Collision and Transition-State Theory

1. According to Collision Theory, what needs to happen in order for a reaction to occur?

a) Reactants need to slam into each other with sufficient force

b) Reactants need to slam into each other at the right angle

c) Reactants need to be at the right temperature

d) Both a & b

e) Both b & c

Collision Theory states that reactants need to do two things for a reaction to occur: they need to slam into one another with sufficient force and at the right angle. An increase in in temperature will increase the chances of those two things happening, but a specific temperature is not necessary for a reaction to happen. Humans aren't so different. If the floor is cold, no one wants to get up in the middle of the night to go to the bathroom, but if you have to go bad enough, it doesn't matter what temperature the floor is.

2. Activation energy is _______.

a) the maximum energy required for a reaction to occur.

b) the minimum energy required for a reaction to occur.

c) the amount of energy released during a chemical reaction.

d) the amount of energy absorbed during a chemical reaction.

e) None of the above

The activation energy of a reaction is the minimum amount of energy needed to get the reactants mingling with one another and excited enough to react. Depending on the reactants, they will require more or less activation energy. Again, like people, some are party animals ready to react at the slightest prompting, while others are shy little wallflowers that need to be coaxed onto the dancefloor.

3. An increase in temperature __________ the number of collisions that occurs during a chemical reaction:

a) decreases

b) increases

c) doubles

d) does nothing to

e) None of the above

An increase in temperature excites atoms and molecules, making them vibrate or fly around more quickly, and therefore increases the number of collisions that occur in a chemical reaction. So basically, temperature is sort of like sugar for little kids. Give them some candy and they're bouncing off the walls.



5. In general as the temperature of a reaction increases, the _______ the reaction will occur.

a) slower

b) faster

c) hotter

d) colder

e) None of the above

As the temperature of a reaction increases, the number of collisions between particles increases. Collision Theory states that particles must collide in order for a reaction to occur, like in a demolition derby. Therefore, in general, as the temperature of a reaction increases, the faster the reaction will occur.

Factors Influencing Reaction Rates

1. What three factors can affect the rate of a reaction?

Concentration, temperature, and pressure all affect the rate of a reaction. Why? Because it's all about collisions. An increase in the concentration of reactants, an increase in the temperature of the reaction, or an increase the pressure of starting materials, all cause an increase in collisions. An increase in collisions means an increased likelihood for a reaction to occur, not to mention higher care insurance rates. Ouch.



2. Higher concentrations ________ the likelihood of a collision occurring.

a) increase

b) decrease

c) have no effect on

d) All of the above

e) None of the above

Higher concentrations increase the likelihood of a collision occurring. Picture 10 people walking around in a gymnasium. Now picture those same 10 people walking around in a porta-potty. In which scenario are collisions more likely to occur? The elevator, of course, because the concentration of people (# of people/area) is higher, not to mention the concentration of stink.

3. Higher temperatures _______ the likelihood of a collision occurring.

a) increase

b) decrease

c) have no effect on

d) All of the above

e) None of the above

Higher temperatures increase the likelihood of a collision occurring. Increasing the temperature of a system makes the particles move more quickly, turning them into mad hooligans in a mosh pit that can't help but ram into each other.

4.Higher pressures ________ the likelihood of a collision occurring.

a) increase

b) decrease

c) have no effect on

d) All of the above

e) None of the above

Higher pressures increase the likelihood of a collision occurring, because increasing the pressure of a system effectively increases the concentration of particles. More particles packed in a tighter space means more crashing into each other means more reactions. The little guys just can't keep their cool under pressure.

5. Which of the following is incorrect?

a) Catalysts increase the rate of reaction.

b) Catalysts bring two reactants closer together.

c) Catalysts participate in the reaction itself.

d) All of the above

e) None of the above

Catalysts increase the rate of reaction but they do not actually participate in the reaction itself. They're like molecular matchmakers, bringing the reactants together so they can get to know each other, get married, buy a minivan, and have bratty little kids, otherwise known as products.

