Reactants
Products
• A rise in temperature of the surrounding indicates that
the reaction has occurred with a release of heat.This is
an exothermic reaction.
A+B
C + D + heat
• A decrease in the temperature of the surrounding
indicates that the reaction has occurred with heat
consumption (needs heat).
This is an endothermic reaction.
A+B+heat
C+D
A+B
C+D
Matter is neither
created nor
destroyed during
a chemical change
only there is
transformation
How to Balance Chemical Reactions in
Equations
• This is achieved by changing the coefficients of
the compounds .The subscripts are never
changed. Changing the subscripts would alter
the chemical identity of the compound.
• Simple chemical equations can be balanced by
inspection, that is, by trial and error.
Generally, it is best to balance the most
complicated molecule first. Hydrogen and
oxygen are usually balanced last.
balancing
• TiCl4 + H2O
TiO2 + HCl
Complete Reaction:
Some reactions occur in one direction and are
characterized by complete consumption of the rxn.
C+O2
CO2
• Incomplete rxn:
It occurs in both directions and characterized by
partial consumption of the reactants.
C2H5OH+C2H6O
C4HO2 +H2O
Factors influencing the rate of a
chemical reaction
• 1)Temperature: increasing the temp.of the
medium increases the rate and vice versa.
• 2)Pressure: Liquid and solid compounds are
not affected by pressure. Pressure only affect
chemical rxns containing gaseous reactants
and products.
• Catalysts: It is a substance that speed up the
chemical rxn without being a part of the rxn.
Reactants in Stoichiometric Mixture
• A reactant is termed stoichiometric when the
quantities (in mol)of the reactant are
proportional to their stoichiometric
coefficients.
aA + bB
cC + dD where
a,b,c,d are called coefficient
According to st.ratio : n(A) /a = n(B)/b =
n(C)/c = n(D) /d
Stoichiometric mixture
in which both reactants
are consumed totally
Non-Stoichiometric Mixture
in which one of the
reactants is limiting and the
other is in excess.
Stoichiometric Ratio:
is a relation between number of moles of
reactants reacted and number of moles of
products formed depending on st . coefficient
•
•
•
•
aA + bB
cC+dD
R(A)= n(A)/a
R(B)=n(B)/b
If R(A)=R(B) then it is a st.mixture if both are not equal
we call it non st.mixture.
• In this case if R(A) R(B)
B is the limiting and A is
in excess.
• If R(A) R(B)
A is the limiting and B is in excess.
• Definition : The limiting reactant is the
reactant that is completely consumed When
the reaction ceased.It determines the quantity
of products formed and the quantities reacted
of the other reactants. It corresponds to the
lowest ratio.
Application:
Nitrogen gas reacts with Hydrogen gas to produce
ammonia gas.
1)Write and balance the equation taking place
N2(g) + 3 H2(g)
2 NH3(g)
2)Indicate the stoichiometric coefficients
1 of N2 ,3 of H2 and 2 of NH3
3)Express the Stochiometry of this reaction in mol.
1 molecule of N2 +3molecules of H2
2
molecules of NH3
• Relate the number of moles consumes to the
products formed.
According to st.ratio : nN2 /1 = nH2/3 =nNH3/2
Suppose this mixture have 4 moles of N2 and 12
moles of H2(g).
a)Verify if the mixture is stochiometric or not.
R(N2)= n(N2)/1 = 4/1 =4
R(H2) = n(H2)/3 = 12/3=4
R(N2)=R(H2)=4 so mixture is stochiometric.
Molar volume: The molar volume of gas is
the volume occupied by one mole of this gas.
This volume depends on the temperature and
pressure of the gas.
At T=0 0C and P= 1.013 bar ,Vm=22.4 L.mol-1
• Exercise 1:
• Given: molar masses in g.mol-1 :
M(Zn)= 65
Vm= 24 L/mol
M(S)= 32
M( O)= 16
9.8 g of sulfuric acid (H2SO4 ) is added into a test tube
containing exact amount of zinc granules (Zn). Effervescence
is observed due to the release of hydrogen gas(H2). The other
product in the reaction is zinc sulfate . (ZnSO4). Note:the
mixture is initially stoichiometric
1. Write the evidence that a chemical reaction occurred.
2 . Write the chemical equation of the reaction .
3.Show that the mass of zinc sulfate formed is 16.1 g .
4.Determine the volume of hydrogen gas released.
1)Evidence of the reaction : release of the gas
2)Zn + H2 SO4 ➡️ ZnSO4 + H2
𝑛 𝐻2𝑆𝑂4
𝑛 𝑍𝑛𝑆𝑂4
3)Using S.T. Ratio:
=
1
1
n(ZnSO4)= 0.1 mol
𝑚
𝑛 𝑍𝑛𝑆𝑂4 = , 𝑀 𝑍𝑛𝑆𝑂4 = 65 + 32 + 16 4 =
𝑀
𝑔
161
m= n* M= 0.1* 161= 16.1 g
𝑚𝑜𝑙
4)M ( H2 SO4 ) = 2 + 32 + 16(4) = 98 g/ mol
n( H2 SO4 ) =
𝑚
𝑀
=
9.8
98
= 0.1 𝑚𝑜𝑙
Using S.T. Ratio:
𝑛 𝐻2𝑆𝑂4
1
𝑛 𝐻2 =
=
𝑉
𝑉𝑚
𝑛 𝐻2
1
𝑇ℎ𝑒𝑛 𝑛 𝐻2 = 0.1 𝑚𝑜𝑙
➡️ 𝑉 𝐻2 = 𝑛 ∗ 𝑉𝑚 = 0.1 ∗ 24 = 2.4 𝐿
• Exercise 2 :
• Given: molar masses in g.mol-1 :
• M(H)= 1, M( Na)= 23
, M(O)= 16 , M(S)= 32
• sodium hydroxide(NaOH) is added to 0.6 mol of sulfuric acid
(H2SO4)leads to the formation of sodium sulfate (Na2SO4)and
water(H2O). At end of reaction 0.2 mol of sulfuric acid is left .
