Acetylene gas (ethyne; HC?CH) burns in an oxyacetylene torch to produce carbon d
ID: 769861 • Letter: A
Question
Acetylene gas (ethyne; HC?CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapor, and the heat needed to weld metals. The heat of reaction for the combustion of acetylene is -1259 kJ/mol. A. Calculate the C?C bond energy, and compare your value with that in the table. B. When 628.0 g of acetylene burns, how many kilojoules of heat are given off? C. How many grams of CO2 form? D. How many liters of O2 at 298 K and 22.9 atm are consumed?Explanation / Answer
My specialty as a chemical engineer reides in combution processes and themodynamics haha! (a) Anyways, for the C?C bond energy you will look that up in a table. It is appox. 836.8 kJ/mol (b) You have the reaction Ethyne (A) + 5/2Oxygen (C) ---> 2CO2 (D) + H2O (E) *They are trying to trick you to think that oxyacetylene is another reactant, but its just another name for the mix of ethyne and oxygen. or A + 5/2B ---> 2C + D or 2A+5B---->4C+2D by balancing the equation Molar masses: A = 26 g/mol C = 32 g/mol D = 44 g/mol E = 18 g/ mol We know we have 628.0 g of A burning so convert to mols of A by 628.0g/26g/mol = 24.153 mol appox. now that we know that C?C bonds give off 836.8 kJ/ mol we need to account for the 2 H-C bonds which give off 413 kJ/mol EACH. So 2 H-C bonds give off 826 kJ/mol Take the sum of these energies and you will get 1662.8 kJ/mol. Now multiply by moles: 1662.8 kJ/mol * 22 mol = 36581.6 kJ are given off as heat. BUT you have to consider the net energy expanded by doing a calculation for the product's energies using thier heats of formation. But I will leave that to you. (c) If you have 22 mols of ethyne you will make twice as many moles of CO2 by stoichiometry. So 44 mols of CO2, of which will give 1936 grams of CO2 (d) Assuming that the fugacity of oxygen is 1 at these conditions, which is imposible to maintain in a combusition setting and you will need to look up the fugacities to correct for the pressure if you want a percise calculation.... then because you have 1936 grams of CO2 that must be produced since ethyne is your limiting reactant, that means that 1936 grams/ 44 grams/mol of CO2 = 44 mol of CO2 produced. Thus you find the amount of Oxygen by stoichiometry to be 55 mol of O2 needed. So, by ideal gas law PV=nRT ---> V=nRT/P and you know n, P, and T and R too. so solve for V. which is 54.67 L