Question

The molar heat of vaporization for methane, CH4, is 8. 53 kJ/mol. How much energy is absorbed when 54. 8 g of methane vaporizes at its boiling point? Use q equals n delta H. 6. 42 kJ 29. 1 kJ 137 kJ 467 kJ.

Answer 1

Molar heat of vaporization is defined as the heat absorbed by one mole of substance to convert from liquid to gas.

How do you calculate the heat of vaporization?

The formula used to calculate the heat of vaporization is:

$\rm Q &= n \times \Delta H$

Where,

Q = Amount of Heat

n = number of moles of a substance

$\Delta H =$ molar enthalpy of fusion

Now, to calculate the moles of methane:

$\rm Moles &= \dfrac{Mass \;of\; methane }{Molar\; Mass\; of \; Methane} &= \dfrac{54.8\;g}{16\;g/mol}$

Moles = 3.425 mol

Now, 1 mol of methane absorbs = 8.53 KJ

3.425 mol of methane absorbs = $3.424 \times 8.53 &= 29.1 KJ$

Thus, the energy is absorbed till the methane vaporizes at its boiling point is 29.1 KJ.

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