Question

What amount of heat, in KJ, would be involved in condensing 12.9 g of ch3OH ?(hvap=38.0 kj/mol

Answer 1

Answer:

$Q=-15.3kJ$

Explanation:

Hello!

In this case, since the enthalpy of vaporization is the contrary of the enthalpy of condensation, we have:

$\Delta _{cond}H=-\Delta _{vap}H=-38.0kJ/mol$

Now, we convert the grams of CH3OH to moles:

$n=12.9g*\frac{1mol}{32.05g}=0.402mol$

Thus, we compute the energy as shown below:

$Q=n*\Delta _{cond}H=0.402mol*-38.0\frac{kJ}{mol}\\\\Q=-15.3kJ$

Which means 15.3 kJ of energy must be back down from the system.

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