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The standard enthalpy of formation of H2O liquid is -68.0 kilo calorie the standard heat of formation of H2O gaseous is likely to be A-68.0 kilo calorie B -67.4 kilo calorie C -80.0 kilo calorie D -58.3 kilo calorie?
the correct answer is B.Explain the reason.?
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The standard enthalpy of formation of H2O liquid is -68.0 kilo calorie...
Enthalpy of Formation

The enthalpy of formation (ΔHf) is the change in enthalpy that occurs when one mole of a compound is formed from its constituent elements in their standard states. It is typically measured under standard conditions, which include a temperature of 25°C and a pressure of 1 atm.

Standard Heat of Formation of H2O Liquid

Given that the standard enthalpy of formation of H2O in the liquid state is -68.0 kilocalories, this means that when one mole of liquid water is formed from its constituent elements (hydrogen gas and oxygen gas), 68.0 kilocalories of energy is released.

Standard Heat of Formation of H2O Gaseous

To determine the standard heat of formation of H2O in the gaseous state, we can use the fact that the difference in enthalpy between the liquid and gaseous states is the heat of vaporization (ΔHvap). The heat of vaporization represents the amount of energy required to convert one mole of a substance from a liquid to a gas at its boiling point under standard conditions.

Since the enthalpy change for vaporization is positive (energy is required to break the intermolecular forces and convert the substance into a gas), we can say that:

ΔHvap = ΔHg - ΔHl

where ΔHg is the enthalpy of formation of H2O in the gaseous state and ΔHl is the enthalpy of formation of H2O in the liquid state.

Applying the Equation

Using the given values, we have:

ΔHvap = ΔHg - ΔHl
ΔHvap = ΔHg - (-68.0 kcal/mol)
ΔHvap = ΔHg + 68.0 kcal/mol

Since the standard heat of formation of H2O in the liquid state is -68.0 kilocalories, the enthalpy change for vaporization (ΔHvap) is equal to the standard heat of formation of H2O in the gaseous state (ΔHg) plus 68.0 kilocalories.

Therefore, the standard heat of formation of H2O in the gaseous state is -68.0 kcal/mol - 68.0 kcal/mol = -67.4 kilocalories.

Conclusion

The correct answer is B) -67.4 kilocalories. This value represents the standard heat of formation of H2O in the gaseous state and is determined by subtracting the enthalpy change for vaporization from the standard heat of formation of H2O in the liquid state.
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The standard enthalpy of formation of H2O liquid is -68.0 kilo calorie the standard heat of formation of H2O gaseous is likely to be A-68.0 kilo calorie B -67.4 kilo calorie C -80.0 kilo calorie D -58.3 kilo calorie? the correct answer is B.Explain the reason.?
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The standard enthalpy of formation of H2O liquid is -68.0 kilo calorie the standard heat of formation of H2O gaseous is likely to be A-68.0 kilo calorie B -67.4 kilo calorie C -80.0 kilo calorie D -58.3 kilo calorie? the correct answer is B.Explain the reason.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The standard enthalpy of formation of H2O liquid is -68.0 kilo calorie the standard heat of formation of H2O gaseous is likely to be A-68.0 kilo calorie B -67.4 kilo calorie C -80.0 kilo calorie D -58.3 kilo calorie? the correct answer is B.Explain the reason.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The standard enthalpy of formation of H2O liquid is -68.0 kilo calorie the standard heat of formation of H2O gaseous is likely to be A-68.0 kilo calorie B -67.4 kilo calorie C -80.0 kilo calorie D -58.3 kilo calorie? the correct answer is B.Explain the reason.?.
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