Thermodynamics Flashcards
The _________________________ (Q) is a quantity that provides the relativeamounts of products to reactants during a reaction at any given point in time. It is calculated like the equilibrium constant (K) but does not necessarily or likely involved equilibrium amounts.
Reaction quotient
The equilibrium constant tells us whether _______ (K>1) or reactants (K<1) are favored at equilibrium.
Products
Products are favored at equilibrium when K is ________ than 1.
Greater than one
Reactants are favored at equilibrium with K is __________ than one.
Less than one
How is Keq determined?
Remember that for calculations of K or Q, the concentration must be in molarity; however, units are not included.
Units matter; we just don’t include them
If Q>K, what direction does the reaction move to reach equilibrium?
To the left
Toward reactants
If Q<k></k>
To the right
Toward the products
If Q=K, what direction does the reaction move to reach equilibrium?
None; the reaction is at equilibrium
The Gibbs free energy change tells us if a reaction is spontaneous at a specific condition. The standard Gibbs free energy change tells us if the reaction is spontaneous at ___________ conditions, and the Gibbs free energy change (sometimes called the actual) tells us if a reaction is spontaneous at ___________ conditions.
Standard
Any
What are standard conditions?
Standard conditions specifies at 1 atm pressure liquids and solids be pure and that solutions be at 1 M concetrations
Biochemists define a special “biochemical standard state” that includes….?
pH = 7
[Mg2+] = 1 mM
The biochemical standard state is denoted by a prime symbol for ∆G°’
If a reaction is at standard conditions, what is the value of the reaction quotient?
Q = 1
All concentrations are at 1 M
If the same reaction at standard conditions has a K = 5 x 105, what direction does the reaction spontaneously move to reach equilibrium?
Q = 1
Therefore Q < K
Reaction will move toward the products or to the right
Recall that for a reaction at standard conditions that is spontaneous in the forward direction, the ∆G° is negative and is called an ____________ reaction.
Exergonic
For the same reaction under standard condition with a K = 5 x 105, what is the sign of ∆G°?
Because the question before stated that the reaction proceeded toward the products spontaneously, the ∆G° must be negative
Mathematically, we can convert between ∆G°’ and K using the equation
∆G°’ = -RT ln K
In the equation ∆G°’ = - RT ln K, what is the value of R?
R = 8.134 J/mol*K
Calculate the standard Gibbs free energy change ∆G° for the reaction above with a K = 5 x 105 if the temperature is 37 degrees celsius.
∆G° = - RT ln K
-3.4 x 104 J/mol
For a given reaction, ∆G°’ = + 25.2 kJ/mol. What is the K at 37 degrees celsius?
5.7 x 10-7
You must use “e” to get rid of the ln
All temperatures must be in Kelvin (37 degrees celsius is 273 K)
Many reactions occur under non-standard conditions. For those reactions, we can determine the spontaneous direction of the reaction by comparing Q and K and/or calculating the actual ∆G’ using what equation?
∆G’ = ∆G°’ + RT lnQ
What do the variables refer to in the free Gibbs energy equation below?
∆G’ = ∆G°’ + RT lnQ
∆G = Gibbs free energy (non-standard conditions)
∆G° = Standard Gibbs free energy
R = 8.314 J/mol*K
T = Kelvin
If Q<k></k>
Negative
If Q>K, what is the sign of ∆G’?
Positive
If Q = K, what is the value of ∆G’?
Zero
What thermodynamic parameter (∆G°’ or ∆G’) ultimately defines the spontaneous direction of a reaction?
∆G’
Is it possible for a reaction with a postive ∆G°’ to be spontaneous? If so, how? Provide at least one reason.
Yes, when there are no products
Q>1 then lnQ = +
Q<1 then lnQ = -
When can a spontaneous reaction have a positive ∆G°’?
- When Q<<<<<<<<k>
<li>Coupling</li>
</k>
The oxidation of malate to form oxaloacetate occurs in the final step of the citric acid cycle (aka the Krebs cycle). The ∆G°’ for the reaction shown below is +29.7 kJ/mol:
Malate + NAD+ <—-> oxaloacetate + NADH
Is this reaction spontaneous under standard conditions? How do you know?
