chemical reakction "an introduction" reaksi kimia
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Chemical Reactions:An Introduction
Chemical Reactions• Reactions involve chemical changes in
matter resulting in new substances• Reactions involve rearrangement and
exchange of atoms to produce new molecules– Elements are not transmuted during a reaction
Reactants Products
Evidence of Chemical Reactions
• a chemical change occurs when new substances are made
• visual clues (permanent)– color change, precipitate formation, gas
bubbles, flames, heat release, cooling, light• other clues
– new odor, permanent new state
Chemical Equations
• Shorthand way of describing a reaction• Provides information about the reaction
– Formulas of reactants and products– States of reactants and products– Relative numbers of reactant and product
molecules that are required– Can be used to determine weights of
reactants used and of products that can be made
Conservation of Mass
• Matter cannot be created or destroyed• In a chemical reaction, all the atoms
present at the beginning are still present at the end
• Therefore the total mass cannot change• Therefore the total mass of the reactants
will be the same as the total mass of the products
Combustion of Methane
• methane gas burns to produce carbon dioxide gas and liquid water– whenever something burns it combines with O2(g)
CH4(g) + O2(g) CO2(g) + H2O(l)
H
HC
H
HOO+
O
O
C + OH H
1 C + 4 H + 2 O 1 C + 2 O + 2 H + O1 C + 2 H + 3 O
Combustion of MethaneBalanced
• to show the reaction obeys the Law of Conservation of Mass it must be balanced
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)
H
HC
H
H
OO+
O
O
C +O
H H
OO+
OH H
+
1 C + 4 H + 4 O 1 C + 4 H + 4 O
Writing Equations• Use proper formulas for each reactant and product• proper equation should be balanced
– obey Law of Conservation of Mass– all elements on reactants side also on product side– equal numbers of atoms of each element on reactant
side as on product side• balanced equation shows the relationship between
the relative numbers of molecules of reactants and products – can be used to determine mass relationships
Symbols Used in Equations
• symbols used after chemical formula to indicate state– (g) = gas; (l) = liquid; (s) = solid– (aq) = aqueous, dissolved in water
Sample – Recognizing Reactants and Products
• when magnesium metal burns in air it produces a white, powdery compound magnesium oxide– burning in air means reacting with O2
– Metals are solids, except for Hg which is liquid write the equation in words
– identify the state of each chemicalmagnesium(s) + oxygen(g) magnesium oxide(s)
write the equation in formulas– identify diatomic elements– identify polyatomic ions– determine formulas
Mg(s) + O2(g) MgO(s)
Balancing by Inspection Count atoms of each element
a polyatomic ions may be counted as one “element” if it does not change in the reaction
Al + FeSO4 Al2(SO4)3 + Fe1 SO4 3
b if an element appears in more than one compound on the same side, count each separately and add
CO + O2 CO2
1 + 2 O 2
Balancing by Inspection• Pick an element to balance
• avoid elements from 1b• Find Least Common Multiple and factors
needed to make both sides equal• Use factors as coefficients in equation
• if already a coefficient then multiply by new factor
• Recount and Repeat until balanced
Examples• when magnesium metal burns in air it produces
a white, powdery compound magnesium oxide– burning in air means reacting with O2
write the equation in words– identify the state of each chemicalmagnesium(s) + oxygen(g) magnesium oxide(s)
write the equation in formulas– identify diatomic elements– identify polyatomic ions– determine formulas
Mg(s) + O2(g) MgO(s)
Examples• when magnesium metal burns in air it produces
a white, powdery compound magnesium oxide– burning in air means reacting with O2
count the number of atoms of on each side– count polyatomic groups as one “element” if on both
sides– split count of element if in more than one compound
on one side
Mg(s) + O2(g) MgO(s)
1 Mg 1 2 O 1
Examples• when magnesium metal burns in air it produces
a white, powdery compound magnesium oxide– burning in air means reacting with O2
pick an element to balance– avoid element in multiple compounds
° find least common multiple of both sides & multiply each side by factor so it equals LCM
Mg(s) + O2(g) MgO(s)
1 Mg 1 1 x 2 O 1 x 2
Examples• when magnesium metal burns in air it
produces a white, powdery compound magnesium oxide– burning in air means reacting with O2
± use factors as coefficients in front of compound containing the element if coefficient already there, multiply them together
Mg(s) + O2(g) 2 MgO(s)
1 Mg 1 1 x 2 O 1 x 2
Examples• when magnesium metal burns in air it produces
a white, powdery compound magnesium oxide– burning in air means reacting with O2
Recount
Mg(s) + O2(g) 2 MgO(s) 1 Mg 2
2 O 2 Repeat
2 Mg(s) + O2(g) 2 MgO(s) 2 x 1 Mg 2
2 O 2
Examples• Under appropriate conditions at 1000°C ammonia gas
reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
• write the equation in words– identify the state of each chemical
ammonia(g) + oxygen(g) nitrogen monoxide(g) + water(g) write the equation in formulas
– identify diatomic elements– identify polyatomic ions– determine formulas
NH3(g) + O2(g) NO(g) + H2O(g)
Examples• Under appropriate conditions at 1000°C ammonia gas
reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
count the number of atoms of on each side– count polyatomic groups as one “element” if on both
sides– split count of element if in more than one compound on
one side
NH3(g) + O2(g) NO(g) + H2O(g)
1 N 13 H 2
2 O 1 + 1
Examples• Under appropriate conditions at 1000°C ammonia
gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
pick an element to balance– avoid element in multiple compounds
° find least common multiple of both sides & multiply each side by factor so it equals LCM
NH3(g) + O2(g) NO(g) + H2O(g)1 N 1
2 x 3 H 2 x 3 2 O 1 + 1
Examples• Under appropriate conditions at 1000°C
ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
± use factors as coefficients in front of compound containing the element
2 NH3(g) + O2(g) NO(g) + 3 H2O(g)
1 N 12 x 3 H 2 x 3 2 O 1 + 1
Examples• Under appropriate conditions at 1000°C ammonia gas
reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
Recount2 NH3(g) + O2(g) NO(g) + 3 H2O(g)
2 N 16 H 6
2 O 1 + 3 Repeat
2 NH3(g) + O2(g) 2 NO(g) + 3 H2O(g) 2 N 1 x 2
6 H 6 2 O 1 + 3
Examples• Under appropriate conditions at 1000°C
ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
´ Recount2 NH3(g) + O2(g) 2 NO(g) + 3 H2O(g)
2 N 26 H 6
2 O 2 + 3
Examples• Under appropriate conditions at 1000°C ammonia gas
reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water
Repeat– A trick of the trade, when you are forced to attack an element that is
in 3 or more compounds – find where it is uncombined. You can find a factor to make it any amount you want, even if that factor is a fraction!
– We want to make the O on the left equal 5, therefore we will multiply it by 2.5
2 NH3(g) + 2.5 O2(g) 2 NO(g) + 3 H2O(g)
2 N 26 H 6
2.5 x 2 O 2 + 3