Study Guide for CHM 116 Exam I
1) Stoichiometry of Reactions
a) Cancel out units until you get to the desired ending i) Can only cancel by dividing
b) Balancing equations
i) Start with things that do not appear often in the equation c) When doing stoichiometry of a carbohydrate reaction balance Carbon first, Hydrogen, then Oxygen
a) The experimental way- measure the time dependent concentration and fit the data to determine the reaction rate b) Rate laws predict the entire reaction, so when the rate increases because concentration increased of one of the reactants in the rate law then the entire rate has sped up. c) For a reaction A B
i) Rate1 = change in concentration / change in time ii) Rate 2 = change in concentration / change in time iii)-change in [A] / change in time = change in [B] / change in time
d) For a reaction 3A + 2B 4C
i) -change in[A] / change in time = -3change in [A] / 2 change in time = 3 change in [C] / 4 change in time e) Reaction rates depend totally on concentration We also discuss several other topics like What is the content of minersville (p.a.) vs. gobitis?
f) Correct units for the rate constant k, mol^-1 s^-1 g) Reaction orders
i) On this exam we will only have orders 0-2.
ii) The rate of a zero order reaction is independent of concentration of reactants or products; all other rates are dependent on each of these.
(1) For zero order: Rate=-kt + [A]
(2) For 1st order: ln(k) = -kt + ln[A]
(3) For 2nd order: 1/k = kt + 1/[A]
iii)Find these by setting:
(i) [A2]^n[B2]^m=rate2 over [A1]^n[B1]^m = rate1 (2) If you are given a table of information, find two concentrations of A that are the same, so that you can solve for m. This way, [A2]^n/[A2]^n will cancel. And
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you can solve for m by plugging in what B1, B2, rate1 and rate 2 are.
(3) Find n by holding [B] constant
h) At K(eq) the forward and reverse reactions are equal
a) Concentration will never go to zero, but each half-life the concentration will be halved.
b) 1st order: [A]t = [A]0 * e^(-kt)
c) k = ln(2)/t
i) t is the time it takes for one half-life
d) The decay of C14 is how we are able to determine how much time has passed.
e) % change
i) -ln(100-percent change) = kt
a) Used when there is a reaction with a large excess of some product [X].
b) If the rate law was previously written as k[X][Z], we would now write is as:
i) Rate = kobs’=[X]^n
c) This is because the other product [Z] will become extremely small compared to [X], and will become not important to the rate law anymore
d) Rate = k[A]^n[B]^m
i) Large excess of B
ii) So therefore k’ = k[B]^m
e) Steps to use in this case:
i) Determine which of the reactants (if any) is in large excess and therefore will not change significantly in concentration during the reaction
ii) Combine the fixed concentration with underlying rate constant to get k’ = k[B]^m
iii)Use plots with respect to T or ratios to determine n and k’ iv)Repeat step above for two or more starting concentrations of [B] Don't forget about the age old question of What is the difficulty in producing or comprehending speech not produced by deafness or a simple motor deficit?
v) Determine m and k from k’
5) Activation Energy
a) Energy of a collision has to exceed the activation energy otherwise it will stay as a reactant.
b) An increase in temperature increases the reaction rate because more collisions will have enough energy to exceed the activation energy.
i) Activation energy decreases in this case
6) Plotting concentration verses time
a) y = m x + b
b) k = A * e^(-Ea/(RT)) – Arrhenius Equation
d) ln(k) = -Ea/R * (1/T) + ln[A] – first order We also discuss several other topics like Should religion be taught in schools?
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e) –Ea/R = m (slope) If you want to learn more check out Who is the target audience of newspapers?
f) [A] or the y-intercept can be found by taking e^(ln[A]) g) Whichever graph gives the slope closest to the calculated slope will be the order of the reaction. This will also be the straightest graph.
h) Temperature is measured in K
i) A represents the probability that when products collide that they will have a correct orientation in order to react. 7) Reaction Mechanisms
a) Rate determining step
b) Intermediates are produced in one step and then consumed in another.
i) These are not present until something happens in the reaction
c) Catalysts are in the first step and get consumed in another and then are produced again later on.
d) Neither catalysts or intermediates will change the rate determining step
e) If the first step is not the slow step:
i) Write down the balanced new equation
ii) Write down the rate equation for the rate-limiting step iii)Identify reaction intermediates
iv)Remove intermediates by substitution
(1) Start with slow step, find the rate equation. Replace intermediates by substituting in what the intermediate equals in another step. Simplify until the experimental rate is correct.
8) Chemical Kinetics Labs
a) Steps to the lab
b) CV+(aq) + OH-(aq) CVOH(aq)
c) Rate = k[CV+]^n [OH-]^m
i) In the first lab we determined k
ii) Studied with multiple initial concentrations of OH-, at the same temperature
iii)We must repeat the standard curve (calibration plot) for crystal violet each week because:
(1) The spectrophotometer may not be the same (2) The cuvette will not be the same
(3) And The spectrophotometer could give different readings on different days.
d) The study of Chemical Kinetics provides information about: i) Rates of chemical reactions
ii) Reaction Mechanisms
iii)Factors that influence rates of chemical reactions 9) According to the Collision Model of Chemical Kinetics, in order for a reaction to occur between two colliding reactant molecules, they mist be oriented correctly and have enough energy to overcome the activation energy.
10) The rate of disappearance of a reactant is not generally constant over time, and typically it will slow down.