Molar Enthalpy

Molar enthalpy, ΔHrxn, is the enthalpy change associated with a physical, chemical, or nuclear change involving one mole of a substance, where x is a letter or combination of letters to indicate the type of change that occurs.

Type of molar enthalpy, ΔHrxn

Example of change

solution (ΔHsol)

NaBr (s) → Na+ (aq)  +  Br- (aq)

combustion (ΔHcomb)

CH4  (g) + O2 (g)  →  CO2  (g) + H2O (g)

vapourization (ΔHvap)

C2H5OH (l)  →  C2H5OH (g)

neutralization (ΔHneut)*

2 NaOH (aq)  + H2SO4  (aq) →  Na2SO4  (aq)  + 2 H2O (l)

neutralization (ΔHneut)*

   NaOH (aq)  + ½ H2SO4  (aq) →  ½ Na2SO4  (aq)  +  H2O (l)

* Enthalpy of neutralization can be expressed per mole of either base or acid consumed.

∆Hrxn = ∆H ±l q l ±l mc∆T l    
x x x    


ΔH is the change in enthalpy of the system (under the restrictions mentioned above), and
q is the energy added to the system through heat.
n is the moles of the reactant consumed that the molar enthalpy is to be expressed as

 

Example: 9.0 grams of charcoal (C) were completely consumed in a bomb calorimeter. If we assume that the 2.0 L of water absorbed all of the heat released by the charcoal, and if the temperature of the water increased from 20.25oC to 56.04oC, what is the molar enthalpy of combustion for carbon?

System - C Surrounding - H2O
m = 9.0 g C
MC = 12.01 g/mol
 m = 2000 g
c = 4.184 J/goC
∆T = 56.04 oC – 20.25oC = 35.79 oC

exothermic
∆H = - q

 

n∆Hcomb = ∆H = - q 
      n∆Hcomb = - q = - mc∆T 
  (9.0 g C / 12.01 g/mol)∆Hcomb = - 2000g(4.19 J/(g oC))(35.79 oC)
                            ∆Hcomb = - 299.92 kJ / 0.7492 mol

                  ∆Hcomb = -400 kJ / mol C      

 

Example: CS2, a very flammable liquid, has a molar enthalpy of -1028 kJ/mole. What do you expect aluminum's final temperature to be if 1.0 kg of Al is initially at 20.0oC, and it absorbs all the heat from the following sample of CS2:

System - CS2 Surrounding - Al  
intial mass CS2 = 22.6 g
final mass CS2 = 11.6 g
m CS2 = 11.0 g
M = 76.13 g / mol
∆Hcomb = -1028 kJ/mol CS2

m = 1.00 kg
c = 0.900 kJ / kg o
∆T
T1 = 20oC

  
  exothermic
∆H = - q		  

         n∆Hcomb = ∆H = - q
(11.0 g CS2) / (76.13 g CS2/mol CS2))(-1028 kJ / mol CS2) = - (1000 g)(0.900 J / goC) ∆T                                       

∆T = 165oC

∆T = 165.0oC = T2 – T1
T2 = 165.0oC + T1 = 165.0oC + 20.0oC = 185.0oC

Example: 300 mL of 0.2 M aqueous KOH neutralizes 150 mL of aqueous 0.2 M H2SO4. We go from an average initial temperature of 22.3 oC to a maximum of 29.2o C. Calculate the molar heat(enthalpy) of neutralization of KOH. \


2KOH + H2SO4 → K2SO4 + 2 H2O

System - KOH Surrounding - H2O  
VKOH = 0.300 L
[KOH] = 0.200 mol KOH / L
∆Hneut =

m = m1 + m2 = 300 g + 150 g = 450
m = 0.450 kg
c = 4.184 kJ / kg oC

 
    exothermic
∆H = - q		  

n∆Hneut = ∆H = - q                     

                                n∆Hneut = mc∆T = (0.450 kg)(4.184 kJ/kgoC)(6.9oC)

(0.300 L)(0.200 mol /L)∆Hneut = -13.01 kJ / 0.0600 mol KOH                        

 ∆Hneut = - 216. 8kJ / mol KOH