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At 300k,the standard enthalpy of formation of C6H5COOH(s),CO2(g) and H2O(l) are -408,-393 and -286kj/mol . Calculate the enthalpy of combustion of benzoic acid and constant pressure? Here i am facing issue that everywhere solve this question as heat of product minus heat of reactant but in combustion we use heat of reactant minus heat of product. Explain me this please?
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At 300k,the standard enthalpy of formation of C6H5COOH(s),CO2(g) and H...
Enthalpy of Formation and Combustion:
Enthalpy of formation (ΔHf) is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. It is usually given as a negative value, indicating that energy is released during the formation process.
Enthalpy of combustion (ΔHc) is the enthalpy change when one mole of a substance undergoes complete combustion in excess oxygen. It is also given as a negative value, as energy is released during combustion.
Calculation of Enthalpy of Combustion:
To calculate the enthalpy of combustion of benzoic acid (C6H5COOH), we need to use the enthalpy of formation values for the products (CO2 and H2O) and reactant (benzoic acid) to determine the overall enthalpy change.
The balanced equation for the combustion of benzoic acid is:
C6H5COOH(s) + 9O2(g) → 6CO2(g) + 3H2O(l)
Step 1: Calculate the enthalpy change for the formation of products:
The enthalpy change for the formation of CO2 and H2O can be calculated using their standard enthalpies of formation:
ΔHf(CO2) = -393 kJ/mol
ΔHf(H2O) = -286 kJ/mol
ΔHf(products) = 6ΔHf(CO2) + 3ΔHf(H2O)
= 6(-393 kJ/mol) + 3(-286 kJ/mol)
= -2358 kJ/mol - 858 kJ/mol
= -3216 kJ/mol
Step 2: Calculate the enthalpy change for the formation of reactant:
The enthalpy change for the formation of benzoic acid can be calculated using its standard enthalpy of formation:
ΔHf(C6H5COOH) = -408 kJ/mol
Step 3: Calculate the enthalpy of combustion:
The enthalpy of combustion can be calculated by subtracting the enthalpy change for the formation of reactant from the enthalpy change for the formation of products:
ΔHc = ΔHf(products) - ΔHf(C6H5COOH)
= -3216 kJ/mol - (-408 kJ/mol)
= -3216 kJ/mol + 408 kJ/mol
= -2808 kJ/mol
Therefore, the enthalpy of combustion of benzoic acid is -2808 kJ/mol at constant pressure.
Summary:
To calculate the enthalpy of combustion, we use the enthalpy of formation values for the products and reactant. By subtracting the enthalpy change for the formation of reactant from the enthalpy change for the formation of products, we can determine the overall enthalpy change during combustion. In this case, the enthalpy of combustion of benzoic acid is -2808 kJ/mol at constant pressure.
Enthalpy of formation (ΔHf) is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. It is usually given as a negative value, indicating that energy is released during the formation process.
Enthalpy of combustion (ΔHc) is the enthalpy change when one mole of a substance undergoes complete combustion in excess oxygen. It is also given as a negative value, as energy is released during combustion.
Calculation of Enthalpy of Combustion:
To calculate the enthalpy of combustion of benzoic acid (C6H5COOH), we need to use the enthalpy of formation values for the products (CO2 and H2O) and reactant (benzoic acid) to determine the overall enthalpy change.
The balanced equation for the combustion of benzoic acid is:
C6H5COOH(s) + 9O2(g) → 6CO2(g) + 3H2O(l)
Step 1: Calculate the enthalpy change for the formation of products:
The enthalpy change for the formation of CO2 and H2O can be calculated using their standard enthalpies of formation:
ΔHf(CO2) = -393 kJ/mol
ΔHf(H2O) = -286 kJ/mol
ΔHf(products) = 6ΔHf(CO2) + 3ΔHf(H2O)
= 6(-393 kJ/mol) + 3(-286 kJ/mol)
= -2358 kJ/mol - 858 kJ/mol
= -3216 kJ/mol
Step 2: Calculate the enthalpy change for the formation of reactant:
The enthalpy change for the formation of benzoic acid can be calculated using its standard enthalpy of formation:
ΔHf(C6H5COOH) = -408 kJ/mol
Step 3: Calculate the enthalpy of combustion:
The enthalpy of combustion can be calculated by subtracting the enthalpy change for the formation of reactant from the enthalpy change for the formation of products:
ΔHc = ΔHf(products) - ΔHf(C6H5COOH)
= -3216 kJ/mol - (-408 kJ/mol)
= -3216 kJ/mol + 408 kJ/mol
= -2808 kJ/mol
Therefore, the enthalpy of combustion of benzoic acid is -2808 kJ/mol at constant pressure.
Summary:
To calculate the enthalpy of combustion, we use the enthalpy of formation values for the products and reactant. By subtracting the enthalpy change for the formation of reactant from the enthalpy change for the formation of products, we can determine the overall enthalpy change during combustion. In this case, the enthalpy of combustion of benzoic acid is -2808 kJ/mol at constant pressure.
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At 300k,the standard enthalpy of formation of C6H5COOH(s),CO2(g) and H...
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At 300k,the standard enthalpy of formation of C6H5COOH(s),CO2(g) and H2O(l) are -408,-393 and -286kj/mol . Calculate the enthalpy of combustion of benzoic acid and constant pressure? Here i am facing issue that everywhere solve this question as heat of product minus heat of reactant but in combustion we use heat of reactant minus heat of product. Explain me this please?
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At 300k,the standard enthalpy of formation of C6H5COOH(s),CO2(g) and H2O(l) are -408,-393 and -286kj/mol . Calculate the enthalpy of combustion of benzoic acid and constant pressure? Here i am facing issue that everywhere solve this question as heat of product minus heat of reactant but in combustion we use heat of reactant minus heat of product. Explain me this please? 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 enthalpy of formation of C6H5COOH(s),CO2(g) and H2O(l) are -408,-393 and -286kj/mol . Calculate the enthalpy of combustion of benzoic acid and constant pressure? Here i am facing issue that everywhere solve this question as heat of product minus heat of reactant but in combustion we use heat of reactant minus heat of product. Explain me this please? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 300k,the standard enthalpy of formation of C6H5COOH(s),CO2(g) and H2O(l) are -408,-393 and -286kj/mol . Calculate the enthalpy of combustion of benzoic acid and constant pressure? Here i am facing issue that everywhere solve this question as heat of product minus heat of reactant but in combustion we use heat of reactant minus heat of product. Explain me this please?.
At 300k,the standard enthalpy of formation of C6H5COOH(s),CO2(g) and H2O(l) are -408,-393 and -286kj/mol . Calculate the enthalpy of combustion of benzoic acid and constant pressure? Here i am facing issue that everywhere solve this question as heat of product minus heat of reactant but in combustion we use heat of reactant minus heat of product. Explain me this please? 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 enthalpy of formation of C6H5COOH(s),CO2(g) and H2O(l) are -408,-393 and -286kj/mol . Calculate the enthalpy of combustion of benzoic acid and constant pressure? Here i am facing issue that everywhere solve this question as heat of product minus heat of reactant but in combustion we use heat of reactant minus heat of product. Explain me this please? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 300k,the standard enthalpy of formation of C6H5COOH(s),CO2(g) and H2O(l) are -408,-393 and -286kj/mol . Calculate the enthalpy of combustion of benzoic acid and constant pressure? Here i am facing issue that everywhere solve this question as heat of product minus heat of reactant but in combustion we use heat of reactant minus heat of product. Explain me this please?.
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