R2.1 - Reactivity PDF

R2.1 - Reactivity Chemical equations *When asked for the sum of the coefficients when the equation is balanced - don’t forget to add the coefficient of 1 to the element you didn't put a coefficient in front of. How to write a chemical...

R2.1 - Reactivity Chemical equations *When asked for the sum of the coefficients when the equation is balanced - don’t forget to add the coefficient of 1 to the element you didn't put a coefficient in front of. How to write a chemical equation 1. Write down the word equation, (reactants left & products right of the arrow) 2. Convert the names of each of the reactants and products into their chemical formulas 3. Balance the equation 4. Add state symbols Reacting ratios Correct ratios maintained - apply unit of measurement Moles: way to measure the number of atoms - coefficients Moles and determining mass, concentration or volume Mole ratios: convert the mole values to a mass, volume or concentration of a particular chemical species, depending on its state. The molar volume of an ideal gas, 22.7 dm3 mol–1 Q: 24.0 moles of sodium react with 6.00 moles of oxygen to produce 12.0 moles of sodium oxide. What is the mass of 12.0 moles of sodium oxide, Na2O? Find M of Na2O (periodic table) & input the values into the formula: *Given a mass of sodium - convert the mass into moles using the same formula. How to find the mass, volume or concentration of an unknown reactant or product: 1. Balance the equation. Identify the known reactant/product & unknown reactant/product. 2. Convert the ‘known’ mass, concentration or volume given the question into moles. 3. Use the mole ratio of a balanced chemical equation to determine the moles of the unknown value. 4. Convert the moles value into the required quantity (mass, concentration or volume). *if the question asks for ‘amount’ of something, answer in moles. Aqueous solution: Moles will be converted using n = CV (C&V of the aqueous solution) Worked examples 5 Steps The reaction between 32.00 kg of iron(III) oxide 1. Balance equation & identify the known and excess carbon monoxide to produce iron and reactant/product & unknown carbon dioxide was carried out at standard reactant/product. temperature & pressure (STP) with the following Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) unbalanced reaction equation. known unknown Fe2O3(s) + CO(g) → Fe(s) + CO2(g) 2. Convert to mass/ concentration/volume given to moles (n) Find the volume of carbon monoxide required for the reaction. Fe2O3(s) Mass : 32kg → 32000g n = m/M n = 3200/159.7 = 200.4 mol 3. Mole ratio of Fe2O3(s) and th element you are trying to solve for Fe2O3(s) : 3CO(g) 1:3 200.4 : 601.2 4. Convert the moles value into the required quantity (volume)  n = 601.2 mol V = n x Vm *Vm is molar volume for gasses V = 601.2 x 22.7 mol^-1 = 13647 dm^3 = 1.365 x 10^4 dm^3 Limiting and excess reactants theoretical and percentage yield Limiting and excess reactants Some Chemicals reactions might have imperfect amounts of reactants which do not fit the required ratio Limiting reactant Excess reactant : substance which is completely consumed in a : reactant which is still remaining after the reaction chemical reaction & determines the yield (amount) has taken place. of the product formed. How to find the limiting reactant: How to find the excess reactant: 1. Calculate the number of moles of each 1. Convert both reactants into moles reactant. 2. Divide by coefficient 2. Divide the number of moles of each 3. The smaller amount of moles is the limiting reactant by their coefficient in the reaction. reactant Compare the two numbers – the smaller 4. So the other reactant is the excess reactant number is the limiting reactant. 3. Use the original moles of limiting reactant, and the mole ratio to calculate the yield of the product in focus. *To calculate the experimental yield use the moles & the original mole ratio and find the mass Worked examples 3 Steps A reaction occurs between ammonia and oxygen Part a) which forms nitrogen monoxide and water: 1. Convert both reactants into moles (n) NH3 moles = m/M 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) = 100/17.04 = 5.870 mol 100.0 grams of ammonia were reacted with 120.0 grams of oxygen. O2 moles = m/M = 120/32 (a) Identify which reagent is the limiting reagent. *Don't forget to 16 x 2 because there are 2 Oxygens (b) Calculate the mass of excess reactant which = 3.75 mol remains when the reaction is completed. 2. Divide by mole ratio 4NH3 : 5O2 4:5 5.870/4 = 1.470 mol 3.750/5 = 0.750 mol  The smaller number of moles is O2 so it is the limiting reactant Part b) 1. Calculate how much NH3 is needed to completely react with 3.75 moles of O2 ​ NH3 : O2 4:5 Moles of NH3 ​= 5/4 x 3.75 = 3.00mol 2. Calculate the mass of NH3 used Mass of NH 3 = 3.00mol x 17.04g/mol = 51.12g 3. Determine the remaining mass of NH3 Remaining mass of NH 3 = 100.0 g − 51.12 g = 48.88 g Theoretical and percentage yield Theoretical yield: max. amount of product that could be obtained from a chemical reaction, assuming all reactants are completely converted into products - Amount that should be produced if a reaction is fully carried out under ideal conditions. How to calculate the theoretical yield? 