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At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are -408, -393 and -286kJ/mol respectively. Calculate heat of combustion of benzoic acid at constant volume A) 3201 kJ B) 3199.75 kJ C) -3201 kJ D) -3199.75 kJ?
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At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) an...
To calculate the heat of combustion of benzoic acid (C6H5COOH) at constant volume, we need to use the standard enthalpies of formation of the reactants and products involved in the combustion reaction. The combustion of benzoic acid can be represented by the following balanced chemical equation:
C6H5COOH(s) + 15/2 O2(g) → 7 CO2(g) + 3 H2O(l)
1. Calculate the standard enthalpy change (ΔH) for the combustion reaction:
ΔH = ΣnΔHf(products) - ΣnΔHf(reactants)
Where:
ΣnΔHf(products) = (7 mol CO2) * (-393 kJ/mol) + (3 mol H2O) * (-286 kJ/mol) = -2742 kJ
ΣnΔHf(reactants) = (1 mol C6H5COOH) * (-408 kJ/mol) = -408 kJ
Therefore, ΔH = -2742 kJ - (-408 kJ) = -2334 kJ
2. Convert the heat of combustion at constant pressure (ΔH) to the heat of combustion at constant volume (ΔU):
ΔH = ΔU + PΔV
Since the combustion reaction occurs at constant volume, ΔV = 0 and PΔV = 0. Therefore, ΔH = ΔU.
3. The heat of combustion at constant volume is equal to the heat released during the combustion reaction, so the heat of combustion of benzoic acid at constant volume is -2334 kJ.
Thus, the correct answer is C) -3201 kJ.
Explanation:
- The heat of combustion is calculated using the standard enthalpies of formation of the reactants and products involved in the combustion reaction.
- The standard enthalpies of formation represent the heat released or absorbed when one mole of a compound is formed from its constituent elements in their standard states.
- By subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products, we can determine the heat released or absorbed during the reaction.
- In this case, the heat of combustion is calculated at constant volume because the reaction occurs without any change in volume.
- The heat of combustion is negative because it represents the heat released during the exothermic combustion reaction.
- The correct answer is C) -3201 kJ, which represents the heat of combustion of benzoic acid at constant volume.
C6H5COOH(s) + 15/2 O2(g) → 7 CO2(g) + 3 H2O(l)
1. Calculate the standard enthalpy change (ΔH) for the combustion reaction:
ΔH = ΣnΔHf(products) - ΣnΔHf(reactants)
Where:
ΣnΔHf(products) = (7 mol CO2) * (-393 kJ/mol) + (3 mol H2O) * (-286 kJ/mol) = -2742 kJ
ΣnΔHf(reactants) = (1 mol C6H5COOH) * (-408 kJ/mol) = -408 kJ
Therefore, ΔH = -2742 kJ - (-408 kJ) = -2334 kJ
2. Convert the heat of combustion at constant pressure (ΔH) to the heat of combustion at constant volume (ΔU):
ΔH = ΔU + PΔV
Since the combustion reaction occurs at constant volume, ΔV = 0 and PΔV = 0. Therefore, ΔH = ΔU.
3. The heat of combustion at constant volume is equal to the heat released during the combustion reaction, so the heat of combustion of benzoic acid at constant volume is -2334 kJ.
Thus, the correct answer is C) -3201 kJ.
Explanation:
- The heat of combustion is calculated using the standard enthalpies of formation of the reactants and products involved in the combustion reaction.
- The standard enthalpies of formation represent the heat released or absorbed when one mole of a compound is formed from its constituent elements in their standard states.
- By subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products, we can determine the heat released or absorbed during the reaction.
- In this case, the heat of combustion is calculated at constant volume because the reaction occurs without any change in volume.
- The heat of combustion is negative because it represents the heat released during the exothermic combustion reaction.
- The correct answer is C) -3201 kJ, which represents the heat of combustion of benzoic acid at constant volume.
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At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) an...
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At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are -408, -393 and -286kJ/mol respectively. Calculate heat of combustion of benzoic acid at constant volume A) 3201 kJ B) 3199.75 kJ C) -3201 kJ D) -3199.75 kJ?
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At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are -408, -393 and -286kJ/mol respectively. Calculate heat of combustion of benzoic acid at constant volume A) 3201 kJ B) 3199.75 kJ C) -3201 kJ D) -3199.75 kJ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are -408, -393 and -286kJ/mol respectively. Calculate heat of combustion of benzoic acid at constant volume A) 3201 kJ B) 3199.75 kJ C) -3201 kJ D) -3199.75 kJ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are -408, -393 and -286kJ/mol respectively. Calculate heat of combustion of benzoic acid at constant volume A) 3201 kJ B) 3199.75 kJ C) -3201 kJ D) -3199.75 kJ?.
At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are -408, -393 and -286kJ/mol respectively. Calculate heat of combustion of benzoic acid at constant volume A) 3201 kJ B) 3199.75 kJ C) -3201 kJ D) -3199.75 kJ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are -408, -393 and -286kJ/mol respectively. Calculate heat of combustion of benzoic acid at constant volume A) 3201 kJ B) 3199.75 kJ C) -3201 kJ D) -3199.75 kJ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 300K the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are -408, -393 and -286kJ/mol respectively. Calculate heat of combustion of benzoic acid at constant volume A) 3201 kJ B) 3199.75 kJ C) -3201 kJ D) -3199.75 kJ?.
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