17.4. Hess’s Law http://www.ck12.org
Applying the equation from the text:
∆Hf° = [2 mol NO 2 (33.85 kJ/mol)] - [2 mol NO (90.4 kJ/mol) + 1 mol O 2 (0 kJ/mol)] = -113 kJ
The standard heat of reaction is−113 kJ.
Step 3: Think about your result.
The reaction is exothermic, which makes sense, because combustion reactions typically release heat.
Practice Problem- Calculate∆H° for the following reactions.
a. CaCO 3 (s)→CaO(s)+CO 2 (g)
b. 7O 2 (g)+4NH 3 (g)→4NO 2 (g)+6H 2 O(l)
Lesson Summary
- Hess’s law of heat summation allows chemical reactions to be added together in such a way as to produce
a final desired equation. The heats of reaction for each individual reaction are also added, resulting in the
enthalpy change for the final reaction. - The standard heat of formation is the enthalpy change that occurs when a compound is formed from its
elements in their standard states. - The standard enthalpy change for any reaction is equal to the sum of the heats of formation of the products
minus the sum of the heats of formation of the reactants.
Lesson Review Questions
Reviewing Concepts
- What are two reasons why the heat of reaction may not be able to be determined directly?
- What are standard conditions for thermochemical problems?
- What is the standard heat of formation of an element in its standard state?
- For which of the following is the standard heat of formation equal to zero?
a. F(g)
b. F 2 (g)
c. He(l)
d. Hg(l)
e. CO(g)
f. Co(s) - In general, the more negative the value of the∆Hf° for a compound, the more stable the compound. Explain
why this is true. Use the table above (Table17.3) to choose the more stable substance from each pair below.
a. CO(g) or CO 2 (g)
b. graphite or diamond
c. Al 2 O 3 or Fe 2 O 3
d. O 2 (g) or O 3 (g)