Answer:
[tex]Q=-15.3kJ[/tex]
Explanation:
Hello!
In this case, since the enthalpy of vaporization is the contrary of the enthalpy of condensation, we have:
[tex]\Delta _{cond}H=-\Delta _{vap}H=-38.0kJ/mol[/tex]
Now, we convert the grams of CH3OH to moles:
[tex]n=12.9g*\frac{1mol}{32.05g}=0.402mol[/tex]
Thus, we compute the energy as shown below:
[tex]Q=n*\Delta _{cond}H=0.402mol*-38.0\frac{kJ}{mol}\\\\Q=-15.3kJ[/tex]
Which means 15.3 kJ of energy must be back down from the system.
Best regards!
Enthalpy is the measure of the energy that is not available to work. 15.27 kJ of energy is needed to condence 12.9 g of methenol.
Enthalpy:
It is the measure of the energy that is not available to work. It is denoted by [tex]\bold{ \Delta H}[/tex].
The energy of a system can be calculated by the formula,
[tex]\bold{Q = n \times \Delta H}[/tex]
Where,
Q - energy( in kJ)
n - Number of moles
[tex]\bold{ \Delta H}[/tex] - enthalpy
Given here,
[tex]\bold{CH_3OH = 12.9 g}[/tex]
[tex]\bold{\Delta Hvap = 38.1 kJ/mol}[/tex]
molar mass of the [tex]\bold{CH_3OH}[/tex] -
The number of moles of [tex]\bold{CH_3OH}[/tex]
[tex]\bold{n = \frac{w}{m}}\\\\\bold{n = \frac{12.9}{32.03}} \\\\\bold{n = 0.402}[/tex]
Put the values in the formula,
[tex]\bold{Q = 0.402\times 38.0}\\\\\bold{Q =15.27 }[/tex]
Therefore, the 15.27 kJ of energy is needed to condense 12.9 g of methenol.
To know more about Enthalpy, refer to the link:
https://brainly.com/question/9444545
