Chemistry Notecard - Chapter 15 / Rate of rxn= [ ] / t /...

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Chapter 15 // Rate of rxn=∆ [ ] / ∆ t // Use the rxn to find gen rate law A →B+C {-∆[A]/∆t→ ∆[B]/∆t + ∆[C]/∆t } Use fracts for stoich coeffs // Rxn rate = k[A]^a[B]^b// To find the powers that the rate equation is raised to we look at how the ∆[ ] affects the rate. Divide the changed [ ] by the original, and the changed rate by the original and then divide the ∆rate/∆concentration to get the power // Order of rxn is exps on rxn rate (←) added togeth// 1 st order rxns ln*([R]t/[R]0)=-kt ln is the fract at what is left at time, t, vs orig and the right side is the rate constant and elapsed time // 2 nd Order rxns (1/[R]t)-(1/[R]0)=kt) // 0 Order ([R]t)-([R]0)= kt // ½ life: t1/2= (0.693/k) // Arrhenius eq lnk=lnA+ (-Ea/RT) // To find Ea ln(k2/k1)=- (Ea/R)*[(1/t2)-(1/t1)] // Slow steps are rate determining.// Rearrange equations into the y=mx+b format and graph [ ] vs t –slope for 1 st order equations and 2 nd order equations have a pos slope. // Units of k 1 st order: time^-1, 2 nd order: [L/(mol*time)] // Chapter 16 // A+B →2C so, K=[C]²/ ([A][B]) Use ICE charts to help figure out these problems* solids do not factor in! Quad eq may be needed. The same equation can be used for pressures, Kp. // Q is used with the same equation when not at equilibrium. {Q>K Reac→Prod, Q=K equilibrium, Q<K Reac←Prod} // Chapter 17 // Kw=[H3O+][OH-]=1.0x10^-14 at 25C // Neutral [H3O+]=[OH-], Acidic [H3O+]>1.0x10^-7>[OH-], Basic [H3O+]<1.0x10^-7<[OH-] // pH=- log[H3O+], pOH=-log[OH-], at 25C pH+pOH=14 // Use pH to find out conc [H3O]=10^-pH // Ka=[H3O+][A-]/[HA], Kb= [BH+][OH-]/[B]// SAs
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This note was uploaded on 06/08/2008 for the course CH 223 taught by Professor Carney during the Spring '08 term at Linn Benton Community College.

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Chemistry Notecard - Chapter 15 / Rate of rxn= [ ] / t /...

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