Calculating Reaction Rates

1. The Powerpuff Girls were created by reacting "sugar, spice, and everything nice…and CHEMICAL X". Professor Utonium, being a scientist, studied the kinetics of Powerpuff formation. He found the rate equation to be first order in sugar, second order in CHEMICAL X, and zero^th order in spice and everything nice. What is the rate law?

a) Rate = k[sugar]¹[spice]⁰[everything nice]⁰[CHEMICAL X]²

b) Rate = [sugar]¹[spice]⁰[everything nice]⁰[CHEMICAL X]²

c) Rate = k[1 × sugar]¹[0 x spice]⁰[0 × everything nice]⁰[2 × CHEMICAL X]²

d) Rate = k[sugar]¹[spice]¹[everything nice]¹[CHEMICAL X]²

e) Rate = k[sugar]¹[spice]¹[CHEMICAL X]²

The rate order tells us what number to use for the coefficients. Therefore, sugar should have a coefficient of one, chemical X a coefficient of two, and the other reactants a coefficient of zero. This eliminates answers C, D, and E. There is also one more important ingredient, and no, it's not ice cream. It's the rate constant, k. That knocks out answer B, leaving us with A.

2. Mojo tried a similar experiment to create the Rowdyruff Boys. He used underarm hair, snails, and the tail of the talking dog. Gross. Anywho, Mojo ran some kinetics experiments. His results are given below. Concentrations in moles/L (M).

Exp.	Underarm Hair	Snails	Talking Dog Tail	Rate (mol/L sec)
1	1M	1M	1M	0.0025
2	1M	0.5M	1M	0.0025
3	1M	1M	2M	0.0050
4	0.5M	1M	1M	0.00125

What are the reaction orders for the reactants?Underarm Hair - 2; Snails – 1; Talking Dog Tail – 0

a) Underarm Hair - 1; Snails – 1; Talking Dog Tail - 1

b) Underarm Hair - 1; Snails – 0; Talking Dog Tail - 1

c) Underarm Hair - 1; Snails – 0; Talking Dog Tail - 0

d)Underarm Hair - 1; Snails – 1; Talking Dog Tail – 1

Experiments 1 and 2 have the same rate, while the snails concentration goes down by ½. This means that the reaction order for snails must be zero. This eliminates A, B, and E. Experiment 3 has twice the rate as compared to experiment 1, while the concentration of 'talking dog tail' also has doubled. This means that 'talking dog tail' has a reaction order of 1. This eliminates answer D, leaving answer C as the only possibility.

3. Using the data from question two, what is the value of the rate constant k?

a) 1

b) 0.0025

c) 0.0050

d) 0.00125

e) 0

From question 2, we know that the rate = k[underarm hair][talking dog tail]. Using the data from experiment 2, we can plug in the rate, underarm hair concentration, and talking dog tail concentration. 0.0025 = k[1][1], or k = 0.0025. Nice. We're constantly amazed how that works.

For questions 4-5, we are actually going to give you a real reaction.

2 CO + O₂→ 2 CO₂. And real-ish data.

	[CO] (mol/L)	[O2] (mol/L)	Rate (M/s)
Exp. 1	0.013	0.0089	0.0027
Exp. 2	0.022	0.0089	0.0046
Exp. 3	0.013	0.016	0.0095

4. What is the correct rate law for this reaction?

a) Rate = k [CO]²[O₂]

b) Rate = k [CO][O₂]

c) Rate = k [CO]²[O₂]²

d) Rate = k [CO]²[O₂]²[CO₂]

e) Rate = k [CO][O₂]²

In experiments 1 and 2 the O₂ concentration remains constant and the CO concentration goes from 0.013 to 0.022 mol/L. Dividing 0.022/0.013 gives us 1.7, meaning that the concentration of CO in experiment 2 is 1.7 times larger than in experiment 1. Dividing the rates in the same order (exp2/exp1, or 0.0046/0.0027) gives us a rate increase of 1.7. From this, we can state that the reaction order for CO is 1. Repeating the same process in experiments 3 and 1, we find the ratios for O₂ and the rate to be 1.8 and 3.5, respectively. The rate change is approximately twice the change in the oxygen concentration (3.5/1.8 = 1.9 ≈ 2). This means that the reaction order for O₂ is 2, and that the answer must be E. Now say all that ten times real fast.

5. What is the rate constant for this reaction?

a) 3.8 × 10^-4

b) 0.013

c) 1.7 × 10⁴

d) 2600

e) There is insufficient data to answer.

The rate constant can be obtained by plugging in the concentrations from one experiment into the rate law. Using experiment 1, we get 0.0027 = k [0.013][0.0089]². k = 2622, which rounds to 2600. Now, time to plug our brains in to recharge.

