Issue 33: Chemistry: Exothermic/Endothermic (Hoyle's Freshman Honor Chem Class)

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Exothermic: Heat released. Heat is product.

Endothermic: Heat absorbed. Heat is reactant.

Calorimetry can be used to find out how much heat is released:

1.Using q(water)=-q(rxn)

2.Then use q=mCΔT

Heat is in proportion to coefficients

Enthalpy-heat of a rxn(ΔH)

Sign of ΔH shows if a reaction is endothermic or exothermic. If ΔH is positive, the rxn is endothermic, but if ΔH is negative, the rxn is exothermic.

Potential Energy is stored in bonds, so heat in rxns comes from the breaking of bonds.

Potential Energy Diagram

If reactants are higher than products, the reaction is exothermic. If the products are higher than the reactants, the reaction is endothermic.

ΔH=potential energy of products-potential energy of reactants

Universe favors lower energy systems

Activation energy-energy needed to get a reaction started

Activation energy is the potential energy of the highest point on the potential energy diagram

Activated complex-unstable transition state

Catalyst-activation energy decreases

Entropy-measure of disorder (ΔS)

Systems tend towards higher disorder so systems tend towards higher entropy

Gases have high entropy, solids have low entropy

Higher entropy is probably because there are more ways to be disordered than to be ordered

How is entropy measured

To start off, entropy is the measure of system’s disorder. The higher is the disorder, the higher is entropy, and vice versa. Although determining entropy seems quite impossible, in fact, scientists derived formula for measuring change in disorder.

It is   , in which S stands for increase in disorder, q is the heat absorbed by substance (In J, of course).To understand this, we must understand that solid has lower entropy (less disorder) than gas, because solid’s atoms have less combinations, compared to gas. Also, the higher is the temperature, the more energy is applied to the atoms, thus canceling out the IMFs between the atoms, and pulling them apart, thus creating a gas. So, at higher temperatures, the substance has higher change in entropy.

On the other hand, T is the temperature at which the absorption takes place (again, in K). This characteristic of entropy is a little bit more confusing. To understand this, let’s make an analogy:

If a 10-year-old boy is allowed to play in his bedroom for half an hour, the increase in disorder is  lower, because the room is already disordered (i.e., at a higher “temperature”). By contrast, if the boy is let loose for half an hour in a neat and tidy living room (i.e., at a lower “temperature”), the effect is much more dramatic.

So, the higher is the temperature at which the absorption is taking place, the lower is the change in disorder, because the substance is already in disorder (again, gas has higher entropy than solid).

Now, let’s see this equation in action!

In this problem, we have an a substance that absorbed 1 kJ (1000 J), at different temperatures. First is 3 K.

Second is 300 K.

Third is 3000 K.

Ready? Set. Chemistry!

a) At 3 K we have:

           

As you can see, at lower temperatures, the change in disorder is extremely high.

b) At 300 K, similarly,

           

c) At 3000 K

           

Now, we notice that change in disorder is extremely low. That is due to the temperature at which this absorption is taking place. The substance is already at disorder, thus it is pretty hard for it to become more disorderly.

Calorimetry

To start off, let’s define what do we mean by calorimetry. Usually, it is the energy that is located inside the substance. However, in the classroom, most of the time, we don’t really talk about calorimetry. Rather, when we use the word “calorimetry,” we mean heat of capacity. Heat of capacity is the amount of heat (energy, usually in J or kJ, sometimes in Kilocalories) required to raise the temperature of one gram of substance by 1 K/C. Remember that during these reactions, volume and pressure must remain constant (thus, if this is not stated in the problem, assume it does). Now, let’s go from theory to actual Math!

Necessary Equations:

           q water = -q rxn

           q=mC△T    for water, C=4.18J=1 Cal     

First of all, we must remember the heat capacity of water. It is 4.18 J per gram/mL of water. Thus, to raise a gram of water by 1 C/K, we need to apply 4.18 J of energy on it. This is arch-important! Now, if the amount of water doubles (2 mL instead of 1, let’s say) then the energy that is required to raise its temperature by 1 mL doubles. Now, let’s put it to practice!

