law can be written for each step if we know the molecules that collide in that step. In general – steps will proceed at different rates Rate of an overall reaction will be rate of its slowest steprate limiting steprate-determining step When we experimentally determine the rate law of the overall reaction, we are determining the rate law of its slowest step.From that rate law, we an write the slowest step and propose a mechanism for the overall reaction.Example F2+ 2 ClO2àฏ2 FClORate = k (F2) (ClO2) Experimentally determined rate law What is the reaction mechanism:Certainly not: F2+ ClO2+ ClO2•this is an unlikely termolecular Rx. •not consistent with the rate law Possibly: F2+ ClO2––––>F2ClO2 intermediatF2ClO2+ ClO2––––>2 FClO• The two equations add up to net equation • Slow step is consistent with rate law • Possible experimental support of mechanism: Spectroscopic observation (detection) of F2ClO2 slow fast Does not appear in net equation Can never absolutely prove a mechanism; but can disprove one

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2/12/12 19 Another Example 2 NO + 2 H2–––> N2+ 2 H2O Rate = k [NO]2[H2] Possible Mech.: NO + NO + H2––––>N2O + H2ON2O + H2––––>N2+ H2O slow fast However, termolecular 1ststep Consider another more likely mechanism Consider another more likely mechanism for 2 NO + 2 H2–––> N2+ 2 H2O