بوابة الثانوية العامة المصرية - شرح كيمياء thermochemistry

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شرح كيمياء thermochemistry

السلام عليكم

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The Time-Temperature Graph

We are going to heat a container that has 72.0 grams of ice (no liquid water yet!) in it. To make the illustration simple, please consider that 100% of the heat applied goes into the water. There is no loss of heat into heating the container and no heat is lost to the air.

Let us suppose the ice starts at minus 10.0 °C and that the pressure is always one atmosphere. We will end the example with steam at 120.0 °C. There are five major steps to discuss in turn before this problem is completely solved. Here they are:

1) the ice rises in temperature from -10.0 to 0.00 °C.
2) the ice melts at 0.00 °C.
3) the liquid water then rises in temperature from zero to 100.0 °C.
4) the liquid water then boils at 100.0 °C.
5) the steam then rises in temperature from 100.0 to 120.0 °C

Each one of these steps will have a calculation associated with it. WARNING: many homework and test questions can be written which use less than the five steps. For example, suppose the water in the problem above started at 10.0 °C. Then, only steps 3, 4, and 5 would be required for solution.
http://dbhs.wvusd.k12.ca.us/webdocs/...Graph-1to5.GIF

To the right is the type of graph which is typically used to show this process over time.
The ChemTeam hopes that you can figure out that the five numbered sections on the graph relate to the five numbered parts of the list just above the graph.
Also, note that numbers 2 and 4 are phases changes: solid to liquid in #2 and liquid to gas in #4.
Here are some symbols that will be used, A LOT!!

1) Dt = the change in temperature from start to finish in degrees Celsius (°C)
2) m = mass of substance in grams
3) Cp = the specific heat. Its unit is Joules per gram-degree Celsius (J / g °C is one way to write the unit; J g¯1 °C¯1 is another)
4) q = the amount of heat involved, measured in Joules or kilojoules (symbols = J and kJ)
5) mol = moles of substance.
6) DHfus is the symbol for the molar heat of fusion and DHvap is the symbol for the molar heat of vaporization.

We will also require the molar mass of the substance. In this example it is water, so the molar mass is 18.0 g/mol.

By the way, the p means the specific heat is measured at constant pressure; there is a related specific heat we will not discuss (yet) which is measured at constant volume. Not too suprisingly (I hope), it has the symbol Cv.

Step One: solid ice rises in temperature

http://dbhs.wvusd.k12.ca.us/webdocs/...mp-Graph-1.GIF

As we apply heat, the ice will rise in temperature until it arrives at its normal melting point of zero Celsius.
Once it arrives at zero, the Dt equals 10.0 °C.
Here is an important point: THE ICE HAS NOT MELTED YET.
At the end of this step we have SOLID ice at zero degrees. It has not melted yet. That's an important point.

Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount of energy is called specific heat and has the symbol Cp.
.

Step Two: solid ice melts

http://dbhs.wvusd.k12.ca.us/webdocs/...mp-Graph-2.GIF

Now, we continue to add energy and the ice begins to melt.
However, the temperature DOES NOT CHANGE. It remains at zero during the time the ice melts.

Each mole of water will require a constant amount of energy to melt. That amount is named the molar heat of fusion and its symbol is DHfus. The molar heat of fusion is the energy required to melt one mole of a substance at its normal melting point. One mole of solid water, one mole of solid benzene, one mole of solid lead. It does not matter. Each substance has its own value. During this time, the energy is being used to overcome water molecules' attraction for each other, destroying the three-dimensional structure of the ice.

The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion between calories and Joules is 4.184 J = 1.000 cal.
Sometimes you also see this number expressed "per gram" rather than "per mole." For example, water's molar heat of fusion is 6.02 kJ/mol. Expressed per gram, it is 334.16 J/g.
Notice how I shifted to Joules instead of kilojoules. This was done to keep the number within the ranges of ones to hundreds. Writing the value using kJ would require I write 0.33416. It is more understandable to write 334.16.
Typically, the term "heat of fusion" is used with the "per gram" value.

