Wize High School Grade 12 Chemistry Textbook > Energy Changes

Molar Enthalpy

Calculate the molar enthalpy of crystallization/freezing (∆Hfreezing) when 10.00kJ of energy are lost as 30.00g of water are frozen at 0oC. (Ans in kJ/mol)

What variables do we know?

q=-10kJ

m=30g

T=0oC

What variable are we being asked to solve for?

∆Hfreezing=?

What equation should we use to solve for this?

$q=n\Delta H_{\left(freezing\right)}$

Rearrange to solve for the variable we want:

$\Delta H_{(freezing)}=\frac{q}{n}$

Solve for n:

Use the equation:

$n=\frac{m}{M}$

$n=\frac{30g}{\frac{18.02g}{mol}}$

n=1.665 moles

$\Delta H=-\frac{10kJ}{\ 1.665\ moles}$

∆Hfreezing=-6.01 kJ/mol

Methane (CH4) has a normal boiling point of -161.6oC. At this temperature, the ΔHcondensation = -8.17kJ/mol. If 16.5g of liquid methane vapourizes, how much energy is absorbed?

+2.4 kJ

+6.2 kJ

-8.4 kJ

+8.4 kJ

I don't know

Molar Enthalpy

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