Molar Enthalpies

remember all those pesky chemical equations and how to balance them out? 
let's take a look at how they're used in thermochemistry!

hot

1 H₂(g) + ½ O₂ (g) → 1 H₂O(g) + 241.8 kJ

this is an example of a "thermochemical equation"! it's a little different from what u might be used to for 2 reasons:

1. what is this "241.8" kJ nonsense!?! that's not any chemical I recognize!
2. there's a fractional term (½)!! what's up with that!?!?

addressing the "241.8 kJ": 
this is a measure of the enthalpy change (ΔH), or the change in the system's total energy that happened during the reaction! 
remember that chemical reactions always involve the making and breaking of bonds - this requires a change in the system's energy. for our purposes, energy change will always be measured in joules or kilojoules (1000 J = 1 kJ)!
the fact that the ΔH is listed on the "products" side of the equation means that the system produced or released 241.8 kJ of heat energy.
mini-quiz: what type of energy change is happening in this example equation?

CH4 (g) + O2 (g) --> CO2 (g) + H2O (g) + 882.0 kJ

exo thermic!

addressing the "½" term in the eq'n

you might be thinking, "this is outrageous!!! how can you react half an oxygen molecule!?!?"
remember: the coefficients in an equation can also represent the number of moles of particles! 
one mole = 6.02 * 10^23 particles or roughly 602 000 000 000 000 000 000 000 particles... so it's very possible to have ½ mole of a substance in a reaction XD

1 mole

tying concepts together

why would we want to balance an equation in a way that leaves fractions? fractions are kind of ugly after all...

the reason we have to use 1/2 mol of O2 in the example equation is that we want to know the enthalpy change per 1 mole of H2(g) combusted... also known as the molar enthalpy (ΔHx)! to find the molar enthalpy of combustion for H2(g), we need to combust exactly 1 mol of H2(g), even if it means using fractional moles for other substances!

molar enthalpy, ΔHx: the enthalpy change associated with a change in matter involving one mole of a substance

a problem with finding molar enthalpy

molar enthalpy values have to be collected experimentally in specific conditions (temperature and pressure), then listed in reference tables. 
the molar mass of H₂(g) is 2 g/mol.
this means an experimenter would have to combust exactly 2 grams of H2(g) to find its molar enthalpy of combustion... it might be a little hard to combust such a tiny sample of matter!

to solve this, scientists came up with an easier way to determine the molar enthalpy of a reaction:
ΔH = nΔHx
in which...

• ΔH is the overall enthalpy change (kJ)
• n is the number of moles (mol)
• ΔHx is the molar enthalpy change (kJ/mol)
• x represents any type of change in matter - it'll usually be replaced with a word like "combustion" or "fusion" depending on the problem

this formula lets experimenters use any amount of a substance they want - they just need to know the number of moles they're using!

Example problems

chilly!

1. 13 moles of water are frozen and 78.3 kilojoules of energy are released. what's the molar enthalpy of freezing water? 

givens and unknowns:

n = 13 molΔH 

n = 78.3 kJ

ΔHfusion = ???

we can solve for ΔHfusion by using the molar enthalpy formula! XD

first, we need to rearrange it to isolate ΔHfusion

nΔHfusion = ΔH

(nΔHfusion)/n = ΔH/n

ΔHfusion = ΔH/n

then, we can sub in our given values!

ΔHfusion = (78.3 kJ)/(13 mol)

= 6.02 kJ/mol

freezing water releases energy from the system (i.e. is exothermic), so the enthalpy change would have to be negative to reflect the loss in the system's energy!
therefore the molar enthalpy of freezing water is -6.02 kJ/mol! 

love these things!

2. some fog machines work by vaporizing something called "propylene glycol" and releasing the vapour into the surroundings!
if a machine used 490. kilojoules of energy to vaporize 550. grams of propylene glycol (C3H8O2), what would the molar enthalpy be? 

givens and unknowns: 

mass (m) = 550. g

ΔH = 490. kJ

ΔHvaporization = ???

we need the number of moles (n) of propylene glycol - let's convert from mass to moles using the molar masses from the periodic table!

molar mass (M) = 3 MC + 8 MH + 2 MO 

molar mass (M) = [3(12) + 8(1) + 2(16)] g/mol

molar mass (M) = 76 g/mol

n = m/M

n = (550.0 g)/(76 g/mol)

n = 7.23 mol

now that we have n, we can sub our values into the molar enthalpy formula! i'm going to skip straight to the rearranged formula

ΔHvaporization = ΔH/n

ΔHvaporization = (490. kJ)/(7.23 mol)

ΔHvaporization = 67.8 kJ/mol

vaporizing requires the addition of energy to the system(i.e. endothermic), so the enthalpy change would be positive! 
therefore the molar enthalpy of propylene glycol's vaporization is +67.8 kJ/mol!

hot

3. this is a bit of a thinking question - i haven't had to do something like this for school yet XD
after a long day of performing, the members of exo come back to their apartment to relax. they decide that a relaxing thing to do would be to set some 70% vol/vol vodka on fire, so they completely combust 30.0 mL of vodka!
if ethanol's molar enthalpy of combustion is -1370 kJ/mol, what was the total enthalpy change in exo's experiment?
note: ethanol is C2H6O

givens and unknowns

V/V concentration (c) = 70%

volume (VVodka) = 30.0 mL

ΔHcomb = -1370 kJ/mol

we need to get the number of moles of ethanol - how can we use our given info to get n?

volume of ethanol (V) = 0.70(VVodka)

volume of ethanol (V) = 0.70(30.0 mL)

volume of ethanol (V) = 21 mL

we can't directly convert from volume to moles, so let's start by using density to convert volume to mass! i just googled the density LOL

d = 0.789 g/mL

m = dV

m = (0.789 g/mL)(21 mL)

m = 16.6 g

now we can convert mass to moles using molar mass!

M = 2 MC + 6 MH + MO 

M = [2(12) + 6(1) + 16] g/mol

M = 46 g/mol

n = m/M

n = (16.6 g)/(46 g/mol)

n = 0.361 mol

sub in our values into the molar enthalpy formula

ΔH = nΔHcomb 

ΔH = (0.361 mol)(-1370 kJ/mol)

ΔH = -494.6 kJ

this means -494.6 kJ was the total enthalpy change in exo's experiment!

 good night!