Let’s say we have 10 mL of water. It is at 20 C. However, we don’t want to drink water that cold! No, we want water of room temperature (33 C)! So, how much heat should we apply to it in order to heat it to 33 C?

Remember:

q=mC△T   

So, we substitute the mass of the sample for m. 10 mL = 10 g

For C, as it was mentioned before, we put 4.18 J

△T is the change in temperature. Now, here is the tricky part. Sometimes, when we use Celsius, one temperature value is negative. Thus, in order to not get confused, we should transfer it into Kelvin.

C = K - 273.

So 20 C = 293 K, and 33 C = 306 K

So, △T is = 306 - 293, which is 13 K

Now, let’s substitute!

q = (10. g)(4.18 J)(13 K)

q = 543.4 J (Don’t forget to round to sig. figs!)

q = 540 J

Now, this is where is gets tricky! Time to introduce calorimeter itself. Calorimeter is used to determine heat of capacity of any substance. Sometimes, it is even used to determine heat of solidification and fusion. In order to use calorimeter, we might want to introduce new equation, which comes out of first 2 equations. Here it is:

mC△T (for water) = mC△T (for substance, heat of capacity of which we want to know)

Although it looks quite complicated, with lots of variables, it is actually quite simple. In order to fully explain this, let’s put it to practice first.

Let’s say we had 10 g metal at 50 C (room temperature, let’s say). We then place it into a cup filled with 100 mL of water at 30 C. Then, we record the final temperature of water. It is 40 C. Lots of info, right. So, let’s plug it in!

Let’s start with water!

mC△T

What is the mass of water? Volume is 100 mL, so mass must be 100 g.

What is the Heat of capacity? Well, we know it, It’s 4.18 J

What is the change in temperature? 30 - 40 must be 10 (remember, change in temperature is an absolute value)!

Let’s plug it in and solve it:

(100 g)(4.18 J)(10 C)

You can do this in your head, you say!?

4180 J, right?

Now, it’s time for the right side of the equation, the metal itself

We know the mass, we know the initial temperature, we know the final temperature (which is always the temperature of the water). The only thing we have to figure out is the heat of solidification. Let’s plug the values in now!

(10 g)(x J)(50 C - 40 C)

X joules x 100

Now, let’s equate those

4180 = 100 multiplied by X joules. Divide by 100 on both side, and we get

41.8 = J/g *Now, I know that sig. figs are so wrong, but you get the idea, right?

1. Which one of the following pairs of samples has higher entropy?

a. Br2 (l) or Br2 (g)

b. KOH (s) or KOH (aq)

2. Predict the entropy change for the following processes:

a. O2 (g) → 2O (g)

b. 2O3 (g) → 3O2 (g)

c. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

d. NaCl (s) → Na+ (aq) + Cl- (aq)

e. C2H5OH (l) → C2H5OH (g)

f. Ag+ (aq) + Cl- (aq) → AgCl (s)

3. Of the following reaction,

which are spontaneous at any temperature,

which are never spontaneous regardless of the temperature,

which are spontaneous only at high temperature

and which are spontaneous only at low temperature

*Note: I won’t add the reactions themselves (too lazy). Just going to add the △H & △S

△H △S Temp △G Spontaneous

a. - + any - yes

b. + - any + no

c - - high - yes

d. - - low + no

e. + + high + no

f. + + low - yes

Ammonium nitrate spontaneously dissolves in water at room temperature and the process causes the solution to become quite cold. Which of the following is TRUE about the dissolution of ammonium nitrate?

a) The process is exothermic. False

b) Its solubility will be greater in warmer water. True

c) △S for the reaction is negative. False

d) All the solutions of ammonium nitrate are supersaturated. False

e) All the solutions of ammonium nitrate are cold. False

Examples

Thermodynamic Equation: CH4(g) + 2O2 (g) ---> CO2(g) + 2H2O(l) –890.4

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Exothermic: CH4(g) + 2O2 (g) ---> CO2(g) + 2H2O(l)          delta H= -890.4

Endothermic: N2(g) + O2(g)-----> 2NO(g)                             delta H=182.6

Created by Hoyle's Chem Class 2012-2013  (Freshman Honor Chem)