No, it is not spontaneous under standard conditions because ∆G° > 0
Calculate the equilibrium constant for the reaction at 37 degrees celsius. Are products or reactants favored at equlibirum? How do you know?
∆G°’= -RT lnK
K = 9.9 x 10-6
Reactants are favored because the K is very small
Is this reaction spontaneous under cellular conditions where the concentration of malate is 0.20 mM, oxaloacetate is 11 nM, NAD+ is 0.30 mM, and NADH is 30 µM? The temperature is 37 degrees celsius.
Q = 5.5 x 10-6
From the problem before, we know that K = 9.9 x 10-6
Therefore Q<k></k>
<p>The reaction is spontaneous</p>
<p>(You could also do the math)</p>
</k>
The phosphorylation of glucose is the first reaction in the pathway called glycolysis:
Glucose + phosphate <—-> Glucose-6-phosphate + H20
∆G° = +13.8 kJ/mol
Is this reaction spontaneous under standard conditions?
No, because ∆G°’ > 0
Is this reaction spontaneous under cellular conditions if the concentration of glucose-6-phosphate is 19 mM, glucose is 5.0 mM, and phosphate is 1.6 mM? The temperature is 37 degrees celsius. (Water is not included in Q or K)
Q = 2.3 x 103
∆G’ = ∆G° + RT lnQ
∆G’ = +3.4 x 104 J/mol
Therefore, the reaction is not spontaneous
If we couple the phosphorylation of glucose to the hydrolysis of ATP where ∆G°’ = -30.5 kJ/mol, what is the overall reaction and the corresponding ∆G°’?
Glucose + phosphate/Pi –> Glucose-6-phosphate + H20
ATP + H2O –> ADP + Pi
Glucose + ATP –> Glucose-6-phosphate
Using the calculations from the prior problem, we know that ∆G°’ = +13.8 kJ/mol for the formation of G6P. We can add the two ∆G°’ together, resulting in ∆G°’ of -13.7 kJ/mol
What is the ∆G’ for the overall reaction if the concentration of ATP is 2.25 mM and ADP is 0.25 mM?
Q = 0.42
∆G’ = ∆G°’ + RT lnQ = -18.9 kJ/mol
In glycolysis, fructose-1-6-bisphosphate is converted to two products with a standard free energy change ∆G°’ of 23.8 kJ/mol. Under what conditions encountered in a normal cell will the free energy change ∆G’ be negative, enabling the reaction to proceed spontaneously to the right?
A.) Under standard conditions, enough energy is released to drive the reaction to the right
B.) The reaction will not go to the right spontaneously under any conditions because the ∆G°’ is positive
C.) The reaction can proceed spontaneously to the right if there is a high concentration of products relative to the concentration of reactants
D.) The reaction can proceed spontaneously to the right if there is a high concentration of reactants relative to the concentration of products
E.) None of the above conditions is sufficient
D
Which of the following statements about spontaneity is always correct?
A.) A reaction with a ∆G° = -48.4 kJ/mol is spontaneous
B.) Reactions that have a positive ∆H are not spontaneous
C.) Reactions in which Keq is large are spontaneous
D.) Reactions in which ∆G is negative are spontaneous
E.) MOre than one of the above statements is correct
D
A reaction with an equilibrium constant of 5.0 x 10-5 is moving toward the products to reach equilibrium. Select all true statements based on the information given.
A.) At equilibrium, products are favored
B.) At equilibrium, reactants are favored
C.) Q<k></k>
D.) K<q></q>
E.) Q=K
F.) ∆G=0
G.) ∆G>0
H.) ∆G<0
I.) ∆G°=0
J.) ∆G°>0
K.) ∆G°<0
L.) The reaction is at equilibrium
We know that K favors the reactants; it is move toward the products, which means that Q < K. The relationship between Q and K can only tell us about ∆G not ∆G°. When Q < K, ∆G is spontaneous, meaning it is less than 1. If K<1 we know that ∆G° must be greater than 0.
Statements B, C, H, and J are true
If K>1 then ∆G° must be _______ than zero.
Greater
If K>1 then ∆G° must be _________ zero.
Less
If Q<k then is __________.></k>
Negative
If Q>K the ∆G is _______.
Positive
