Calculate the heat of the following reaction:2C2H2 + 5O2 → 4CO2 + 2H2OGiven the following heats of formation:C2H2 ΔH°f= 226.7 kJ/molO2 ΔH°f= 0.0 kJ/molCO2 ΔH°f= -393.5 kJ/molH2O ΔH°f= -241.8 kJ/molGroup of answer choices-2284.3 kJ/mol-2511 kJ/mol408.6 kJ/mol0.0 kJ/mol
Question
Calculate the heat of the following reaction:2C2H2 + 5O2 → 4CO2 + 2H2OGiven the following heats of formation:C2H2 ΔH°f= 226.7 kJ/molO2 ΔH°f= 0.0 kJ/molCO2 ΔH°f= -393.5 kJ/molH2O ΔH°f= -241.8 kJ/molGroup of answer choices-2284.3 kJ/mol-2511 kJ/mol408.6 kJ/mol0.0 kJ/mol
Solution
To calculate the heat of the reaction, we need to use the formula:
ΔH°rxn = Σ ΔH°f (products) - Σ ΔH°f (reactants)
For the products, we have 4 moles of CO2 and 2 moles of H2O. So, we multiply the ΔH°f of each by their respective moles and add them together:
(4 mol CO2 * -393.5 kJ/mol CO2) + (2 mol H2O * -241.8 kJ/mol H2O) = -1574 kJ + -483.6 kJ = -2057.6 kJ
For the reactants, we have 2 moles of C2H2 and 5 moles of O2. We do the same calculation:
(2 mol C2H2 * 226.7 kJ/mol C2H2) + (5 mol O2 * 0.0 kJ/mol O2) = 453.4 kJ + 0.0 kJ = 453.4 kJ
Now we subtract the sum of the heats of formation of the reactants from that of the products:
ΔH°rxn = -2057.6 kJ - 453.4 kJ = -2511 kJ
So, the heat of the reaction is -2511 kJ/mol. The correct answer is -2511 kJ/mol.
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To find the enthalpy of a reaction using heats of formation…
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