Endothermic and Exothermic Reactions
Every single chemical reaction involves some kind of change in energy which fall into one of two categories Endothermic or Exothermic.
An endothermic reaction is one that absorbs energy. For example: instant ice packs absorb energy and are therefore endothermic.
An exothermic reaction is one that releases energy. For example explosions release energy and are therefore exothermic.
At an atomic level... molecules are held together by chemical bonds. These bonds can break if energy is added and they can form together if energy is given off to form bonds.
In an endothermic reaction it takes more energy to break the bonds then the reaction gives off. But in an exothermic reaction it takes less energy to break the bonds then the reaction gives off to form bonds.
For an in-classroom demonstration of types of endothermic and exothermic reactions watch this video!
Wednesday, 1 February 2012
Friday, 20 January 2012
Solubility and Net Reactions
Solubility and Net Reactions
Net Reactions or Net Ionic Equations are just a simplified version of a reaction involving the forming of a precipitate. Note: A net Ionic Equation can only occur if one of the products in a double replacement reaction is a solid.
A Net Ionic Equation is written using the following steps:
1. Write, predict and balance the complete reaction with state subscripts written in.
2. If a solid is formed, write out the complete equation separating all ions from one another except the ones forming the solid with the charges and state subscripts written in.
3. Cross out any ions found on both sides of the equation.
4. Write out the remaining ions and compounds in a reaction.
Example:
Step 3 and 4:
In order to tell whether or not a chemical reaction can occur, one must determine the solubility of the products.
Use the following link to determine whether or not a compound or soluble or not / barely soluble.
http://dl.clackamas.edu/ch105-03/solubili.htm
If one of the products is not or barely soluble then a precipitate or solid forms meaning that a reaction has occurred.
If both the products are soluble meaning they are aqueous then a reaction has not occurred.
Examples:
Net Reactions or Net Ionic Equations are just a simplified version of a reaction involving the forming of a precipitate. Note: A net Ionic Equation can only occur if one of the products in a double replacement reaction is a solid.
A Net Ionic Equation is written using the following steps:
1. Write, predict and balance the complete reaction with state subscripts written in.
2. If a solid is formed, write out the complete equation separating all ions from one another except the ones forming the solid with the charges and state subscripts written in.
3. Cross out any ions found on both sides of the equation.
4. Write out the remaining ions and compounds in a reaction.
Example:
| Step 1: |
| ___BaCl2 (aq) | + | ___Na2CO3 (aq) |  | ___BaCO3 (s) | + | _2__NaCl (aq) |
| Step 2: |
| _Ba2+(aq) | + | 2Cl - (aq) | + | 2Na +(aq) | + | CO3 2-(aq) | | ___BaCO3 (s) | + | 2Na +(aq) | + | 2Cl- (aq) |
Ba 2+ (aq) | + | CO32- (aq) | | BaCO3(s) |
In order to tell whether or not a chemical reaction can occur, one must determine the solubility of the products.
Use the following link to determine whether or not a compound or soluble or not / barely soluble.
http://dl.clackamas.edu/ch105-03/solubili.htm
If one of the products is not or barely soluble then a precipitate or solid forms meaning that a reaction has occurred.
If both the products are soluble meaning they are aqueous then a reaction has not occurred.
Examples:
| NaOH | Fe2S3 | MgSO4 | PbCl2 | Ba(NO3)2 | MgCO3 |
| soluble | insoluble | soluble | insoluble | soluble | insoluble |
Double Replacement, Combustion and Neutralization Reactions
Double Replacement, Combustion and Neutralization Reactions
Double Replacement Reactions: When two compounds swap components in one of two ways... The anion of one compound may re- attach to the other compound and vice versa OR the cation of one compound may re-attach to the other compound and vice versa.
This reaction can be further simplified using a dancing analogy:
Two couples, comprised of two partners each (a man and a woman), are taking ballroom lessons. The instructor later calls "switch" and the female partners switch and begin dancing with the other male partner.
