Thermodynamics and Kinetics - AP Chemistry

A reaction is spontaneous if and only if . The sign of only tells us if a reaction is exothermic or endothermic. The rate constant only tells us the rate at which a given chemical reaction proceeds based on the reactants and the products. The first law of thermodynamics only tells us that energy is conserved, and can neither be created nor destroyed. The second law of thermodynamics states that the entropy of the universe is spontaneously increasing, however, under the right conditions, these reactions may proceed.

Recall how to find the change in free energy for a reaction from using free energies of formation:

Where is the number of moles of the molecule.

Plug in the given values to find the change in free energy for the reaction.

In this question, we're told of a reaction that is run at a given temperature and with a certain equilibrium constant. We're asked to evaluate the standard change in free energy for this reaction.

First, right off the bat, we can see that the equilibrium constant for the reaction is greater than one. As a result, we know that under standard conditions, equilibrium of this reaction will favor products. Hence, we know right away that will be negative, which allows us to rule out any answer choices that are positive.

To actually calculate the value of , we'll need to use an equation that relates this quantity with temperature and the equilibrium constant. We can do so by using the following expression.

If we plug in the values that we know, we can solve for the correct answer.

For this question, we're asked to determine the temperature needed for a reaction to be in a state of equilibrium. We are also provided with the entropy and enthalpy changes for this reaction.

Remember that for a reaction to be in a state of equilibrium, the value for its change in free energy must be equal to zero. Furthermore, we can state the relationship between the change in free energy with the change in entropy, the change in enthalpy, and the temperature as follows.

Because the term in the above expression must be equal to zero, we can plug this value in and then isolate the term for temperature.

Now that we have this expression, we just need to plug in the two values given to us in the question stem. Remember to make the units match though!

The entropy cannot decrease in an isolated system because the energy can only be degraded. Since the system is isolated, no higher-grade energy—or any energy at all—is being introduced into the system. As a result, the entropy cannot decrease. The other answer choices relate to the other laws of thermodynamics.

Standard states are defined as a specific set of conditions, such as when a gas is at , concentration, and .

Standard enthalpy of formation, the energy required for form 1 mole of a compound from its constituent elements, occurs when elements are in their standard states.

is positive because heat flows into the system to raise the temperature of the water.

Use the equation that relates heat, mass, specific heat, and change in temperature:

There are four main laws of thermodynamics, which describe how temperature, energy, and entropy behave under various circumstances. The zeroth law of thermodynamics helps to define temperature; it states that if two systems are each in thermal equilibrium with a third system, they must be in thermal equilibrium with each other. The first law of thermodynamics negates the possibility of perpetual motion; it states that when energy passes into or out of a system, the system's internal energy changes in accord with the law of conservation of energy. The second law of thermodynamics also negates the possibility of perpetual motion; it states that in a natural thermodynamic process, the sum of the entropies of the interacting systems increases. Lastly, the third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature nears absolute zero.

The fully filled in reaction coordinate diagram is displayed below.

The arrow marked in the question represents the activation energy, which is the energy barrier that must be overcome in order for the reactants to form products.

This reaction is also exothermic because the energy of the products is lower than that of the reactants.

Start by drawing and labeling the reaction coordinate diagram.

Since we know that is negative, the products must be lower in energy than the reactants.

Now, the question wants you to find the activation energy if we were to reverse the reaction. In other words, we want to start at the products and end up with the reactants. From the diagram, you can see that the reverse activation energy is just the sum of the activation energy for the forward reaction and the .

In this question, we're presented with a free energy diagram for a chemical reaction occurring under standard conditions. We're asked to identify a true statement with regards to this diagram.

Right off the bat, we can see that the free energy of the products is greater than that of the reactants. Thus, this reaction has a positive change in free energy and so it is an endergonic reaction. Recall that one expression for the value of free energy change is as follows.

From this equation, we can see that the equilibrium constant is related to the standard change in free energy. Moreover, we can see that if the standard free energy change is positive, then the value of must be less than . Thus, we know that the reaction is favored towards the reactants.

Additionally, we cannot determine whether the entropy increases or decreases based on the diagram. The reason for this is that entropy is not the only factor that determines the change in free energy of a reaction. The change in enthalpy is also a factor that must be considered.

For this question, we need to determine a true statement with regard to reactions that are catalyzed by enzymes.

To answer this, it's important to distinguish between the thermodynamics of a reaction and the kinetics. When adding an enzyme, the activation energy for the reaction is lowered. What this means is that it becomes easier for the reactants to achieve the high-energy transition state on their way to becoming product. Thus, enzymes affect the kinetics of a reaction by increasing both the forward rate and the reverse rate.

But if we look at the thermodynamics of an enzyme-catalyzed reaction, there is no change. In other words, the difference in free energy between the reactants and the products remains the same, regardless of the presence of enzyme. Thus, an enzyme will not cause a reaction to change its equilibrium position; it cannot shift a reaction towards the left or towards the right.

Because the thermodynamics of the reaction remain unchanged, the reaction will not become more exergonic, nor will it become more endergonic. Furthermore, the equilibrium of the reaction will not shift.

For this question, we're asked to select a true statement regarding reactions that are influenced by enzymes.

When thinking about how enzymes affect reactions, its important to draw a distinction between kinetics and thermodynamics. Enzymes increase a reaction's rate, but they don't change the equilibrium of a reaction. In other words, they make reactions go faster, but they don't make reactions go.

Since enzymes don't affect equilibrium, we can rule out two of the answer choices. Moreover, we know that enzymes increase the reaction rate by lower the activation energy. Consequently, this raises the rate constant for the reaction.

Increasing the amount of reactant increases the rate of reaction only if the concentration of each reactant increases and if the reactants are dissolved in a solution. This is not the case for all reactions. Heating the reaction and introducing a catalyst enable the reactants to reach the activation energy needed to proceed into the reaction. Increasing the surface area between the reactants enables more of each reactant to interact with one another.

The correct answer is because the differential rate law for a zero order reaction is . Therefore, the plot has a linear relationship to the concentration of hydrogen.

Because the differential rate law for a first order reaction is the graph will have a linear relationship to .

Because the differential rate law for a second order equation is the graph will have a linear relationship to .

In order to solve this problem we must compare the initial concentrations of our two reactants between equations with how they effect the rate of the reaction. When comparing experiment 1 and 2, we see that the concentration of A was doubled. This resulted in a quadrupling of the rate. Therefore the reaction is second order with respect to . When comparing experiments 1 and 3, we see that the concentration of B was doubled, and this resulted in a doubling of the rate. Therefore, the reaction is first order with respect to . Then, we plug in all of this information into the rate law, and we get the correct answer of .