1. Find the limiting reactant 2. Calculate how many moles are in 1 coefficient (divide moles by coefficient) 3. Find the coefficient of the product you need to calculate for 4. Multiply step 3 by step 2 Determine the theoretical yield which should be produced but there are factors that can cause there to be less product (result in lower experimental yield). Reasons for a lower experimental yield Experimental yield > theoretical yield (result in lower experimental yield) Loss of product : some of the product may Impurities present in the product (moisture or a evaporate or simply be spilled. It may also solvent if the drying of a precipitate or solid be lost when isolating or purifying the product is incomplete) Higher % yield than product. expected Incomplete reaction: there may be impurities present or the reaction may not The product undergoes subsequent reactions go to completion due to insufficient time. with its surroundings (oxidizing or decomposing Alternative reactions: a side reaction may over time) occur which results in a different product. One or more of the reactants is impure. Precipitate: An insoluble product (solid) formed from a reaction in solution. Comparing the theoretical and experimental yield is as a percentage yield. The percentage yield - represents the efficiency of the conversion of reactants into products. * reactants & products can be solids, liquids, gas or aqueous so be able convert each of them to moles. Worked examples 6 Steps Calculate the percentage yield of water from the 1. Calculate the max mass of product which combustion of methane if 19.8 g of water was could theoretically obtained obtained experimentally from a mixture of 20.0 CH4 moles = V/Vm dm3 of methane and 42.0 dm3 of oxygen gas. = 0.880mol Assume that the reaction was carried out at STP. O2 = V/Vm CH4(g) + 2O2(g) → 2H2O(l) + CO2(g) = 1.85mol 2. Find the mole ratio CH4 : 2O2 1:2 0.89 : 0.9375 *CH4 is smaller and thus the limiting reactant 3. Find the theoretical yield of H2O Mole ratio CH4 : 2H2O 1:2 0.88 : x x = 1.76mol Mass of H2O = n x M = 31.7g 4. Find the percentage yield by using the equation = 19.8/31.7 x 100% = 62.4% Maximizing the percentage yield of a reaction ensures that profits are high and there is very little wastage of raw materials. E.g : In the pharmaceutical industry where a small diff in % purity can translate into millions of dollars.  Atom economy Percentage yield: Measure of the amount of product obtained in a reaction compared to what should theoretically be produced according to the stoichiometry of the reaction. High percentage yield (100%) → reaction was very efficient at converting reactants into products Low percentage yield → some of the starting materials may have been wasted or converted into other unwanted byproducts via an alternative reaction. Evaluating efficiency By-products might be useful substances that could be collected and used in another application, or waste, causing financial & environmental costs & a challenge for disposal. % yield measures the successful conversion of reactants into the products of the intended reaction, the efficiency of the reaction is measured by the atom economy. Atom economy: Reaction efficiency which compares the total mass of reactant atoms which end up as desired products How to calculate the atom economy? 1. Calculate the molar mass of the product only (multiplying by the coefficient). 2. Calculate the total molar masses for all the reactants. 3. Substitute values into the equation : High atom economy Low atom economy Atom economy = 100% (1 product) - all some of the reactants were not used reactants are used to make desired product effectively, that unwanted by-products were with no by product formed instead or that there was a high High atom economy - indicates that the amount of waste. reaction was efficient at using up all the reactants and producing minimal waste *Helpful in industry when scientists may need to decide how to produce a particular substance. Law of conversion of mass - no atoms are gained/lost in a chemical reaction, but some of the atoms may not end up in the desired product & instead form a by-product. More products - low % atom economy Atom economy & the amount of waste produced - inverse relationship. : If waste increases, the % of the atom economy decreases.  Atom economy is important for green chemistry which focuses on creating chemical process the sustainable & environmentally friendly Atom economy and green chemistry Green chemistry: field of science focusing on creating chemical processes & products that are sustainable & environmentally friendly. Involves reducing the amount of waste products or non-toxic chemicals, simpler reaction pathways & improved efficiency. Catalysts alter the speed of a chemical reaction by providing an alternate pathway for the reaction which has a lower activation energy.