Plotting Kinetics Experiments

1. What is the half-life for a first-order chemical reaction or process?

a) t_1/2 = 1/k[A_o]

b) t_1/2 = [A_o]/2k

c) t_1/2 = 5k

d) t_1/2 = ln2/k

e) t_1/2 = k[A₀]²

First-order reactions can be described mathematically by plotting ln concentration vs. time. A half-life examines the time required for half of a reactant to disappear. Consider that [A] = [A]₀ exp(-kt), which rearranges to [A]/[A]₀ = exp(-kt). Since [A]/[A]₀ = ½, this equation becomes ½ = exp(-kt). Take the natural log of both sides, and this becomes ln(1/2) = -kt. We can rearrange this to be ln(1) – ln(2) = -kt. Ln(1) = 0, so this becomes -ln(2) = -kt. We divide both sides by –k to get the final equation: t_1/2 = ln2/k. A bit of work, sure, but it's still easier than dividing a cookie and sharing when you want that cookie all to yourself.

2. A linear plot of 1/concentration vs. time describes a _______ reaction.

a) 0^th order

b) 1^st order

c) 2^nd order

d) 3^rd order

e) None of the above

A plot of concentration vs. time describes a 0^th order reaction, while a plot of ln concentration vs. time describes a 1^st order reaction. Third order reactions are nonexistent. However, 1/concentration vs. time does describe a second-order reaction. Therefore, the answer is C, and cookies start with the letter C, so we want a cookie. Now, please.

3. Given the kinetics data below, what is the experimental rate constant?

Time (min)	[X] (M)
0	0.28
5	0.20
10	0.13
15	0.06

a) 0.015 M/min

b) 0.0023 M/min

c) 0.10 M/min

d) 230 M/min

e) 12 M/min

We first must identify what type of kinetics are involved in this reaction. Plotting concentration vs. time gives us a straight line.


The slope of the line is -0.0146, which is our rate constant k for this reaction.

4. What is the rate law for a reaction that is first order in potassium and second order in oxygen? K + O₂→ KO₂

a) Rate = k[K][O₂]

b) Rate = k[K][O]²

c) Rate = k[K][O₂]²

d) Rate = k[K][O₂]^1/2

e) Rate = k[K]²[O₂]

The rate law is first order in potassium, meaning [K]¹, and second law in oxygen, meaning [O₂]². This eliminates answers A, D, and E. Answer B is not correct because O₂ is a reactant, not O. That leaves us with ole reliable, letter C.

5. A given first-order reaction has a rate constant of 0.012 M/sec and a [X]₀ = 1.1 M. What is the half-life for this reaction?

a) 1.3 seconds

b) 2.8 hours

c) 0.96 minutes

d) 508 seconds

e) None of the above

Since this reaction is first order, we know that the half-life is ln2/k, which our trustworthy calculator tells us is 58 seconds. That answer choice isn't there, but being the crafty Shmooperinos we are, we can easily convert the units from seconds to minutes: 58 seconds/60 seconds = 0.96 minutes.

Arrhenius Equation

Use the following data for questions 1-3. Kinetics experiments were performed at various temperatures, and rate constants were calculated for each experiment.

Temperature (K)	k (L/mol*s)
300	0.00387
325	0.00499
350	0.00620
375	0.00747

1. What is the activation energy for this reaction?

a) 3.02 kJ/mol

b) 3.56 kJ/mol

c) 8.20 kJ/mol

d) 13.1 kJ/mol

e) 13.8 kJ/mol

To determine the activation energy for this reaction, we must graph log k vs. 1/T. If you need to stretch first, no problem. The resulting graph is given below.



The slope of the line is –E_a/2.303R. Multiplying the slope of the line by -19.14 gives us the activation energy in joules/mole. (-428.55)*(-19.14) = 8200 J/mol, which is 8.20 kJ/mol.

2. Calculate the rate constant k at 400K.

a) -2.055

b) .0108

c) 0.00881

d) 0.00338

e) None of the above

The plot given above is log k vs. 1/T, and has the formula y = -428.55x – 0.9835. The variable x should be replaced with 1/400, and the equation solved for y. In this case, the variable y = log k at 400K, which is found to be -2.055. We then must perform an inverse log function on our y-value to obtain the rate constant k. A backflip into handspring is optional. With or without the gymnastics, the correct answer is 0.00881 or 8.81 × 10^-3.