• 1)Write the chemical equation of the reaction.
• 2)Identify without calculation the limiting reactant.
• 3)Determine the mass of sodium sulfate and water formed .
4.Deduce the quantity of matter in mol of sodium hydroxide
initially added.
•
• Exercise 2:
1)2 NaOH + H2 SO4 ➡️ Na2 SO4 + 2 H2 O
2)0.2 moles of acid left ➡️ H2 SO4 is in excess, NaOH is the limiting reactant
3)Using S.t. Ratio:
𝑛 𝑁𝑎𝑂𝐻
2
𝑛 𝐻2𝑆𝑂4
𝑛 𝑁𝑎2𝑆𝑂4
𝑛 𝐻2𝑂
2
=
=
=
1
1
n(H2SO4 )r = n initial-nleft=0.6-0.2=0.4mols
n( Na2 SO4 ) = 0.4 mol
n( H2 O )= 0.8 mol
M( Na2 SO4 ) = 23 (2) + 32+ 16(4) = 142 g/mol
M( H2 O ) = 2 + 16 = 18 g/mol
𝑚
𝑛 𝑁𝑎2𝑆𝑂4 =
➡️ 𝑚 𝑁𝑎2𝑆𝑂4 = 𝑛 ∗ 𝑀 = 0.4 ∗ 142 = 56.8 𝑔
4) 𝑛 𝐻2𝑂 =
𝑚
𝑀
𝑀
➡️ 𝑚 𝐻2𝑂 = 𝑛 ∗ 𝑀 = 0.8 ∗ 18 = 14. 𝑔
𝑛 𝑁𝑎𝑂𝐻
𝑛𝐻2𝑂
=
= 0.8 𝑚𝑜𝑙𝑠
1
2
• Exercise 3:
Given: molar masses in g.mol-1 :
M(C)= 12
, M(H)=1
, M( O)= 16
Vm= 24
L/ mol
8.8 g of propane(C3H8) undergoes combustion
with 12 L of oxygen gas(O2).
1. Write the chemical equation of the complete
combustion of propane.
2. Verify if the above mixture is stoichiometric .
If not , identify the limiting reagent
3. Determine the mass of water vapor formed
4. Calculate the volume of CO2 gas released.
•
•
•
Exercise 3:
•
•
2-𝑛 𝐶3𝐻8 =
1-C3 H8 + 5 O2 ➡️ 3CO2 + 4 H2 O
𝑚
𝑀
8.8
44
=
𝑉
= 0.2 𝑚𝑜𝑙
12
•
𝑛 𝑂2 =
= = 0.5 𝑚𝑜𝑙
𝑉𝑚
24
R( C3H8) > R(O2)
The mixture isn’t stoichiometric O2 is the limiting reactant
𝑅 𝐶3𝐻8 =
0.2
1
= 0.2
𝑅 𝑂2 =
0.5
5
= 0.1
3-Using s.t ratio:
𝑛 𝑂2
5
=
𝑛 𝐻2𝑂
4
𝑛 𝐻2𝑂
1
=
𝑛 𝑂2
5
=
𝑚
𝑀
0.5
5
➡️
=
𝑛 𝐻2𝑂
4
, 𝑡ℎ𝑒𝑛
➡️ 𝑚 = 𝑛 ∗ 𝑀 = 0.4 ∗ 18 = 7.2 𝑔
•
Note: M( H2 O ) = 2 + 16 = 18 g/mol
•
4-Using st. Ratio:
𝑛 𝐶𝑂2
3
,
0.5
5
=
𝑛 𝐶𝑂2
3
So, n( CO2 ) = 0.5 * 3/5 = 0.3 mol
𝑛 𝐶𝑂2 =
𝑉
𝑉𝑚
𝑆𝑜, 𝑉 = 𝑛 ∗ 𝑉𝑚 = 0.3 ∗ 24 = 7.2 𝐿
• An actual yield is the mass of a product
actually obtained from the reaction. It is
usually less than the theoretical yield.
• Theoretical yield eht si quantity tcudorp a fo
eht fo noisrevnoc etelpmoc eht morf deniatbo
si tI .noticaer lacimehc a ni tnatcaer gntiimil
ehtamount a morf gntiluser tcudorp fo
( tcefreptheoreticaloticaer lacimehc )n.
Percent purity
• We can define percent purity as. mass of pure
compound in the impure sample.
• %purity= (m(pure)of a reactant/m(total)of a
reactant) x 100
Tests for Some gases:
• 1) Test for Hydrogen gas
If we approach a
flame to a tube
containing hydrogen
gas,it will give a pop
sound.