6.4.1 Hess's Law

Question 1

Hess’s Law

Answer 1

• Hess’s law states that the enthalpy change ( H) for a multistep process is the sum of the H values of the individual steps.

Question 2

note

Answer 2

• 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 (ΔAlt) between Salt Lake City and Denver, you can calculate ΔAlt directly (5280 ft – 4266 ft), or determine ΔAlt 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 (ΔH) is also a state function. Hess’s law states that ΔH for a multistep process is the sum of the ΔH values of the individual steps.
• Problem: What is the standard molar enthalpy change ( H°) for the combustion of one mole of methane gas (CH 4 ) to give gaseous carbon dioxide (CO 2 ) and liquid water (H 2 O)?
• Hess’s law can be used to determine this enthalpy change. ΔH° for the combustion of CH 4 to give gaseous CO 2 and gaseous H 2 O is –802 kJ. ΔH° for the evaporation of liquid H 2 O into gaseous H 2 O is +44 kJ.
• Since liquid water is a product in the desired reaction, the evaporation of H 2 O must be reversed before the reactions are added together. Therefore, ΔH° for the evaporation of H 2 O is multiplied by negative one. Also, since two moles of H 2 O are required, a stoichiometric coefficient of two is required on both sides of the evaporation of water. Therefore, ΔH° for the evaporation of H 2 O is multiplied by two.
• Summing the ΔH° values for the two reactions yields ΔH° 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.

Question 3

When one mole of ice melts, the change in enthalpy is +6.0 kJ. What is the enthalpy change for the freezing of one mole of water?

Answer 3

−6.0 kJ

Question 4

From the given information, calculate the enthalpy change for the formation of sodium chloride.
Na+ (g) + Cl − (g) → NaCl(s) ΔH = ?

Na(g) + Cl(g) → Na+ (g) + Cl − (g) ΔH = +147 kJ
Na(s) + ½ Cl2 (g) → Na(g) + Cl(g) ΔH = +230 kJ
Na(s) + ½ Cl2(g) → NaCl(s) ΔH = −411 kJ

Answer 4

−788kJ

Question 5

Hess’ Law can be applied to which of the following?

Answer 5

All chemical reactions

Question 6

Which of the following statements is true?

Answer 6

The reverse reaction of an exothermic reaction has a positive ΔH.

Question 7

How does the enthalpy change for the reverse reaction compare to the enthalpy change of the forward reaction?

Answer 7

The changes in enthalpy are the same in magnitude but different in sign.

Question 8

The combustion of methane is given by the following equation:
CH4(g) + 2 O2(g) → CO2(g) + H2O(g) ΔH = −890 kJ
What is the enthalpy change for 5 moles of methane?

Answer 8

−4450 kJ

Question 9

Use the data given to calculate the ΔH for the following reaction:
K(g) + Cl(g) -> K+(g) + CL-(g) ΔH = ?
KCl(s) -> K+(g) + Cl-(g)
ΔH = 718+718 kJ
KCl(s) -> K(s) + 1/2Cl2(g)
ΔH = +436 kJ
K(s) + 1/2Cl2(g) -> K(g) + Cl(g) 
ΔH = +211 kJ

Answer 9

+71 kJ

Question 10

Which of the following does not change the ΔH of a reaction?

Answer 10

Changing the path the system took to get to its current state

Question 11

Which of the following statements is not true about Hess’ Law?

Answer 11

The change in enthalpy depends on the route taken to get to the final state of the chemical reaction.