Step Three: liquid water rises in temperature

http://dbhs.wvusd.k12.ca.us/webdocs/...mp-Graph-3.GIF

Once the ice is totally melted, the temperature can now begin to rise again.
It continues to go up until it reaches its normal boiling point of 100.0 °C.
Since the temperature went from zero to 100, the Dt is 100.
Here is an important point: THE LIQUID HAS NOT BOILED YET.
At the end of this step we have liquid water at 100 degrees. It has not turned to steam yet.
Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount of energy is called specific heat and has the symbol Cp. There will be a different value needed, depending on the substance being in the solid, liquid or gas phase.

Step Four: liquid water boils

http://dbhs.wvusd.k12.ca.us/webdocs/...mp-Graph-4.GIF

Now, we continue to add energy and the water begins to boil.
However, the temperature DOES NOT CHANGE. It remains at 100 during the time the water boils.

Each mole of water will require a constant amount of energy to boil. That amount is named the molar heat of vaporization and its symbol is DHvap. The molar heat of vaporization is the energy required to boil one mole of a substance at its normal boiling point. One mole of liquid water, one mole of liquid benzene, one mole of liquid lead. It does not matter. Each substance has its own value. During this time, the energy is being used to overcome water molecules' attraction for each other, allowing them to move from close together (liquid) to quite far apart (the gas state).

The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion between calories and Joules is 4.184 J = 1.000 cal.
Sometimes you also see this number expressed "per gram" rather than "per mole." For example, water's molar heat of vaporization is 40.7 kJ/mol. Expressed per gram, it is 2261 J/g or 2.26 kJ/g.
Typically, the term "heat of vaporization" is used with the "per gram" value. .

Step Five: steam rises in temperature

http://dbhs.wvusd.k12.ca.us/webdocs/...mp-Graph-5.GIF

Once the water is completely changed to steam, the temperature can now begin to rise again.
It continues to go up until we stop adding energy. In this case, let the temperature rise to 120 °C.
Since the temperature went from 100 to 120, the Dt is 20. Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount of energy is called specific heat and has the symbol Cp. There will be a different value needed, depending on the substance being in the solid, liquid or gas phase

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Time-Temperature Calculation #1:
Solid Water Changing in Temperature

72.0 grams of ice (no liquid water yet!) has changed 10.0 °C. We need to calculate the energy needed to do this.
This summarizes the information needed:
Dt = 10 °C
The mass = 72.0 g
Cp = 2.06 Joules per gram-degree Celsius

The calculation needed, using words & symbols is:
q = (mass) (Dt) (Cp)
Why is this equation the way it is?
Think about one gram going one degree. The ice needs 2.06 J for that. Now go the second degree. Another 2.06 J. Go the third degree and use another 2.06 J. So one gram going 10 degress needs 2.06 x 10 = 20.6 J. Now we have 72 grams, so gram #2 also needs 20.6, gram #3 needs 20.6 and so on until 72 grams.
I hope that helped.
With the numbers in place, we have:
q = (72.0 g) (10 °C) (2.06 J/g °C)
So we calculate and get 1483.2 J. We won't bother to round off right now since there are four more calculations to go. Maybe you can see that we will have to do five calculations and then sum them all up.
One warning before going on: three of the calculations will yield J as the unit on the answer and two will give kJ. When you add the five values together, you MUST have them all be the same unit. In the context of this problem, kJ is the preferred unit. You might want to think about what 1483.2 J is in kJ.

