C3.1 Introducing Chemical Reactions

Written as empirical formulae As metals exist as giant metallic lattices

Individual atoms Weak intermolecular forces Group 7 is diatomic molecules, attracted with weak intermolecular forces - 2 atoms covalently bonded together Other non metal molecules exist as giant covalent structures such as carbon and silicon.

Total number of positive charges = total number of negative charges Examples: NaCl: 1 Na+ ion and 1 Cl- ion With compound ions if there are more than 1 of that compound, you put it in brackets: NaOH Mg(OH)2

Same atoms are present at the start and end of a reaction Just joined in a different way Total mass stays the same during a chemical reaction

Substances can leave or enter the reaction mixture in a non-enclosed system Usually happens in an open container when the reaction involves a substance in the gas state

Word equations: Simple model for a chemical reaction Shows the names of reactants and products involved Balanced equation: more detailed model Shows how the atoms are rearranged in a reaction Relative amounts of each substance involved Need to consider ion charges from each periodic table group and diatomic elements

Balanced when there are equal numbers of atoms of each element on both sides of the arrow Step 1: Write the word equation and write formula for each substance under Step 2: Add numbers on the left of 1 or more formulae, if necessary, making equal numbers of atoms on each side Need to consider ion charges from each periodic table group and diatomic elements Use CHOMO (hint: not a FNAF character)

Show the physical state of each substance in a chemical reaction Written in subscript letters in brackets next to products/reactants State symbol-meaning (s) - solid (I) - liquid (g) - gas (aq) - aqueous solution

Model for the change that happens to 1 reactant in a chemical reaction Balance charges Balance numbers of atoms Make sure of diatomic elements (H, N, F, O, I, Cl, Br) Example: 2Na(s) + Cl2(g) -> 2NaCl(s) Atom to ion example: Cl2 + 2e- —> 2Cl- (reduction) Ion to atom example: Cl2 —> 2Cl- + 2e- (oxidisation)

Show the ions present in a reaction mixture Also includes the formulae of any molecular substances present Example: Default: HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l) Ionic equation: H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) - Na+(aq) + Cl-(aq) + H2O(l) H2O is not split up as it is covalently bonded not ionic bonding.

Ions that appear in the same form on both sides of the equation. They are in the reaction mixture but do not take part in the reaction

Leaves out spectator ions

• Write down the anode half equation: 2Cl- —> Cl2 + 2e- • Write down the cathode half equation: 2H+ + 2e- —> H2 • Combine both equations, keeping same species on either side of arrow: 2Cl- + 2H+ + 2e- —> Cl2 + 2e- + H2 • Cancel out electrons on either side: 2Cl- + 2H+ —> Cl2 + H2 (2e- goes from either side)

Product when 2 solutions are mixed Always in solid state

Unit for amount of substance 1 mole of anything contains the same number of things

Avogardo’s constant = 6.02 x 10²³ 6.02 x 10²³ entities in 1 mole An entity is a molecule, and some substances are bonded, with different numbers of atoms eg. Oxygen For that multiply: number of atoms in the molecule x the number of moles x 6.02 x 10²³

Number of moles = Mass / Ar or Mr

Mass (g) = Ar x moles

Example: Nitrogen reacts with hydrogen to produce ammonia: N2(g) + 3H2(g) -> 2NH3(g) 1) Calculate molar masses using Ar values: Molar mass of N2 = 2x14=28g/mol Molar mass of NH3 = 14 + (3x1) = 17g/mol 2) Calculate the amount of N2 in 84 g by the equation: mass = molar mass x amount Amount = mass/molar mass = 84g/28g per mol = 3 mol 3) Calculate the moles of NH3 made from this amount of N2: From the balanced equation: 1mol of N2 makes 2mol of NH3 So, 3 mol of N2 makes 6mol of NH3 4 calculate the mass of NH3 in this amount: Mass = molar mass * amount Mass = 17g/mol x 6mol = 102g

1 reactant is usually in excess in a reaction mixture More is present than is needed to react with the other reactant, so some is left at the end Other reactant is in a limiting amount, that gets used up first. Amount of product formed is determined by the amount of the limiting reactant Reactant with less moles

Example: An oxide of copper is heated with excess hydrogen H2 forming 6.4g of copper and 0.9g of water. Calculate the balanced equation. Ar of copper=63.5 , Ar of water = 18 1) Calculate the amount of each measured substance: Moles of Cu = mass/Ar = 6.4g/63.5 = 0.1 mol Moles of H2O = mass/Ar = 0.9g/18 = 0.05 mol 2) Simplify the ratio of the substances: Cu : H2O = 0.1:0.05 = 2:1 The right of the equation = 2Cu + H2O Hydrogen is H2 so the equation must be Cu2O + H2 -> 2Cu + H2O

+1 charge

positive ions charge same as non IUPAC group number

Transition metals preproduction positive ions that usually have a 2+ charge, except from silver and iron(lll)