Question

butanol thermodynamics
spontaneous conditions
consider the thermodynamics of the
combustion of butanol to determine the
conditions in which the reaction is
spontaneous or thermodynamically
favored.

\ce{c4h9oh(l) + 6o2(g) -> 4co2(g) + 5h2o(g)}

$\delta h^\circ_{\text{comb}} = -2670\\ \text{kj/mol}_{\text{rxn}}$

$\delta s > 0$

under what set of conditions is the
reaction thermodynamically favorable or
spontaneous?

• it is favorable at all temperatures.
• it is favorable at low temperatures.
• it is never favorable.
• it is favorable at high temperatures.

Explanation:

To determine the spontaneity of a reaction, we use the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$. For a reaction to be spontaneous, $\Delta G < 0$.

Given:

• $\Delta H^\circ_{\text{comb}} = -2670 \, \text{kJ/mol}_{\text{rxn}}$ (negative, exothermic).
• $\Delta S > 0$ (positive, increase in disorder).

Substitute into $\Delta G$:
$\Delta G = \text{(negative)} - T(\text{positive})$.

Since $\Delta H$ is negative and $\Delta S$ is positive, the term $-T\Delta S$ will be negative (because $T$ is always positive in Kelvin, and we subtract a positive value multiplied by $T$). Thus, $\Delta G$ will be negative (spontaneous) for all temperatures (because both contributions to $\Delta G$ favor spontaneity: negative $\Delta H$ and positive $\Delta S$).

Answer:

It is favorable at all temperatures.