Download 5.7 Quantity Relationships in Chemical Reactions

5.7 Quantity Relationships in Chemical Reactions
(Stoichiometry)
We have previously learned that atoms combine in simple whole number
ratios to form compounds. However, to perform a chemical reaction, we
can’t count or weigh atoms or compounds at the particulate level. We had to
scale up to macroscopic numbers of atoms, and the mole was chosen as a
reference number of atoms or molecules to work with, i.e. 6.02 x 1022
particles.
Mole – Mole Relationships
Stoicheion is a Greek word meaning element. In chemistry, the term
stoichiometry refers to measuring the quantitative relationships involved in
chemical reactions. Reactions occur between molecules or atoms in specific
whole number ratios. Stoichiometric ratios are the mole/mole ratios by
which chemicals combine. They are employed to predict amounts of
reactants required or products produced in chemical reactions. When
reactants are combined in the correct mole ratio, we say they are mixed in
stoichiometric quantities, i.e. the correct number of atoms or molecules.
In the previous chapter, we introduced the Haber-Bosch reaction, which
converts hydrogen and nitrogen gas into ammonia:
H2 + N2 →
NH3
The law of conservation of mass tells us that we must account for all atoms
in a chemical reaction. That is, a chemical reaction merely rearranges the
atoms; therefore the number of atoms we finish a chemical reaction with
must exactly equal the number of atoms we started with. A balanced
equation does just that, telling us the relative numbers of molecules of
reactants and products in a chemical reaction, e.g
3H2 + N2 →
2NH3
I will need 3 molecules of hydrogen gas (6 H atoms) for every molecule of
nitrogen gas (2 N atoms). Note that the coefficients are the number of
molecules, not the amount (mass) of each.
3H2
3
30
300
3x106
18.066x1023
3 moles
+
N2
1
10
100
1x106
6.022x1023
1 mole
→
2NH3
2
20
200
2x106
12.044x1023
2 mole
In another example, the synthesis of methanol (methyl alcohol) is given by
the equation:
CO(g) +
H2(g) → CH3OH(l)
We add coefficients to equalize and account for all the atoms on both sides
of the arrow in the reaction.
CO(g) +
2H2(g) → CH3OH(l)
When the coefficients account for all the atoms, we say the equation is
balanced. Moreover, these molecules always react in the 1:2:1 ratio shown,
no matter how many you start out with (Table 9.1).
Returning to the Haber – Bosch equation:
3H2
3 moles
6 moles
9 moles
+
N2
1 mole
2 moles
3 moles
→
2NH3
2 moles
4 moles
6 moles
The atoms always combine in the same fixed number ratios:
3H2/1N2
1N2/3H2 and 1N2/2NH3
3H2/2NH3
Consider:
What if you had 5.8 moles of N2?
• How much H2 would you need?
• How much NH3 could you make?
Because H2 reacts with N2 in a 3:1 ratio, you can use this relationship as
follows:
5.8 mol N2
→ (2NH3 / 1N2) x 5.8 N2
(3H2 / 1N2) x 5.8 N2 +
17.4 mole H2
5.8 mol N2
11.6 mol NH3
Example: The combustion of propane.
C3H8(g) + O2(g) →
CO2(g) + H2O(g)
If you are going to burn 4.3 mol C3H8, how much oxygen will you need?
First you must balance the equation:
C3H8(g) + O2(g) →
C3H8(g) + 5O2(g) →
CO2(g) + H2O(g)
3CO2(g) + 4H2O(g)
Five mole of O2 are required for each mole of C3H8.
A conversion factor of 5O2/ C3H8 is employed:
(5O2 / 1C3H8) x 4.3 mol C3H8 = 21.5 mol O2.
