CHEM 0120 1nd Edition Lecture 4 Outline of Last Lecture I. Dispersion ForcesII. Dipole-Dipole InteractionsIII. Hydrogen Bonding Outline of Current Lecture I. Collision TheoryII. Rate ConstantIII. Arrhenius Collision Theory IV. Similar to Example 13.7V. Rate OrderVI. Initial Rate Method Determines Order of ReactionCurrent Lecture(Continued from January 6th Lecture)B. Increasing Rate of Reaction 3. Surface Area of Reactant 4. TemperatureC. Rate = −1aΔ[ A ]Δt = −1bΔ[B]Δt = 1cΔ[C]Δt = 1dΔ[ D]Δt1. Δ[A] = [A]f – [A]i2. *Always want to convey rate as a positive value*I. Collision Theory – a reaction occurs when a collision occurs at proper orientation and sufficient energy (greater than Ea)II. Rate constant = k = zfpA. z = number of collisions These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.B.f =e−E aRt = fraction of collisions with energy sufficient to overcome activation energy R = 8.31 Jmol∗Kt = temperature (Kelvins)1.  temperature :  exponent :  e :  f :  k2.  stability of activation complex :  Ea (activation energy)III. Arrhenius Collision Theory k = A ×e−E aRtA = zp*Relatively temperature independent*ln k = ln A ×e−E aR(1t)plotting ln k vs. 1t will provide Ea 1t 1−1t 2lnk 2k 1=EaR¿)k1 & k2 = rate constant of respective temperatures t1 &t2 = temperature (Kelvins)IV. Similar to Example 13.7 Temp 1 = 759K k = 1.05× 10−3 Temp 2 = 836K k = 2.14 ×10−2 1759 K−1836 Kln2.14 ×10−21.05× 10−3=Ea8.31 J /mol × K¿)Ea = 2.06 ×105J /molV. aA + bB  cC + dDRate = k [A]m [B]n*m&n are determined experimentally unless in the case of an elementary reaction (1 step to products) when m&n are determined by the stoichiometric coefficients*Rate = k [NO]2[H2]Second order with respect to NOFirst order with respect to H2Third order reaction VI. Initial rate method determines order of reaction (13.54)2NO(g) + O2 (g)  2NO2 (g) Rate = k [NO]m [O2]nTrials [NO]i [O2]i Rate1 0.0125 0.0253 0.02812 0.0250 0.0253 0.1123 0.0125 0.0506 0.0561Rate 2=k 2[NO]m+[O2]nRate 1=k 1[NO]m+[O2]n¿ Always place higher concentration on top∗¿0.112molL× s=(0.0250)m+(0.0253)n0.0281molL× s=(0.0125)m+(0.0253)n 4=2m m=2 for NO0.0125¿¿¿m+(0.0506)n0.0561molL × s=¿¿ 2=2n n=1 for O2NO¿¿O2¿¿Rate=k