Hess's Law (6.4.1) Flashcards
• Hess’s law states that the enthalpy change (∆H) for a multistep process is the sum of the ∆H values of the individual steps.
• Hess’s law states that the enthalpy change (∆H) for a multistep process is the sum of the ∆H values of the individual steps.
State functions are independent of path—it doesn’t
matter how you get from the starting point to the
ending point, only the difference between them
matters. State functions can be determined directly,
or they can be calculated through multiple steps.
Altitude is a state function. If you want to find the
difference in altitude (∆Alt) between Salt Lake City
and Denver, you can calculate ∆Alt directly
(5280 ft – 4266 ft), or determine ∆Alt by adding
together the difference in altitude between Salt Lake
City and San Francisco (–4266 ft) and the difference
in altitude between San Francisco and Denver
(5280 ft). In either case, the result is 1014 ft.
Enthalpy change (∆H) is also a state function.
Hess’s law states that ∆H for a multistep process is
the sum of the ∆H values of the individual steps.
Problem: What is the standard molar enthalpy
change (∆H°) for the combustion of one mole of
methane gas (CH4) to give gaseous carbon dioxide
(CO2) and liquid water (H2O)?
Hess’s law can be used to determine this enthalpy
change. ∆H° for the combustion of CH4 to give
gaseous CO2 and gaseous H2O is –802 kJ. ∆H° for
the evaporation of liquid H2O into gaseous H2O is
+44 kJ.
Since liquid water is a product in the desired
reaction, the evaporation of H2O must be reversed
before the reactions are added together. Therefore,
∆H° for the evaporation of H2O is multiplied by
negative one. Also, since two moles of H2O are
required, a stoichiometric coefficient of two is
required on both sides of the evaporation of water.
Therefore, ∆H° for the evaporation of H2O is
multiplied by two.
Summing the ∆H° values for the two reactions yields
∆H° for the combustion of one mole of methane gas
to give gaseous carbon dioxide and liquid water.
This value is –890 kJ. This value makes sense,
since the combustion of methane gas is
exothermic.
State functions are independent of path—it doesn’t
matter how you get from the starting point to the
ending point, only the difference between them
matters. State functions can be determined directly,
or they can be calculated through multiple steps.
Altitude is a state function. If you want to find the
difference in altitude (∆Alt) between Salt Lake City
and Denver, you can calculate ∆Alt directly
(5280 ft – 4266 ft), or determine ∆Alt by adding
together the difference in altitude between Salt Lake
City and San Francisco (–4266 ft) and the difference
in altitude between San Francisco and Denver
(5280 ft). In either case, the result is 1014 ft.
Enthalpy change (∆H) is also a state function.
Hess’s law states that ∆H for a multistep process is
the sum of the ∆H values of the individual steps.
Problem: What is the standard molar enthalpy
change (∆H°) for the combustion of one mole of
methane gas (CH4) to give gaseous carbon dioxide
(CO2) and liquid water (H2O)?
Hess’s law can be used to determine this enthalpy
change. ∆H° for the combustion of CH4 to give
gaseous CO2 and gaseous H2O is –802 kJ. ∆H° for
the evaporation of liquid H2O into gaseous H2O is
+44 kJ.
Since liquid water is a product in the desired
reaction, the evaporation of H2O must be reversed
before the reactions are added together. Therefore,
∆H° for the evaporation of H2O is multiplied by
negative one. Also, since two moles of H2O are
required, a stoichiometric coefficient of two is
required on both sides of the evaporation of water.
Therefore, ∆H° for the evaporation of H2O is
multiplied by two.
Summing the ∆H° values for the two reactions yields
∆H° for the combustion of one mole of methane gas
to give gaseous carbon dioxide and liquid water.
This value is –890 kJ. This value makes sense,
since the combustion of methane gas is
exothermic.
Heat Capacity (C)
The heat required to raise the temperature of a given quantity of material by one kelvin.
