The reaction rate is defined as the rate of change in the concentration of reactants or products. This means how fast a reactant gets used up and how fast a product gets produced. How fast a reactant disappears is denoted by the formula:
Rate = (– ΔReactant)/(ΔTime)
How fast a product forms is denoted by the formula:
Rate = ΔProduct/ΔTime
The unit for rate is molarity per second or M/s.
The rate law or rate equation for a chemical reaction is an equation that links the reaction rate with the concentrations or pressures of the reactants and constant parameters
aA + bB → cC + dD
If the above reaction is single-step, then:
rate = k[A]a[B]b
If the above reaction is the rate-determining step of a multi-step reaction, then:
rate of the multi-step reaction = k[A]a[B]b
If the above reaction is a multi-step reaction, then:
rate = k[A]x[B]y
where x and y are unknowns that correspond to the rate-determining step.
To determine the rate law, you refer to a table of rates vs reactant concentrations.
|[A] (M)||[B] (M)||[C] (M)||rate (M/s)|
r = k[A]x[B]y[C]z
From this table, a 2x increase in [A] corresponds to a 4x increase in the rate. 2x = 4, so x = 2.
A 2x increase in [B] corresponds to a 2x increase in the rate. 2y = 2, so y = 1.
A 2x increase in [C] corresponds to 1x (no change) in rate. 2z = 1, so z = 0.
r = k[A]2[B]1[C]0
r = k[A]2[B]
The k in the rate law is the rate constant. The rate constant is an empirically determined value that changes with different reactions and reaction conditions.
Reaction order is the sum of all exponents of the concentration variables in the rate law.
Reaction order in A = the exponent of [A]
|Reaction Type||Reaction Order||Rate Law(s)|
|Unimolecular||1||r = k[A]|
|Bimolecular||2||r = k[A]2, r = k[A][B]|
|Termolecular||3||r = k[A]3, r = k[A]2[B], r = k[A][B][C]|
|Zero order reaction||0||r = k|
The slowest step of a multi-step reaction is the rate determining step. The rate of the whole reaction is the rate of the rate determining step. The rate law corresponds to the components of the rate determining step. The reaction rate is dependent on temperature
Activated complex refers to what is present at the transition state. In the transition state, bonds that are going to form are just beginning to form, and bonds that are going to break are just beginning to break.The transition state is the peak of the energy profile. The transition state can go either way, back to the reactants or forward to form the products.You can’t isolate the transition state. Don’t confuse the transition state with a reaction intermediate, which is one that you can isolate
The activation energy is the energy it takes to push the reactants up to the transition state.
ΔH is the difference between the reactant H and the product H (net change in H for the reaction).
H is heat of enthalpy. An exothermic reaction will have a negative ΔH and an endothermic reaction will have a positive ΔH.
Arrhenius equation is denoted by the formula:
k = Ae – Ea/RT
Where, k is rate constant, Ea is activation energy, T is temperature in Kelvins and R is universal gas constant. A is a constant.5 What this equation tells us is that a low Ea, High T results in a lager k which will give a faster reaction.
When the activation energy approaches zero, the reaction proceeds as fast as the molecules can move and collide. When temperature approaches absolute zero, reaction rate approaches zero because molecular motion approaches zero.5
A reaction can have 2 possible products: a kinetic or a thermodynamic product. A kinetic product has a lower activation energy and is formed preferentially at lower temperatures. A thermodynamic product has a lower (more favorable/negative) ΔG and is formed preferentially at higher temperatures. Thermodynamics tells you whether a reaction will occur. In other words, whether it is spontaneous or not. A reaction will occur if ΔG is negative.
ΔG = ΔH – TΔS
Factors favoring a reaction includes it being exothermic (-ΔH) and resulting in an increase in entropy (positive ΔS). Factors disfavoring a reaction includes it being endothermic (+ΔH) and resulting in an decrease in entropy (negative ΔS). Temperature is a double-edged sword as high temperatures amplify the effect of the ΔS term, whether that is favoring the reaction (+ΔS) or disfavoring the reaction (-ΔS). Kinetics tells you how fast a reaction will occur. A reaction will occur faster if it has a lower activation energy.
