Question

$$ce{p_{4}o_{10}(s) + 6h_{2}o(l) -> 4h_{3}po_{4}(aq)}$$considering the thermodynamic data given below, which statement describes the thermodynamic favorability of the system?
$$delta g^circ = -356 \text{kj}$$
$$delta h^circ = -416 \text{kj}$$
$$delta s^circ = -1093 \text{j/k}$$
a. always favorable
b. never favorable
c. favorable at high t
d. favorable at low t
enter the answer choice letter.

Explanation:

To determine thermodynamic favorability, we use the Gibbs free energy equation $\Delta G = \Delta H - T\Delta S$. Here, $\Delta H^\circ = -416\ \text{kJ}$ (exothermic, favorable) and $\Delta S^\circ = -1093\ \text{J/K} = -1.093\ \text{kJ/K}$ (decrease in entropy, unfavorable). The term $-T\Delta S$ becomes positive (since $\Delta S$ is negative) and its magnitude depends on temperature $T$. For $\Delta G$ to be negative (favorable), the magnitude of $\Delta H$ (negative, favorable) must be greater than the magnitude of $-T\Delta S$ (positive, unfavorable).

Let's analyze the temperature dependence: $\Delta G = -416\ \text{kJ} - T(-1.093\ \text{kJ/K}) = -416\ \text{kJ} + 1.093T\ \text{kJ}$. We want $\Delta G < 0$:

$-416 + 1.093T < 0$

$1.093T < 416$

$T < \frac{416}{1.093} \approx 379.6\ \text{K}$ (or ~106.45°C).

At low temperatures, the positive contribution from $-T\Delta S$ (which is $+T|\Delta S|$) is small enough that $\Delta H$ (negative, large magnitude) dominates, making $\Delta G$ negative (favorable). At high temperatures, the positive $-T\Delta S$ term would become large enough to make $\Delta G$ positive (unfavorable). The given $\Delta G^\circ$ is already negative, but we analyze the general case. The key is that with $\Delta H < 0$ and $\Delta S < 0$, the reaction is favorable at low temperatures (since the enthalpy term dominates when $T$ is small, as the entropy penalty is small).

Now, check the options:

• A: "Always favorable" is wrong because at high T, $\Delta G$ becomes positive.
• B: "Never favorable" is wrong because $\Delta G^\circ$ is negative, and at low T it's favorable.
• C: "Favorable at high T" is wrong (high T makes $\Delta G$ positive).
• D: "Favorable at low T" matches our analysis.

Answer:

D