Thermodynamics

Question 1

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.

Answer 1

Reaction quotient

Question 2

The equilibrium constant tells us whether _______ (K>1) or reactants (K<1) are favored at equilibrium.

Answer 2

Products

Question 3

Products are favored at equilibrium when K is ________ than 1.

Answer 3

Greater than one

Question 4

Reactants are favored at equilibrium with K is __________ than one.

Answer 4

Less than one

Question 5

How is K_eq determined?

Answer 5

Question 6

Remember that for calculations of K or Q, the concentration must be in molarity; however, units are not included.

Answer 6

Units matter; we just don’t include them

Question 7

If Q>K, what direction does the reaction move to reach equilibrium?

Answer 7

To the left

Toward reactants

Question 8

If Q<k></k>

Answer 8

To the right

Toward the products

Question 9

If Q=K, what direction does the reaction move to reach equilibrium?

Answer 9

None; the reaction is at equilibrium

Question 10

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.

Answer 10

Standard

Any

Question 11

What are standard conditions?

Answer 11

Standard conditions specifies at 1 atm pressure liquids and solids be pure and that solutions be at 1 M concetrations

Question 12

Biochemists define a special “biochemical standard state” that includes….?

Answer 12

pH = 7

[Mg²⁺] = 1 mM

Question 13

The biochemical standard state is denoted by a prime symbol for ∆G°’

Question 14

If a reaction is at standard conditions, what is the value of the reaction quotient?

Answer 14

Q = 1

All concentrations are at 1 M

Question 15

If the same reaction at standard conditions has a K = 5 x 10⁵, what direction does the reaction spontaneously move to reach equilibrium?

Answer 15

Q = 1

Therefore Q < K

Reaction will move toward the products or to the right

Question 16

Recall that for a reaction at standard conditions that is spontaneous in the forward direction, the ∆G° is negative and is called an ____________ reaction.

Answer 16

Exergonic

Question 17

For the same reaction under standard condition with a K = 5 x 10⁵, what is the sign of ∆G°?

Answer 17

Because the question before stated that the reaction proceeded toward the products spontaneously, the ∆G° must be negative

Question 18

Mathematically, we can convert between ∆G°’ and K using the equation

∆G°’ = -RT ln K

Question 19

In the equation ∆G°’ = - RT ln K, what is the value of R?

Answer 19

R = 8.134 J/mol*K

Question 20

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.

Answer 20

∆G° = - RT ln K

-3.4 x 10⁴ J/mol

Question 21

For a given reaction, ∆G°’ = + 25.2 kJ/mol. What is the K at 37 degrees celsius?

Answer 21

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)

Question 22

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?

Answer 22

∆G’ = ∆G°’ + RT lnQ

Question 23

What do the variables refer to in the free Gibbs energy equation below?

∆G’ = ∆G°’ + RT lnQ

Answer 23

∆G = Gibbs free energy (non-standard conditions)

∆G° = Standard Gibbs free energy

R = 8.314 J/mol*K

T = Kelvin

Question 24

If Q<k></k>

Answer 24

Negative

Question 25

If Q>K, what is the sign of ∆G’?

Answer 25

Positive

Question 26

If Q = K, what is the value of ∆G’?

Answer 26

Zero

Question 27

What thermodynamic parameter (∆G°’ or ∆G’) ultimately defines the spontaneous direction of a reaction?

Answer 27

∆G’

Question 28

Is it possible for a reaction with a postive ∆G°’ to be spontaneous? If so, how? Provide at least one reason.

Answer 28

Yes, when there are no products

Q>1 then lnQ = +

Q<1 then lnQ = -

Question 29

When can a spontaneous reaction have a positive ∆G°’?

Answer 29

1. When Q<<<<<<<<k>
   <li>Coupling</li>

</k>

Question 30

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?

Answer 30

No, it is not spontaneous under standard conditions because ∆G° > 0

Question 31

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

Answer 31

K = 9.9 x 10^-6

Reactants are favored because the K is very small

Question 32

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.

Answer 32

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>

Question 33

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?

Answer 33

No, because ∆G°’ > 0

Question 34

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)

Answer 34

Q = 2.3 x 10³

∆G’ = ∆G° + RT lnQ

∆G’ = +3.4 x 10⁴ J/mol

Therefore, the reaction is not spontaneous

Question 35

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°’?

Answer 35

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

Question 36

What is the ∆G’ for the overall reaction if the concentration of ATP is 2.25 mM and ADP is 0.25 mM?

Answer 36

Q = 0.42

∆G’ = ∆G°’ + RT lnQ = -18.9 kJ/mol

Question 38

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

Answer 38

D

Question 39

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

Answer 39

D

Question 40

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

Answer 40

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

Question 41

If K>1 then ∆G° must be _______ than zero.

Answer 41

Greater

Question 42

If K>1 then ∆G° must be _________ zero.

Answer 42

Less

Question 43

If Q<k then is __________.></k>

Answer 43

Negative

Question 44

If Q>K the ∆G is _______.

Answer 44

Positive