Exercises Chem Eqm
... 7.1(a) K = 2.85 x 10-6; (b) ∆rGo = +240 kJ mol-1; (c) ∆rG = 0 7.4(a) Mole fractions A: 0.087, B: 0.370, C: 0.196, D: 0.348, Total: 1.001; (b) Kx – 0.33; (c) p = 0.33; (d) ∆rGo = + 2.8 x 103 J mol-1. 7.6(a) ∆rHo = +2.77 kJ mol-1, ∆rSo = -16.5 J K-1 mol-1 7.9(a) χB = 0.904, χI = 0.096 7.11(a) ∆rGo = – ...
... 7.1(a) K = 2.85 x 10-6; (b) ∆rGo = +240 kJ mol-1; (c) ∆rG = 0 7.4(a) Mole fractions A: 0.087, B: 0.370, C: 0.196, D: 0.348, Total: 1.001; (b) Kx – 0.33; (c) p = 0.33; (d) ∆rGo = + 2.8 x 103 J mol-1. 7.6(a) ∆rHo = +2.77 kJ mol-1, ∆rSo = -16.5 J K-1 mol-1 7.9(a) χB = 0.904, χI = 0.096 7.11(a) ∆rGo = – ...
ABCT2772
... energy function and equilibrium constants, phase rule, ClausiusClapeyron equation. Chemical Kinetics Rate equations and rate constants, reaction mechanism and elementary reactions. Common reaction types: opposing reactions, consecutive reactions, parallel reactions, chain reactions. Reaction rate th ...
... energy function and equilibrium constants, phase rule, ClausiusClapeyron equation. Chemical Kinetics Rate equations and rate constants, reaction mechanism and elementary reactions. Common reaction types: opposing reactions, consecutive reactions, parallel reactions, chain reactions. Reaction rate th ...
PP - Columbia University
... Now, Vmax = k3Eo So: k3 = Vmax/Eo = the maximum (dP/dt)/Eo, = the maximum (-dS/dt)/Eo k3 = the TURNOVER NUMBER • the maximum number of moles of substrate converted to product per mole of enzyme per second; • the maximum number of molecules of substrate converted to product per molecule of enzyme per ...
... Now, Vmax = k3Eo So: k3 = Vmax/Eo = the maximum (dP/dt)/Eo, = the maximum (-dS/dt)/Eo k3 = the TURNOVER NUMBER • the maximum number of moles of substrate converted to product per mole of enzyme per second; • the maximum number of molecules of substrate converted to product per molecule of enzyme per ...
the original file
... study list for exam 1 1. how to draw resonance structures 2. meaning of conjugated vs isolated pi bonds 3. what an orbital is 4. be able to draw MO diagrams for allyl radical and cation and benzene, such as the one in Fig. 10.2 but you dont need to know how the MOs look, just the relative energy lev ...
... study list for exam 1 1. how to draw resonance structures 2. meaning of conjugated vs isolated pi bonds 3. what an orbital is 4. be able to draw MO diagrams for allyl radical and cation and benzene, such as the one in Fig. 10.2 but you dont need to know how the MOs look, just the relative energy lev ...
Hydrogen Production by Splitting Water in an Electrolyzer
... the knowledge of the mechanism is incomplete or the mechanism involve large no. of intermediate steps. They also include reactions whose mechanisms are too complicated to yield a rate law; those whose rate laws are too cumbersome to be tested experimentally reactions which never attain steadystate u ...
... the knowledge of the mechanism is incomplete or the mechanism involve large no. of intermediate steps. They also include reactions whose mechanisms are too complicated to yield a rate law; those whose rate laws are too cumbersome to be tested experimentally reactions which never attain steadystate u ...
Math 31 Ch. 3 Review notes
... Profit will be maximized when first derivative = 0 means horizontal line, slope of 0 (maximum on a parabola) Related Rates (3.5) We are given the rate of change of one quantity and we are asked to find the rate of change of a related quantity. We find an equation that relates the 2 (or more) quantit ...
... Profit will be maximized when first derivative = 0 means horizontal line, slope of 0 (maximum on a parabola) Related Rates (3.5) We are given the rate of change of one quantity and we are asked to find the rate of change of a related quantity. We find an equation that relates the 2 (or more) quantit ...
REACTION DYNAMICS
... thermal rate constant k for a reaction at a specific temperature obtained from the reaction cross section? ...
... thermal rate constant k for a reaction at a specific temperature obtained from the reaction cross section? ...
ws-8-14-2
... f. t½ = .693/k k. rate = k[A]2 ; 1/[A] = kt +1/[A]0 g. t½ = [A]0/2k m. rate = k[A] ; ln[A] = -kt + ln[A]0 h. t½ = 1/(k[A]0) _____ 4. (T/F) For the reaction aA bB, the rate remains constant over time. Reactant A is therefore a first order reactant. _____ 5. (T/F) Zero order reactions often have the ...
... f. t½ = .693/k k. rate = k[A]2 ; 1/[A] = kt +1/[A]0 g. t½ = [A]0/2k m. rate = k[A] ; ln[A] = -kt + ln[A]0 h. t½ = 1/(k[A]0) _____ 4. (T/F) For the reaction aA bB, the rate remains constant over time. Reactant A is therefore a first order reactant. _____ 5. (T/F) Zero order reactions often have the ...
Oxidation and Reduction - UCLA Chemistry and Biochemistry
... bonds between a carbon and atoms that are less electronegative than carbon (often hydrogen). ...
... bonds between a carbon and atoms that are less electronegative than carbon (often hydrogen). ...
KEY Final Exam Review - Iowa State University
... k=(0.2130)M/s/(0.250M)(0.250M)=3.41M-1s-1 could use any of the five to calculate this. kave=3.408M-1s-1 d. What is the rate when [BF3]=0.100M and [NH3]=0.500M? rate=3.408M-1s1*(0.100M)*(0.500M)=0.170M/s 2a. Write the rate law for a reaction between A, B, and C that is the first order in A, zero orde ...
... k=(0.2130)M/s/(0.250M)(0.250M)=3.41M-1s-1 could use any of the five to calculate this. kave=3.408M-1s-1 d. What is the rate when [BF3]=0.100M and [NH3]=0.500M? rate=3.408M-1s1*(0.100M)*(0.500M)=0.170M/s 2a. Write the rate law for a reaction between A, B, and C that is the first order in A, zero orde ...