3. At what temperature is the rate constant 1.0 × 10^-4?

a) 0.00704 K

b) 1 K

c) 100 K

d) 142 K

e) 188 K

We use the plot from question 1, as well as the equation of that line. In this problem, we take the logarithm of 1.0 × 10^-4 and replace the y-variable with this value (-4). We then solve for x, which is 1/T in Kelvin. 1/T is found to be 0.00704, and T is found to be 142K.

4. What is wrong with the following Arrhenius Plot?



a) Nothing. Great job!

b) The Rate constant

c) The temperature

d) Everything

The plot shown is k vs. T. The Arrhenius plot is supposed to be log k vs. 1/T. As such, everything on the plot is incorrect. It does look nice, though, so cheers to the graphic artist.

5. What is wrong with the following Arrhenius Plot?



a) Nothing. Great job!

b) The Rate constant

c) The temperature

d) Everything

The plot shown is log k vs. 1/T. All set, right? Wrong. The Arrhenius plot is supposed to be log k vs. 1/T, with the temperature in the Kelvin scale. In this plot, temperature was 1/T, with T in the Celsius scale. Lousy Celsius trying to sneak that one by us….

Dynamic Equilibrium

1. What is the definition of equilibrium?

The state that exists during a chemical reaction when the forward reaction and the backwards reaction proceed at equal rates. Or in Star Wars jargon, the Force is in balance.

2. What is a good metaphor to illustrate dynamic equilibrium?

Running on a treadmill. Technically, you are running but you are not moving anywhere. Bonus points if you said running on your hands on a treadmill, eating a burrito while skipping on a treadmill, or any other variation, just because we like originality. And burritos.

3. In a state of dynamic equilibrium the rate of the forward reaction and the rate of the backward reaction are ________.

a) equal

b) not related

c) forward reaction is slower than the backward reaction

d) forward reaction is faster than the backward reaction

e) None of the above

In a state of dynamic equilibrium, the rate of the forward reaction and the rate of the backward reaction are equal. On a large scale the reaction is not changing but on a molecular level both the forward and backward reaction are occurring at the same time at the same rate. They're probably wearing the same outfit, too.

4. Why is a closed system necessary for a state to remain in dynamic equilibrium?

A closed system means that nothing transfers in or out of the system, which is important to maintain dynamic equilibrium. Dynamic equilibrium can be interrupted if an outside influence is applied to the system, such as temperature or pressure. Obnoxious music is okay, though, no matter what our parents might say.

5. What is the definition of dynamic equilibrium?

The state of a dynamic system in which, although shifting and changing, the overall pattern of forces or energy is in a stable, organized configuration. In other words, even though stuff is happening, but at a steady rate in both directions, so it all evens out.

The Equilibrium Constant

1. What is the equilibrium constant expression for the following reaction?

Pb²⁺_(aq) + 2Cl^-_(aq) → PbCl_{2 (s)}

a) K_eq = [PbCl₂]/[Pb²⁺]*[Cl^-]²

b) K_eq = [PbCl₂]/[Pb²⁺]*[2Cl^-]²

c) K_eq = 1/[Pb²⁺]*[2Cl^-]²

d) K_eq = 1/[Pb²⁺]*[Cl^-]²

e) K_eq = 1/[Pb²⁺]*[Cl^-]

The equilibrium constant only includes reagents that are in solution, which eliminates A and B. Answer C is incorrect because [2Cl^-]² should be [Cl]², while E is incorrect because the chloride equation is not squared in it. Answer D is the only to set everything up correctly, and as such, it gets a sparkly dinosaur sticker. Good job, D.

2. For the reaction given in question 1, the concentration of Pb²⁺ = 8.3 × 10^-11 M and Cl^- = 3.2 × 10^-7 M. What is the equilibrium constant for this reaction?

a) 8.5 × 10^-24

b) 3.4 × 10^-23

c) 2.7 × 10^-17

d) 1.1 × 10^-16

e) None of the above

To get started, we do ten jumping jacks, and then substitute the concentrations given in question 2 into the equation K_eq = 1/[Pb²⁺]*[Cl^-]², giving us K_eq= 1/(8.3 × 10^-11)(3.2 × 10^-7)². This simplifies to 8.5 × 10^-24.

