When 92.0 g of ethanol (c2h5oh are vaporized at its boiling point of 78.3°c, it requires 78.6 kj of energy. what is the approxim
2 answers:
Answer:
Molar heat of vaporization of ethanol, 157.2 kJ/mol
Explanation:
Molar heat of vaporization is the amount heat required to vaporize 1 mole of a liquid to vapor.
The equilibrium is represented as:
C2H5OH(l) ↔ C2H5OH(g)
Given:
Mass of ethanol = 92.0 g
Energy required = 78.6 kJ
Calculation:
Molar mass of ethanol = 46 g/mol
Moles of ethanol =
78.6 kJ of energy is required to vaporise 2 moles of ethanol
Therefore, the amount of energy required per mole would be:
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Since each AB molecule forms one of A and one of B,
. Hence: 
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We'll consider that in the beginning there was not A or B. So,
. Furthermore, since the ratio of AB to A and to B is 1:1, 
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Calculating the time by the expression of velocity:
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Now, we'll calculate the number of mols of the products in the gas. Using the Ideal Gas Law:
Since each AB molecule forms one of A and one of B,
We'll consider that in the beginning there was not A or B. So,
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