Catalysts are things which speed up a reaction without getting itself used up. Enzymes are biological catalysts. Catalysts/enzymes act by lowering the activation energy, which speeds up both the forward and the reverse reaction. Catalysts/enzymes alter kinetics, not thermodynamics. Catalysts/enzymes help a system to achieve its equilibrium faster, but does not alter the position of the equilibrium. Catalysts/enzymes also increase the kinetics rate constant but does not alter Keq of the equilibrium.
The Law of Mass Action is the basis for the equilibrium constant. The Law of Mass Action states that the rate of a reaction depends only on the concentration of the pertinent substances participating in the reaction. Using the law of mass action, you can derive the equilibrium constant by setting the forward reaction rate to equal the reverse reaction rate, which is what happens at equilibrium. For the single-step reaction:
aA + bB <–> cC + dD
rforward = rreverse
kforward[A]a[B]b = kreverse[C]c[D]d
kforward /kreverse = [C]c[D]d/[A]a[B]b
Keq = [C]c[D]d/[A]a[B]b
This holds true for single and multi-step reactions.
There are 2 ways of getting Keq. From an equation,:
Keq = [C]c[D]d/[A]a[B]b
ΔG° = -RT ln (Keq)
ΔG = 0 at equilibrium
ΔG = ΔG° + RT ln Q
0 = ΔG° + RT ln Q at equilibrium
ΔG° = -RT ln Q at equilibrium
ΔG = 0
rforward = rbackward
Q = Keq
Keq is a ratio of kforward over kbackward
If Keq is much greater than 1 (For example if Keq = 103), then the position of equilibrium is to the right and more products are present at equilibrium. If Keq = 1, then the position of equilibrium is in the center and the amount of products is roughly equal to the amount of reactants at equilibrium. If Keq is much smaller than 1 (For example if Keq = 10-3), then the position of equilibrium is to the left and more reactants are present at equilibrium. The reaction quotient, Q, is the same as Keq except Q can be used for any point in the reaction, not just at the equilibrium. If Q < Keq, then the reaction is at a point where it is still moving to the right in order to reach equilibrium. If Q = Keq, the reaction is at equilibrium.
If Q > Keq, then the reaction is too far right, and is moving back left in order to reach equilibrium. The reaction naturally seeks to reach its equilibrium
LeChatelier’s principle states that if you knock a system off its equilibrium, it will readjust itself to reachieve equilibrium. A reaction at equilibrium doesn’t move forward or backward, but the application of LeChatlier’s principle means that you can disrupt a reaction at equilibrium so that it will proceed forward or backward in order to restore the equilibrium. Adding or removing solids or liquids to a reaction at equilibrium doesn’t do anything that will knock the system off its equilibrium
Relationship of the equilibrium constant and standard free energy change is shown below:
ΔG = ΔG° + RT ln Q
ΔG = 0 at equilibrium.
Q becomes Keq at equilibrium.
0 = ΔG° + RT ln (Keq)
ΔG° = -RT ln (Keq)
1) Cermak, N. (2009, March 12). Fundamentals of Enzyme Kinetics: Michaelis-Menten and Deviations. Retrieved from http://cermak.scripts.mit.edu/papers/383final_cermak_enzymekinetics_20090312.pdf
2) Aboazma, S. M. (2014). Biological Oxidation. Retrieved from http://www1.mans.edu.eg/FacMed/english/dept/biochemistry/pdf/OXIDATION.pdf
3) Miles, B. (2003, January 17). Biological Redox Reactions. Retrieved from Texas A&M University: https://www.tamu.edu/faculty/bmiles/lectures/Biological%20Redox%20Reactions.pdf
4) Stephen J. Blanksby, G. B. (2002, August 6). Bond Dissociation Energies of Organic Molecules. Retrieved from Michigan State University: http://www2.chemistry.msu.edu/courses/cem850/handouts/Ellison_BDEs.pdf
5) Pearson. (2011). Thermodynamic versus Kinetic control reactions. Retrieved from http://www.chem.mun.ca/courseinfo/c2400/YZ/Chapter-7c.pdf