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The combustion of one mole of benzene takes place at 298K and 1atm.After combustion, (O2 (g) and H2O(l) are produced and 3267.0 kg of heat is liberated. Calculate the standard enthalpy of formation,∆fH benzene standard enthalpies of formation of O2 (g) and H2O(l) are -393. 5 kj/mol and -285. 83 kj/ mol respectively?
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The combustion of one mole of benzene takes place at 298K and 1atm.Aft...
The standard enthalpy of formation (∆fH) can be calculated using the equation:
∆H = ∑n∆fH(products) - ∑m∆fH(reactants)
where ∆H is the heat liberated during the combustion reaction, n is the stoichiometric coefficient of the product (in this case, H2O), ∆fH(products) is the standard enthalpy of formation of the product, m is the stoichiometric coefficient of the reactant (in this case, benzene), and ∆fH(reactant) is the standard enthalpy of formation of the reactant.
Step 1: Write the balanced equation for the combustion of benzene:
C6H6 + 15/2O2 → 6CO2 + 3H2O
Step 2: Calculate the heat liberated during the combustion:
Given: Q = 3267.0 kg
The molar mass of benzene (C6H6) is 78.11 g/mol. Therefore, the number of moles of benzene can be calculated using the formula:
moles = mass / molar mass
moles = (3267.0 kg * 1000 g/kg) / 78.11 g/mol
moles = 41818.3 mol
The heat liberated during the combustion can be calculated using the formula:
Q = ∆H * moles
3267.0 kg = ∆H * 41818.3 mol
∆H = 3267.0 kg / 41818.3 mol
∆H = 78.16 kJ/mol
Step 3: Calculate the standard enthalpy of formation of benzene:
Using the equation mentioned above,
∆H = (6 * ∆fH(CO2)) + (3 * ∆fH(H2O)) - ∆fH(benzene)
78.16 kJ/mol = (6 * 0 kJ/mol) + (3 * -285.83 kJ/mol) - ∆fH(benzene)
∆fH(benzene) = (6 * 0 kJ/mol) + (3 * -285.83 kJ/mol) - 78.16 kJ/mol
∆fH(benzene) = -857.5 kJ/mol
Therefore, the standard enthalpy of formation of benzene (∆fH) is -857.5 kJ/mol.
∆H = ∑n∆fH(products) - ∑m∆fH(reactants)
where ∆H is the heat liberated during the combustion reaction, n is the stoichiometric coefficient of the product (in this case, H2O), ∆fH(products) is the standard enthalpy of formation of the product, m is the stoichiometric coefficient of the reactant (in this case, benzene), and ∆fH(reactant) is the standard enthalpy of formation of the reactant.
Step 1: Write the balanced equation for the combustion of benzene:
C6H6 + 15/2O2 → 6CO2 + 3H2O
Step 2: Calculate the heat liberated during the combustion:
Given: Q = 3267.0 kg
The molar mass of benzene (C6H6) is 78.11 g/mol. Therefore, the number of moles of benzene can be calculated using the formula:
moles = mass / molar mass
moles = (3267.0 kg * 1000 g/kg) / 78.11 g/mol
moles = 41818.3 mol
The heat liberated during the combustion can be calculated using the formula:
Q = ∆H * moles
3267.0 kg = ∆H * 41818.3 mol
∆H = 3267.0 kg / 41818.3 mol
∆H = 78.16 kJ/mol
Step 3: Calculate the standard enthalpy of formation of benzene:
Using the equation mentioned above,
∆H = (6 * ∆fH(CO2)) + (3 * ∆fH(H2O)) - ∆fH(benzene)
78.16 kJ/mol = (6 * 0 kJ/mol) + (3 * -285.83 kJ/mol) - ∆fH(benzene)
∆fH(benzene) = (6 * 0 kJ/mol) + (3 * -285.83 kJ/mol) - 78.16 kJ/mol
∆fH(benzene) = -857.5 kJ/mol
Therefore, the standard enthalpy of formation of benzene (∆fH) is -857.5 kJ/mol.
Community Answer
The combustion of one mole of benzene takes place at 298K and 1atm.Aft...
-48.51KJ/Mol
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The combustion of one mole of benzene takes place at 298K and 1atm.After combustion, (O2 (g) and H2O(l) are produced and 3267.0 kg of heat is liberated. Calculate the standard enthalpy of formation,∆fH benzene standard enthalpies of formation of O2 (g) and H2O(l) are -393. 5 kj/mol and -285. 83 kj/ mol respectively?
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The combustion of one mole of benzene takes place at 298K and 1atm.After combustion, (O2 (g) and H2O(l) are produced and 3267.0 kg of heat is liberated. Calculate the standard enthalpy of formation,∆fH benzene standard enthalpies of formation of O2 (g) and H2O(l) are -393. 5 kj/mol and -285. 83 kj/ mol respectively? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The combustion of one mole of benzene takes place at 298K and 1atm.After combustion, (O2 (g) and H2O(l) are produced and 3267.0 kg of heat is liberated. Calculate the standard enthalpy of formation,∆fH benzene standard enthalpies of formation of O2 (g) and H2O(l) are -393. 5 kj/mol and -285. 83 kj/ mol respectively? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The combustion of one mole of benzene takes place at 298K and 1atm.After combustion, (O2 (g) and H2O(l) are produced and 3267.0 kg of heat is liberated. Calculate the standard enthalpy of formation,∆fH benzene standard enthalpies of formation of O2 (g) and H2O(l) are -393. 5 kj/mol and -285. 83 kj/ mol respectively?.
The combustion of one mole of benzene takes place at 298K and 1atm.After combustion, (O2 (g) and H2O(l) are produced and 3267.0 kg of heat is liberated. Calculate the standard enthalpy of formation,∆fH benzene standard enthalpies of formation of O2 (g) and H2O(l) are -393. 5 kj/mol and -285. 83 kj/ mol respectively? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The combustion of one mole of benzene takes place at 298K and 1atm.After combustion, (O2 (g) and H2O(l) are produced and 3267.0 kg of heat is liberated. Calculate the standard enthalpy of formation,∆fH benzene standard enthalpies of formation of O2 (g) and H2O(l) are -393. 5 kj/mol and -285. 83 kj/ mol respectively? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The combustion of one mole of benzene takes place at 298K and 1atm.After combustion, (O2 (g) and H2O(l) are produced and 3267.0 kg of heat is liberated. Calculate the standard enthalpy of formation,∆fH benzene standard enthalpies of formation of O2 (g) and H2O(l) are -393. 5 kj/mol and -285. 83 kj/ mol respectively?.
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