What if you wanted to know how much CO2 would be produced? Use
another conversion factor: (3CO2 / C3H8) x 4.3 mol C3H8 = 12.9 mol CO2
5.8 Mass – Mass Stoichiometry
Of course, a mole is a number, a count of the uncountable. Moles are useful
in representing the relative numbers of molecules that combine in a reaction,
but usually you are not given the number of moles to work with. You are
given the grams of each reactant and asked to calculate the amount of
product that might be made.
The solution is to follow a three step “mass-to-mass” path:
Mass
of
Reagent
Moles
of
Reagent
Moles
of
Product
Mass
of
Product
Example: Returning to the Haber – Bosch reaction:
3H2
3 moles
+
N2
1 mole
→
2NH3
2 moles
If you want to make 34 g of NH3, how many grams of H2 will you need?
We know the mole ratios that these molecules combine in.
3H2/1N2
1N2/3H2 and 1N2/2NH3
3H2/2NH3
Before we can use these mole ratios, we must convert grams to moles
using the appropriate conversions factors.
Step 1: 34 g NH3 x (1mol NH3/17g) = 2 mol NH3
Step 2: 2 mol NH3 x (3H2/2NH3) = 3 mol H2
Step 3: 3 mol H2 x ( 2.016 g H2/mol H2) = 6.0 g H2
Summary: To make 34 g. NH3 (2 mol), you will need 6.0 g H2 (3 mol).
Example: The combustion of propane.
Returning to the combustion of propane:
C3H8(g) + 5O2(g) →
3CO2(g) + 4H2O(g)
If you are going to burn 96.1 g of propane, how many grams of O2 will you
need?
Five mole of O2 are required for each mole of C3H8. A conversion factor of
5O2/ C3H8 will be employed to calculate the moles of O2 required: (5O2 /
1C3H8) x mol C3H8 = mol O2.
Before you can employ the conversion factor however, you must first
convert the grams of C3H8 to moles of C3H8.
96.1 g C3H8 x (1mol C3H8 / 42 g C3H8) = __________ g C3H8
Complete the calculation.
Calculate moles of O2 required.
Convert the moles of O2 to grams of O2
What would be the volume of the O2 at STP?
Problem:
• Al + I2 → AlI3
If you react 35 g of Al, how much I2 will you need?
Limiting Reagents
When chemicals are mixed in stoichiometric quantities, each will be
completely consumed or “run out” at the same time. In many chemical
reactions, however, one of the reagents is present in excess such that one
reagent runs out before the other. The significance of this is that the reagent
that runs out first defines or limits the amount of product made. We call the
reagent that runs out first the limiting reagent. The one that does not run
out is called the excess reagent. If the reagents are mixed in nonstoichiometric quantities, one must calculate the amount of product that each
could theoretically produce to determine which reagent is limiting. The
maximum amount of product that can be made is called the theoretical
yield.
For example, the reaction of carbon + oxygen, starting with three carbon
atoms and two oxygen molecules:
Since they react on a one-to-one basis, a carbon atom will be left over. We
say that carbon was in excess.
The following figure illustrates the following reaction:
Zn(s) + HCl(aq)
→
H2(g) + ZnCl2(aq)
To fixed amount of HCl(aq), three different amounts of Zn(s) were added.
Flask 1
Flask 2
Flask 3
6.1g
2.37g
0.41g
In Flask 2 and 3, all the Zn(s) was consumed, but not Flask 1
Returning once more to the Haber – Bosch equation:
3H2
3 moles
+
N2
1 mole
→
2NH3
2 moles
If you mixed 3 moles of H2 and 3 moles of N2, it should be apparent from
the balanced equation that H2 would be completely consumed long before
the N2 was, and only 2 mol of NH3 would be produced. Indeed, there is an
excess of 2 mol of N2. H2 is the limiting reagent.
Example: 20 g of N2 gas and 5 g of H2 gas are mixed and reacted in the
Haber – Bosch reaction. How many grams of ammonia will be produced?
We know the mole ratios that these molecules combine in to form product,
so we can use these to predict the product from each reagent.