Time-Temperature Calculation #2:
Solid Water Melting to Liquid

72.0 grams of solid water is 0.0 °C. It is going to melt AND stay at zero degrees. This is an important point. While the ice melts, its temperature will remain the same. We need to calculate the energy needed to do this.
This summarizes the information needed:

DHfus = 6.02 kJ/mol
The mass = 72.0 g
The molar mass of H2O = 18.0 gram/mol

The calculation needed, using words & symbols is:
q = (moles of water) (DHfus)
We can rewrite the moles of water portion and make the equation like this:
q = (grams water / molar mass of water) (DHfus)
Why is this equation the way it is?

Think about one mole of ice. That amount of ice (one mole or 18.0 grams) needs 6.02 kilojoules of energy to melt. Each mole of ice needs 6.02 kilojoules. So the (grams water / molar mass of water) in the above equation calculates the amount of moles.

I hope that helped.
With the numbers in place, we have:
q = (72.0 g / 18.0 g mol¯1) (6.02 kJ / mol)
So we calculate and get 24.08 kJ. We won't bother to round off right now since there are three more calculations to go. We're doing the second step now. When all five are done, we'll sum them all up. One warning before going on: three of the calculations will yield J as the unit on the answer and two will give kJ. When you add the five values together, you MUST have them all be the same unit

Time-Temperature Calculation #3:
Liquid Water Warming up

72.0 grams of liquid water is 0.0 °C. It is going to warm up to 100.0 °C, but at that temperature, the water WILL NOT BOIL. We need to calculate the energy needed to do this.
This summarizes the information needed:

Dt = 100.0 °C
The mass = 72.0 g
Cp = 4.184 Joules per gram-degree Celsius

The calculation needed, using words & symbols is:
q = (mass) (Dt) (Cp)
Why is this equation the way it is?

Think about one gram going one degree. The liquid water needs 4.184 J for that. Now go the second degree. Another 4.184 J. Go the third degree and use another 4.184 J. So one gram going 100 degress needs 4.184 x 100 = 418.4 J. Now we have 72 grams, so gram #2 also needs 418.4, gram #3 needs 418.4 and so on until 72 grams.

I hope that helped.
With the numbers in place, we have:
q = (72.0 g) (100.0 °C) (4.184 J/g °C)
So we calculate and get 30124.8 J. We won't bother to round off right now since there are two more calculations to go. We will have to do five calculations and then sum them all up.
One warning before going on: three of the calculations will yield J as the unit on the answer and two will give kJ. When you add the five values together, you MUST have them all be the same unit. In the context of this problem, kJ is the preferred unit. You might want to think about what 30124.8 J is in kJ.

Time-Temperature Calculation #4:
Liquid Water Boiling to Gas
72.0 grams of liquid water is at 100.0 °C. It is going to boil AND stay at 100 degrees. This is an important point. While the water boils, its temperature will remain the same. We need to calculate the energy needed to do this.
This summarizes the information needed:

DHvap = 40.7 kJ/mol
The mass = 72.0 g
The molar mass of H2O = 18.0 gram/mol

The calculation needed, using words & symbols is:
q = (moles of water) (DHvap)
We can rewrite the moles of water portion and make the equation like this:
q = (grams water / molar mass of water) (DHvap)
Why is this equation the way it is?

Think about one mole of liquid water. That amount of water (one mole or 18.0 grams) needs 40.7 kilojoules of energy to boil. Each mole of liquid water needs 40.7 kilojoules to boil. So the (grams water / molar mass of water) in the above equation calculates the amount of moles.

I hope that helped.
With the numbers in place, we have:
q = (72.0 g / 18.0 g mol¯1) (40.7 kJ / mol)
So we calculate and get 162.8 kJ. We won't bother to round off right now since there is one more calculation to go. We're doing the fourth step now. When all five are done, we'll sum them all up.
One warning before going on: three of the calculations will yield J as the unit on the answer and two will give kJ. When you add the five values together, you MUST have them all be the same unit. In the context of this problem, kJ is the preferred unit.

