Question

question 3
the heat of formation at 25°c of liquid acetone, ch₃c(o)ch₃(l) is -247.5 kj/mol. which of the following is the correct heat of formation reaction for ch₃c(o)ch₃(l)?
select one answer.
5 points
a. o
3 c(s, graphite) + 6 h(g) + o(g) → ch₃c(o)ch₃(l) △h°f = -247.5 kj/mol
b. o
ch₃c(o)ch₃(l) + 4 o₂(g) → 3 co₂(g) + 3 h₂o(g) △h°f = -247.5 kj/mol
c. o
c₃(s, graphite) + 3 h₂(g) + 1/2 o₂(g) → ch₃c(o)ch₃(l) △h°f = -247.5 kj/mol
d. o
6 c(s, graphite) + 6 h₂(g) + o₂(g) → 2 ch₃c(o)ch₃(l) △h°f = -247.5 kj/mol
e. o
3 c(s, graphite) + 3 h₂(g) + 1/2 o₂(g) → ch₃c(o)ch₃(l) △h°f = -247.5 kj/mol

Explanation:

• The heat of formation ($\Delta H_f^\circ$) is the enthalpy change when one mole of a compound is formed from its elements in their standard states.
• For carbon, the standard state is graphite ($C(s, graphite)$), for hydrogen it is $H_2(g)$, and for oxygen it is $O_2(g)$.
• Analyze each option:
• Option A: $H$ and $O$ are not in their standard states ($H_2$ and $O_2$ are the standard states for hydrogen and oxygen respectively).
• Option B: This is a combustion reaction, not a formation reaction.
• Option C: $C_3(s, graphite)$ is not a standard state (carbon's standard state is $C(s, graphite)$).
• Option D: The equation produces 2 moles of acetone, but the heat of formation is defined per mole.
• Option E: The reactants ($C(s, graphite)$, $H_2(g)$, $O_2(g)$) are in their standard states, and 1 mole of $CH_3C(O)CH_3(l)$ is formed.

Answer:

E. $3\ C(s,\ graphite)+3\ H_2(g)+\frac{1}{2}\ O_2(g)\to CH_3C(O)CH_3(l)\ \ \ \Delta H_f^\circ=-247.5\ \frac{kJ}{mol}$