A double replacement reaction can be modeled by this formula:
AB + XY ---> AY + XB
Examples:
KOH + H2SO4 ---> K2SO4 + H2O
FeS + HCl ---> FeCl2 + H2S
NaCl + H2SO4 ---> Na2SO4 + HCl
AgNO3 + NaCl ---> AgCl + NaNO3
Combustion Reactions: Are essentially a kind of single replacement reaction which involves something burning in air.
It can be modeled by the following formula:
AB + O2 ---> CO2 + H2O
Examples:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
Neutralization Reactions: Are essentially a kind of double replacement reaction which includes acids reacting with bases to produce H2o and an ionic salt.
It can be modeled by this formula:
HA + BOH ---> H2O + BA
Examples:
HCl + NaOH --> NaCl + HOH
2 NaOH + H2CO3 --> N2CO3 + 2 NaOH
Now, you must look at the "Activity Series". If the compound replacing the previous one is higher on the list, the reaction will occur. If it is lower, nothing will happen.
Double Replacement Reactions: When two compounds swap components in one of two ways... The anion of one compound may re- attach to the other compound and vice versa OR the cation of one compound may re-attach to the other compound and vice versa.
This reaction can be further simplified using a dancing analogy:
Two couples, comprised of two partners each (a man and a woman), are taking ballroom lessons. The instructor later calls "switch" and the female partners switch and begin dancing with the other male partner.
A double replacement reaction can be modeled by this formula:
AB + XY ---> AY + XB
Examples:
KOH + H2SO4 ---> K2SO4 + H2O
FeS + HCl ---> FeCl2 + H2S
NaCl + H2SO4 ---> Na2SO4 + HCl
AgNO3 + NaCl ---> AgCl + NaNO3
Combustion Reactions: Are essentially a kind of single replacement reaction which involves something burning in air.
It can be modeled by the following formula:
AB + O2 ---> CO2 + H2O
Examples:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
It can be modeled by this formula:
HA + BOH ---> H2O + BA
Examples:
HCl + NaOH --> NaCl + HOH
2 NaOH + H2CO3 --> N2CO3 + 2 NaOH
Now, you must look at the "Activity Series". If the compound replacing the previous one is higher on the list, the reaction will occur. If it is lower, nothing will happen.
Here's a video that demonstrates how to do it:
Monday, 16 January 2012
Word Equations and Naming Coumpounds
We're going back to the very beginning; when we first learned to name compounds. Remember? Well, here's a little refresher...
For example,
If a word equations is:
Aluminium + Hydrochloric Acid = Aluminium Chloride + Hydrogen
The chemical formula is:
Al + HCl = AlCl3 + H2
BUT WAIT! We still have to balance it! So, the proper equation is:
2Al + 6HCl = 2AlCl3 + 3H2
Now that was easy, wasn't it? Here's a chart reminding you of all the prefixes for covalent compounds:
Don't forget about naming acids! To help remember, I "ate" something "ic"y. If you just say that, the other ending of "ous" for "ite" just falls into place.
For example,
If a word equations is:
Aluminium + Hydrochloric Acid = Aluminium Chloride + Hydrogen
The chemical formula is:
Al + HCl = AlCl3 + H2
BUT WAIT! We still have to balance it! So, the proper equation is:
2Al + 6HCl = 2AlCl3 + 3H2
Now that was easy, wasn't it? Here's a chart reminding you of all the prefixes for covalent compounds:
Don't forget about naming acids! To help remember, I "ate" something "ic"y. If you just say that, the other ending of "ous" for "ite" just falls into place.
Here is a little video to clear up any other confusion:
And one more to help with balancing:
Friday, 6 January 2012
Molar Volume of Gas (STP)
STP, or Standard Temperature and Pressure is a set of data we use to compare our measurements to the gas' properties under the temperature of 0 degrees (273 K)
Key Formula: (one mole of gas occupies 22.4 L of gas)
1 mole of gas
22.4L of gas
Putting it Into Use:
Calculate the volume occupied by 4.6g of C (carbon)
4.6 x 1 mole = 0.38moles
12.0g C
This is where the Key formula is plugged into the equation.
molar volume = 22.4 L x 0.38moles = 6.5 L
1 mole
And there you go! A few easy steps.