Thermochemistry Example Problems

Example Number One
This problem is the *****alent of area number 3 on the time-temperature graph.

calculate the heat necessary to raise 27.0 g of water from 10.0 °C to 90.0 °C

The important factor about this problem is that a temperature change is involved. Therefore, the equation to use is:
q = (mass) (Dt) (Cp)
This summarizes the information needed:

Dt = 80.0 °C
The mass = 27.0 g
Cp = 4.184 Joules per gram-degree Celsius

Only one calculation is needed and it is:
q = (27.0 g) (80.0 °C) (4.184 Joules per gram-degree Celsius)
Multiply it out and round off to the proper number of sig figs.

By the way, you might want to think about how I knew to use the specific heat value for liquid water!!

Example Number Two
This problem is the *****alent of area number 3 and number 4 on the time-temperature graph.

How many kJ are required to heat 45.0 g of H2O at 25.0 °C and then boil it all away?

We must do two calculations and then sum the answers.
Calculation Number One uses this equation:
q = (mass) (Dt) (Cp)
This summarizes the information needed:

Dt = 75.0 °C
The mass = 45.0 g
Cp = 4.184 Joules per gram-degree Celsius

We then have:
q = (45.0 g) (75.0 °C) (4.184 Joules per gram-degree Celsius)
This gives an answer of 14121 J
Calculation Number Two uses this equation:
q = (moles of water) (DHvap)
This summarizes the information needed:

DHvap = 40.7 kJ/mol
The mass = 45.0 g
The molar mass of H2O = 18.0 gram/mol

Substituting, we obtain:
q = (45.0 g / 18.0 g mol¯1) (40.7 kJ/mol)

This gives 101.75 kJ. Adding 14.121 kJ and rounding off gives 115.9 kJ
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Example Number Three: 33.3 grams of ice at 0.00 °C has heat added to it until steam at 150.0 °C results. Calculate the total energy expended. (Hint: 4 calculations are needed - melt, raise, boil, raise.) Good luck.

.

Solution
1) Melt
q = (33.3 g / 18.0 g mol¯1) (6.02 kJ / mol)
2) raise in temperature as a liquid
q = (33.3 g) (100.0 °C) (4.184 J/g °C)
3) Boil
q = (33.3 g / 18.0 g mol¯1) (40.7 kJ / mol)
4) raise in temperature as a gas
q = (33.3 g) (50.0 °C) (2.02 J/g °C)
The ChemTeam gets 103.7 kJ as the final answer.

خللي بالك من المسائل دي

الدرس الثاني
Specific Heat

Here is the definition of specific heat:

the amount of heat necessary to move 1.00 gram of a substance 1.00 °C

Note the two important factors:

1) It's 1.00 gram of a substance
2) and it moves 1.00 °C

Keep in mind the fact that this is a very specific value. It is only for one gram going one degree. The specific heat is an important part of energy calculations since it tells you how much energy is needed to move each gram of the substance one degree.
Every substance has its own specific heat and each phase has its own distinct value. In fact, the specific heat value of a substance changes from degree to degree, but we will ignore that.
The units are usually Joules per gram-degree Celsius (J / g °C). Sometimes the unit J/ kg-K is used. This last unit is technically the most correct unit to use, but since the first one is quite common, you will need to know both.
Here are the specific heat values for water: Phase J g¯1 °C¯1 J kg¯1 K¯1 Gas 2.02 2.02 x 103 Liquid 4.184 4.184 x 103 Solid 2.06 2.06 x 103
Notice that one set of values is simply 1000 times bigger than the other. That's to offset the influence of going from grams to kilograms in the denominator of the unit.
Notice that the change from Celsius to Kelvin does not affect the value. That is because the specific heat is measured on the basis of one degree. In both scales (Celsius and Kelvin) the jump from one degree to the next are the same "distance." Sometimes a kid will think that 273 must be involved somewhere. Not in this case. Specific heat values can be looked up in reference books. They are determined by experiment. In the ChemTeam classroom, you will not need to memorize any specific heat values EXCEPT those for the three phases of water.

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