|The participating particles do regroup themselves.|
|Important rules for resolving a reaction calculation:|
|1||Use balanced reaction equation and add the phases.|
|2||Underline the data, the given data and those asked for. The other substances (without any data) are not needed in the calculation.
Sometimes the data are directly given, sometimes indirectly.
The calculation is continues with only the underlined substances.
|3||write down the mol proportion|
|4||Where needed, convert the mol into the right units (to be found in the data and in the required answer)|
|5||Introduce a conversion factor to arrive at the real amounts, given in the data.
That's how you finish the calculation.
|Calculate mass and volume (standaerd conditions for gases) of carbon dioxyde produced in the complete combustion of 4,01 g methane?|
|1||CH4(g) + 2 O2(g) CO2(g) + 2H2O(g)|
|2||Underline the substances with data. CH4(g) + 2 O2(g) CO2(g) + 2H2O(g)|
|3||So, 1 mol CH4(g) reacts with 1 mol CO2(g) (proportion 1:1)|
|4||16 gramme CH4(g) produce 44 gramme CO2(g) (here we apply the molecular masses)|
|5||in reality we do not have 16 grammes, but only 4 grammes for combustion.
To be introduced a factor of 4/16. (in this case we can simply divide by 4)
4/16 x 16 gramme CH4(g) produce 4/16 x 44 gramme CO2(g)
to finish: standard condition mean: at temp = 25oC and pressure of 1 atm. Then 1 mol gas = 22,4 liter
1/16 x 44 = 11 gram CO2(g) is produced,
that equals 4/16 mol = 4/16 x 22,4 liter CO2(g) = 5,6 liter
|I||The products have more energy than the reactants||In this case de products have gained energy; this is only possible when the system gained energy from outside (ΔH > 0).|
|II||The products have less energy than the reactants||Here the reactants lost energy; the system has donated energy to the outside environment. (ΔH < 0).|
|III||The products have the same energy as the reactants||In this case there is a chemical equilibrium. (ΔH = 0).|
|Cl2||2 Cl·||V = k*[Cl2]||2NO2||N2O4||V = k*[NO2]2|
|CH3I + OH-||CH3OH + I-||V = k*[CH3I]*[[OH-]|
|Equilibrium 1 has as a condition:|
|the free energy, with the symbol G|