3. What is the equilibrium constant for the following unbalanced reaction?

Al³⁺_(aq) + SO₄^2-_(aq) → Al₂(SO₄)_3(s)

a) K_eq = 1/[Al³⁺]*[SO₄²⁺]

b) K_eq = 1/[Al³⁺]³*[SO₄²⁺]²

c) K_eq = 1/[Al³⁺]²*[SO₄²⁺]³

d) K_eq = [Al₂(SO₄)₃]/[Al³⁺]²*[SO₄²⁺]³

e) None of the above

The first step in this problem, after rubbing our chin contemplatively, of course, is to balance the equation.

2Al³⁺_(aq) + 3SO₄^2-_(aq) → Al₂(SO₄)_3(s)

There we go. Now we know the correct answer is C because it correctly uses the coefficients in the balanced equation as the exponents in the equilibrium expression and it omits aluminum sulfate because it is a solid.

4. What is the equilibrium constant for the reaction between iron (III) nitrate and sodium carbonate?

a) K_eq = [Fe³⁺]²*[CO₃^2-]³

b) K_eq = [Fe(NO₃)₃]²*[Na₂CO₃]³

c) K_eq = [Fe₂(CO₃)₃]*[NaNO₃]⁶/ [Fe(NO₃)₃]²*[Na₂CO₃]³

d) K_eq = [NaNO₃]⁶/ [Fe(NO₃)₃]²*[Na₂CO₃]³

e) None of the above.

To reach enlightenment, or at least to answer this question, we have to write the balanced net ionic equation to determine what reaction occurred. The spectator ions in this reaction play no part in the reaction.

2Fe³⁺_(aq) + 3CO₃^2-_(aq) → Fe₂(CO₃)_3(s)

From this, we can determine that the equilibrium constant K_eq = 1/[Fe³⁺]²*[CO₃^2-]³, which is not one of our choices. Therefore, the correct answer is "none of the above."

5. Consider the reaction below. If the equilibrium constant is 1.9 × 10^-7, and the concentration of Fe³⁺ is 7.7 × 10^-2 M, then what is the concentration of carbonate ion in the solution?

2Fe³⁺_(aq) + 3CO₃^2-_(aq) → Fe₂(CO₃)_3(s)

a) 0.010 M

b) 0.0010 M

c) 1.0 × 10^-4 M

d) 1.0 × 10^-5 M

e) 1.0 × 10^-6 M

The equilibrium constant expression is K_eq = 1/[Fe³⁺]²*[CO₃^2-]³. We need to solve for [CO₃] = (1/K_eq*[Fe³⁺]²)^1/3 = (1/(1.9 × 10^-7)(7.7 × 10^-2)²)^1/3. After wiggling our fingers and letting the gears in our heads go vroom, this comes to a 0.0010M CO₃^2- concentration, which is answer B.

Le Chatelier's Principle

Questions 1-2 will refer to the reaction: 2NO + O₂ → 2NO_2.

1. What is the equilibrium constant expression for this reaction?
a) [NO₂]*[O₂]*[NO]
b) [NO₂]/[NO]*[O₂]
c) [NO₂]
d) [NO₂]²/[NO]² * [O₂]
e) [NO₂]^{2 *} [NO]² * [O₂]
The equilibrium expression for a reaction is defined as [product]^{product coefficient}/[reactant]^{reactant coefficient}. This leaves only b and d as possibilities, and we can axe b because it does not include the equation coefficients as exponents; d does.

2. How can you drive the reaction above toward the right (products)?
a) Remove NO₂ from the reaction mixture_.
b) Increase the concentration of NO.
c) Increase the concentration of O₂.
d) All of the above
e) None of the above
LeChatelier's principle states, probably with a French accent, that a reaction equilibrium can be disturbed by changing the concentrations of the reactants and/or products. Removing the products of a reaction pushes a reaction to the right. Increasing the concentrations of the reactants also pushes a reaction to the right.

3. Which of the following may not affect the equilibrium of a reaction?

a) Changing the concentration of reactants

b) Addition of a catalyst

c) Changing the temperature of a reaction

d) Decreasing the pressure of gases in a gaseous reaction

e) All of the above

Equilibria can be affected by perturbations in concentrations to either reactants or products. This eliminates choices a and d (and e, by definition). Changing the temperature of a reaction changes the equilibrium as well. The answer is b, because catalysts amp up the reaction rate but not the equilibrium.

