Lecture 1: Thermodynamics and Kinetics (1/4)
Lecture 1: Thermodynamics and Kinetics (1/4) Chem 31B
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This 3 page Class Notes was uploaded by it's lit notes on Monday January 4, 2016. The Class Notes belongs to Chem 31B at Stanford University taught by Jennifer Poehlmann in Fall 2015. Since its upload, it has received 88 views. For similar materials see Principles of Chemistry II in Chemistry at Stanford University.
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Date Created: 01/04/16
Monday January 4, 2016 1 Important Information: PSet #1 due Mon (Jan 11) Lab WriteUp #1 due Fri (Jan 8) Midterm #1 January 20 Lecture 1: Thermodynamics and Kinetics Main Questions of this Course: Thermodynamics: Will we ever get there? Kinetics: ow and when will we get there? Overview ● thermodynamics gives the fundamental laws of nature → it is always right, but it may not be relevant to the question at hand ● if A → B is thermodynamically favored, then A will for B ○ may happen in 30 seconds or 30 million years ● kinetics cannot change thermodynamic outcomes ○ changing the reaction rate or pathways cannot change the thermodynamic product Reaction Rate ● reaction rate tells you how much the reaction has progressed per unit time ○ can measure by looking at how the concentrations of reactants and products change over time (A → B ) ○ reaction rate = change in concentration of A / time or ○ reaction rate = change in concentration of B / time ○ the rate at which A is decreasing is the same rate at which B is increasing Clicker Question #1 Choose the correct statement for the following reaction: A + B → C a.) Reactant A must be consumed at the same rate that product C is formed b.) A is consumed at half the rate of C formation c.) A is consumed at twice the rate of C formation d.) A and B can be consumed at different rates, but the sum of these rates must equal the rate of C formation e.) None of these statements are true solution: the coefficients are all equal, so the rates must be equal Monday January 4, 2016 2 Clicker Question #2 Choose the correct statement for the following reaction: 2A + C → B a.) A and C are consumed at the same rate b.) A is consumed at the same rate that B is formed c.) A is consumed at half the rate that B is formed d.) A is consumed at half the rate that C is consumed e.) A is consumed at twice the rate that B is formed solution: for every B formed, 2A must be consumed so A is being consumed twice as fast Summary of Reaction Rate: ● A + B → C + D ○ reaction rate = (ΔA /Δt) = (ΔB / Δt) = + (ΔC / Δt) = + (ΔD / Δt) ● aA + bB → cC + dD ○ reaction rate = 1/a(ΔA /Δt) = 1/b(Δ B / Δt) = +1/c (ΔC / Δt) =+ 1/d(ΔD / Δt) Average Reaction Rate: ● not at a specific time → between two points in time = − ΔA /Δt (for a reactant ● the reaction rate depends on time and concentration Instantaneous Reaction Rate ● slope of the line at a specific point in time = − ΔA /Δt (for a reactant) Clicker Question #3 PH3→ P4+ H2 If 0.48 moles of PH3are consumed in a 2L container during each second of reaction, what is the H 2 production rate in this experiment? a.) 0.48 M/s b.) 2.88 M/s c.) 0.12 M/s d.) 0.36 M/s e.) 0.08 M/s solution: first balance the equation → 4PH → P + 6H 3 4 2 find molarity of PH3→ .48 mol / 2L = .24M PH 4 rate of consumption of PH = .24 M/s 3 rate of 2formation → rate of PH3= 1/6 (rate of 2→ .24 M/s x 6 = 0.36M/s Monday January 4, 2016 3 Clicker Question #4 H2 2an be used as a disinfectant; it decomposes as: 2HO → 2H O + O 2 2 2 Determine the average rate for H2 2ecomposition during the period 800 to 1200 seconds: TIME (s) HO (M) 2 2 0 2.32 400 1.72 800 1.30 1200 0.98 1600 0.73 a.) 1.2 x 1 M/s 4 b.) 8.2 x 10 M/s c.) 8.0 x 1 M/s d.) 1.6 x 10 M/s e.) 2.4 x 1 M/s solution rate = change in concentration / change in time → (.98 1.30) / (1200 8.0 x 10 M/s What happens to the rate of HO decomposition as [HO ] decreases? 2 2 2 2 a.) The rate stays constant; that is why it can be described by a rate law b.) The rate decreases c.) The rate increases d.) It depends on the initiaO] 2 2 e.) It is impossible to know solution calculate the rate from 1200 to 1600 seconds and see if it is less than or greater t (it .0 x 10 4 is less than 8.0 x , use the same process as the first part of this question)
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