Key Formula: (one mole of gas occupies 22.4 L of gas)
1 mole of gas
22.4L of gas
Putting it Into Use:
Calculate the volume occupied by 4.6g of C (carbon)
4.6 x 1 mole = 0.38moles
12.0g C
This is where the Key formula is plugged into the equation.
molar volume = 22.4 L x 0.38moles = 6.5 L
1 mole
And there you go! A few easy steps.
Wednesday, 4 January 2012
Solution Preparation
Solutions are shipped in concentrated form to minimize shipping costs. Therefore, they must be prepared before they can be used.
The formula M1L1 = M2L2 is used to prepare solutions.
The most important thing to remember is that the MOLES on one side MUST EQUAL the MOLES on the other side!
Example:
I have 2.00 L of 16.0 M of HCl. I need 0.800 L of 2.00 M HCl.
Solve for L1.
(16.0M1)(L1) = (0.800L2)(2.00M2)
L1 = 0.100 L
Let's do one more example:
You have 11.6 mol/L of HCl. How would you make up 250 mL of 0.500 mol/L?
Solve for L1.
(11.6M1)(L1) = (0.25L2)(0.500M2)
L1 = 0.011 L acid
Then, you subtract the original volume by the amount of acid required.
0.25 - 0.011 = 0.239 L
Therefore, you would need 0.011 L of acid and 0.239 L of water.
Here's a video that explains it all step by step...
The formula M1L1 = M2L2 is used to prepare solutions.
The most important thing to remember is that the MOLES on one side MUST EQUAL the MOLES on the other side!
Example:
I have 2.00 L of 16.0 M of HCl. I need 0.800 L of 2.00 M HCl.
Solve for L1.
(16.0M1)(L1) = (0.800L2)(2.00M2)
L1 = 0.100 L
Let's do one more example:
You have 11.6 mol/L of HCl. How would you make up 250 mL of 0.500 mol/L?
Solve for L1.
(11.6M1)(L1) = (0.25L2)(0.500M2)
L1 = 0.011 L acid
Then, you subtract the original volume by the amount of acid required.
0.25 - 0.011 = 0.239 L
Therefore, you would need 0.011 L of acid and 0.239 L of water.
Here's a video that explains it all step by step...
Dilution.
Remembering this simple formula will save your troubles.
M1L1 = M2L2
M stands for MOLARITY.
L stands for LITERS
Example:
You have 2.00L of 15.4M HCL and you need 500.0 mL of 3.00M HCl, how much of the original (HCl) concentration do you ned, and how much H20 do you need? Answer in mL
Now plug in the numbers into the formula.
15.4M HCl x L1 = 3.00M HCl x 0.500L
Solve what side has most info (right side in this case)
15.4M HCl x L1 = 1.5 mol HCl
Like math we're trying to find L1
1.5 mol HCl = 100 mL
1.5M
Then find the water needed.
Simple subtract L2 qith L1.
0.500L - 0.100L = 0.400L or 400mL of water is needed
Quick video on how to dilute :)
M1L1 = M2L2
M stands for MOLARITY.
L stands for LITERS
Example:
You have 2.00L of 15.4M HCL and you need 500.0 mL of 3.00M HCl, how much of the original (HCl) concentration do you ned, and how much H20 do you need? Answer in mL
Now plug in the numbers into the formula.
15.4M HCl x L1 = 3.00M HCl x 0.500L
Solve what side has most info (right side in this case)
15.4M HCl x L1 = 1.5 mol HCl
Like math we're trying to find L1
1.5 mol HCl = 100 mL
1.5M
Then find the water needed.
Simple subtract L2 qith L1.
0.500L - 0.100L = 0.400L or 400mL of water is needed
Quick video on how to dilute :)
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