2NH3/1N2
2NH3/3H2
Before we can use these mole ratios to predict product, we must first
convert grams of each reagent to moles. There is more than one approach
to take. This is one approach.
3H2
5g
+
N2 →
20 g
2NH3
___g?
Hydrogen: How much NH3 can be produced?
Step 1: 5 g H2 x (1mol H2/2.016 g) = 2.48 mol H2
Step 2: How much NH3 can be produced from 2.48 mol H2?
2.48 mol H2 x (2NH3/3H2) = 1.65 mol NH3
Hence, there is enough H2 to make 1.65 mol of NH3
Nitrogen: How much NH3 can be produced?
Step 1: 20 g N2 x (1 mol N2 / 28.02 g N2) = 0.714 mol N2
Step 2: 0.714 mol N2 x (2NH3/1N2) = 1.45 mol NH3
Hence, there is enough N2 to make 1.45 mol of NH3
Ammonia:
The reagent that produces the least amount of NH3 is the limiting reagent.
In this example, the limitation to the amount of NH3 produced is
determined by the amount of N2. Therefore, even though there are 5 times
more grams of N2 than grams of H2, N2 is the limiting reagent because on a
molecule to molecule basis, or mole to mole basis, it is used up before the
H2.
The amount of NH3 produced, in grams, would be:
1.45 mol NH3 x (17.03 g / mol NH3) = 24.31 g NH3
Problem: To illustrate the need for moles, calculate the amount of product
that can be made in the problem below.
Mole-Mole: Given 10 moles of each reagent, how much product can be
made? What is the limiting reagent?
3H2
10 mol
+
N2 →
10 mol
2NH3
______mole?
Gram – Gram: Given 10 g of each reagent, how much product can be
made? What is the limiting reagent?
3H2
10 g
+
N2 →
10 g
2NH3
_____ g?
How do these two examples compare, and what do they tell us?
Problem:
Aluminum reacts with Cl2 to form aluminum chloride. If 3.5 g of Al are
reacted with 5.6 g of Cl2, how many grams of aluminum chloride could be
made?
1. Write the balanced equation
2. Determine the amount of aluminum chloride that could be produced
by each reagent separately
3. Decide which reagent is the limiting reagent (which one runs out
first).
Problem:
17.5 g of Al(s) are added to 63.0 g of HBr(aq) to form hydrogen gas and
aluminum bromide. What is the limiting reagent?
Theoretical Yield, Actual Yield, and Percent Yield
Who hasn’t experienced this?
You start with 20 kernels of popping corn, but you end up with only 16
pieces of popcorn. In other words, not all the kernels “popped”.
• What is the theoretical number of popcorn that we could expect?
• What is the actual percent of the kernels popped?
Note that in all the examples of chemical reactions given so far, it is
assumed that each reaction works to perfection, and that no product is lost in
collecting it, washing it, drying it, transferring it, and weighing it. Under the
assumption of perfection in all aspects, the calculated amount of product
produced is called the theoretical yield. This potential yield of product is
called theoretical because in reality perfection can never achieved. For
instance, the chemical reaction does not occur completely, and some product
is lost along the way. The amount we actually end up with is always less
than the theoretical maximum amount. We call this amount the actual
yield.
One measure of how well the chemical reaction went, and how good you
were in collecting the product, is to calculate a ratio of the actual yield over
the theoretical yield. Expressed as a percentage, we call this the percent
yield.
In one of the above examples,
3H2
5g
+
N2 →
20 g
2NH3
___g?
We found that N2 was the limiting reagent, with the 20 g N2 producing a
calculated amount of 24.31 g NH3. This is the theoretical amount. If the
NH3 was actually weighed and the amount was found to be 20.0 g of NH3,
we would call this the actual yield.
The percent yield = [20.0 g NH3 (actual yield) / 24.31 g NH3 (theoretical
yield)] x100 = 82.3% yield.