4. When a reaction has reached equilibrium, which of the following is true?

a) The forward reaction and reverse reaction are occurring at the same rate.

b) The concentrations of reactants and products are constant, until the equilibrium is disturbed.

c) The reaction stops completely, and static equilibrium is achieved.

d) a & b

e) All of the above

Answer c is incorrect because chemical equilibrium is dynamic, not static. This eliminates selection, e as well. Answers a and b are both true for a reaction in equilibrium; therefore, the best answer is d. The ole process of elimination trick does wonders.

5. Changes in reactant and product concentrations perturb the equilibrium of a reaction. Which of the following would not be affected by these changes?

a) The equilibrium constant

b) The equilibrium expression K_eq

c) All of the above

d) None of the above

The equilibrium constant and the equilibrium expression remain the same regardless of the concentrations of the reactants and products. Heck, they don't even change when you make funny faces at them.

Thermodynamics

For questions 1-3, use the following information:

ΔH° = -138.5 kJ/mol;ΔS° = -27.1 J/mol K

1. What is the value of ΔG° at 298 K, given the above thermodynamic information?

a) -147 kJ/mol

b) -139,000 kJ/mol

c) +147 kJ/mol

d) -130 kJ/mol

e) +7940 kJ/mol

This question requires the use of the free energy equation, ΔG° = ΔH° – TΔS°. Plugging into the equation, ΔG° = -138.5 kJ/mol – 298K(-27.1 J/molK)/1000. We have to divide the TΔS° term by 1000 to convert joules to kilojoules. Completing the problem with a curtsy or bow, we find, ΔG° = -130.4 kJ/mol or -130 kJ/mol.

2. What is the equilibrium constant for this reaction at 298 K?

a) 7.82 × 10²³

b) 7.39 × 10²²

c) 52.7

d) 1.30 × 10⁵

e) 1.05

From question one, we divined that ΔG° = -130.4 kJ/mol. We'll use that value along with our handy dandy equation ΔG = -RTlnK_eq to answer this question. Plugging into this equation, we get (-130.4 kJ/mol *1000) = -(8.31 J/molK)(298K)*lnK_eq. lnK_eq = 52.7. Using the exp function on both sides, we find K_eq = 7.39 × 10²².

3. At what temperature is the reaction at equilibrium?

a) 5110K

b) 3899K

c) 3000K

d) 4837 K

e) This reaction never reaches equilibrium.

By definition tacos are yummy. Also by definition, a reaction is at equilibrium when ΔG° = 0. We can solve this by using ΔG° = ΔH° – TΔS° and solving for temperature. 0 = -138.5 kJ/mol – [T(-27.1J/molK)/1000]. This rearranges to -138.5 kJ/mol = (-0.0271kJ/mol K)T. T = 5110 K at equilibrium.

4.Which thermodynamic quantity describes the spontaneity of a chemical reaction?

a) Internal Energy (U)

b) Entropy (S)

c) Enthalpy (H)

d) Free Energy (G)

e) Activation Energy (E_a)

Enthalpy is the heat absorbed or given off in a reaction. Internal energy describes the potential and kinetic energy of a system. Entropy describes the disorder of a system. Activation Energy describes the energy barrier that must be overcome for a reaction to occur. Free Energy, defined by the equations ΔG = ΔH – TΔS and ΔG = -RTlnK_eq, describes how a reaction progresses under given conditions. A negative free energy describes a spontaneous reaction at a given temperature while a positive free energy describes a non-spontaneous reaction.

5. If the equilibrium constant for a reaction at 400K is found to be 2.3 × 10^-4, what is the free energy of the reaction at that temperature?

a) -33 J/mol

b) -3.3 kJ/mol

c) -28 kJ/mol

d) 0.3 kJ/mol

e) + 38 kJ/mol

This problem is a bit of a sassy pants and refuses to work with anyone except the equation ΔG = -RTlnK_eq. Plug in 8.31 J/mol K for R, 400K for T, and 0.00023 for K_eq. ln(0.00023) = -8.37, so ΔG = -RTlnK_eq = -(8.31J/molK)(400K)(-8.37) = 28,000J or 28 kJ.