Wize High School Grade 12 Chemistry Textbook > Energy Changes

Molar Enthalpy

Popular Courses

Grade 12 Chemistry

Ontario High School

Grade 12 Chemistry

Canada High School

High School Chemistry

US High School

Chemistry 30

Alberta High School

PHYS 111

University of Victoria

Find My Course

0:00 / 0:00

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

Rearrange to solve for the variable we want:

ΔH(freezing)=qn\Delta H_{(freezing)}=\frac{q}{n}ΔH(freezing)​=nq​
 
Solve for n:
Use the equation:
n=mMn=\frac{m}{M}n=Mm​

n=30g18.02gmoln=\frac{30g}{\frac{18.02g}{mol}}n=mol18.02g​30g​

n=1.665 moles          
         
ΔH=−10kJ 1.665 moles\Delta H=-\frac{10kJ}{\ 1.665\ moles}ΔH=− 1.665 moles10kJ​

∆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? 