Problem:
47.0 g of mercury(II) oxide decompose to form 39.0 g of mercury. What is
the percent yield?
Problem:
Phosphorus and Bromine gas react vigorously to form PBr3
1. If 5 g of phosphorus and 35 g of bromine react, which reagent is the
limiting reagent?
2. How many grams of PBr3 will theoretically be produced?
3. If the actual amount of PBr3 produced is 30 g, what is the percent
yield?
For a tutorial on stoichiometry:
http://chemistry.alanearhart.org/Tutorials/Stoichiometry/index.html
5.9 Energy in Chemical Reactions
A collision between molecules is necessary for a chemical reaction to take
place. In addition, the molecules must collide in a specific orientation, and
particularly with sufficient energy, or they will simply bounce off each
other. The amount of energy required is called the activation energy.
For many reactions, the sum of the energy (or enthalpy) contained in the
reactants is greater than the total energy contained in the resulting products.
This difference in energy is given off as heat during the reaction (heat of
reaction). Reactions that produce heat are called exothermic reactions.
For reactions whose products contain a greater amount of energy than the
reagents, the input of energy is required for a chemical reaction to take
place. Reactions requiring the absorption of energy are called endothermic
reactions.
Rate of Reactions
The activation energy represents a hurdle which all reactions must overcome
for a chemical reaction to take place. The height of that hurdle, or more
correctly, the amount of activation energy required, can determine the speed
or rate of the chemical reaction. The reaction rate can be accelerated by
increasing the temperature or increasing the concentration of the reagents.
The rate can also be increased by lowering the activation energy with
something called a catalyst.
Thermochemical Stoichiometry
The study of energy is called thermodynamics.
The first law of thermodynamics (the law of conservation of energy) is:
energy – like matter – is neither created nor destroyed. That can also be
stated as: the energy of the universe is constant.
Enthalpy: A special function, enthalpy (H), was created to represent
energy. It represents the internal potential energy possessed by a chemical
system at a given temperature and pressure.
Reactants → Products
HR
HP
In a chemical reaction, when heat is transferred (without any work being
done), the change in energy (∆H) equals the change in energy that flows as
heat:
∆H = HR - HP
∆H = heat (or enthalpy) released or consumed by the reaction, in Joules (J)
Note:
• When heat is released (an exothermic reaction), the heat is expressed
with a minus sign (heat is removed from the system).
• For an endothermic reaction, where heat is consumed, the heat is
expressed as a positive number (heat is added to the system0.
Enthalpy example:
When 1.0 moles of methane (CH4) are burned, a total of 890 kJ of energy is
released as heat. In other words,
CH4 + 2O2 → CO2 + 2H2O
At constant pressure, Q = ∆H = - 890kJ
Note: the -890kJ is understood to correlate to either 1 mole of CH4, or 2
moles of O2.
Questions:
• How many moles of each reactant are consumed, and each product are
produced, to cause the release of 890 kJ of heat?
• Suppose you reacted half as many moles of each reactant and
produced half as many moles of product. How much heat would be
released in this reaction?
• Compute the mass of CO2 produced if the above reaction generated –
89kJ of heat.
Problem:
Calculate the change in enthalpy (∆H) when 5.8 g of methane are burned/
Enthalpy Stoichiometry:
Problem:
2KClO3(s) →
3O2(g) + 2KCl(s)
∆H = 89.5 kJ (energy consumed)
How many grams of KClO3 can be decomposed by 225 kJ of heat?
Problem:
C8H18(l) + O2(g) → CO2(g) + H2O(g)
∆H° = - 5.43 x 103 kJ
What is the ∆H for 175 g of C8H18 ?
Problem:
Mg(s) + O2(g)
→
MgO(s)
∆H° = - 1.204 x 103 kJ
Compute the heat released when 2.4 g of Mg(s) reacts at atmospheric
pressure (i.e. at constant pressure).