September 02, 2015
∆H
enthalpy change
for a balanced
equation
∆H = n∆rH
∆rH = ∆H/n
n = coefficient
∆rH
∆H = n∆rH
∆rH = ∆H/n
n = moles
∆H
enthalpy change
for an amount of
fuel
Enthalpy Changes and Stoichiometry
-­‐the amount of energy released in a chemical reaction is affected by how much fuel is used
-­‐to calculate the amount of energy released when a particular amount of fuel is burned use ∆H = n∆rH
September 02, 2015
Example 1
C12H22O11(s) + 12 O2(g) -­‐-­‐-­‐> 12 CO2(g) + 11 H2O(l) ∆rH (C12H22O11) = -­‐5640.3 kJ/mol
What is the enthalpy change for the reaction of 20.0 g of sucrose (C12H22O11)?
Example 2
SO2(g) + 2 H2S(s) -­‐-­‐> 3 S(l) + 2 H2O(g) ∆rH (S) = -­‐31.0 kJ/mol
What is the enthalpy change for the reaction of 100 g of S(s)? September 02, 2015
If you are given an enthalpy change for a balanced reaction, and asked to find an enthalpy change for a mass of fuel, you will need to find the molar enthalpy first. ∆H
enthalpy change
for a balanced
equation
∆H = n∆rH
∆rH = ∆H/n
n = coefficient
∆rH
∆H = n∆rH
∆rH = ∆H/n
n = moles
Example 3
2 NaOH(s) + 2 Al(s)+ 2 H2O(l) -­‐-­‐> 2NaAlO2(aq) + 3 H2(g) ∆H = -­‐850 kJ
How much energy is released when 250 g of Al(s) reacts?
∆H
enthalpy change
for an amount of
fuel
September 02, 2015
Using an enthalpy change for an amount of fuel and a molar enthalpy change to find out the moles of fuel that burned.
n=
∆H
∆rH
Example 4
C12H22O11(s) + 12 O2(g) -­‐-­‐-­‐> 12 CO2(g) + 11 H2O(l) -­‐5640.3 kJ/mol
∆rH (C12H22O11) = If 10 000 kJ of energy is released, what mass of sucrose reacted? September 02, 2015
Example 5
2 KCl(s) + 2 H2SO4(l) -­‐-­‐-­‐-­‐> 2 HCl(g) + K2SO4(s) ∆H = + 41.0 kJ
If 300 kJ of energy is absorbed, how many moles of KCl(s) reacted?