Question

what can be said about a reaction with $delta h = - 890$ kj/mol and $delta s=-0.24$ kj/(mol·k)?
a. it is always spontaneous.
b. it is never spontaneous.
c. it is at equilibrium at 371 k.
d. it is spontaneous at 2000 k.

Explanation:

Step1: Recall Gibbs - free energy formula

The Gibbs - free energy change is given by $\Delta G=\Delta H - T\Delta S$, where $\Delta H$ is the enthalpy change, $\Delta S$ is the entropy change, and $T$ is the temperature in Kelvin. A reaction is spontaneous when $\Delta G<0$, non - spontaneous when $\Delta G > 0$, and at equilibrium when $\Delta G=0$.

Step2: Analyze the sign of $\Delta G$ based on given $\Delta H$ and $\Delta S$ values

Given $\Delta H=- 890\ kJ/mol$ and $\Delta S=-0.24\ kJ/(mol\cdot K)$. Substitute into the $\Delta G$ formula: $\Delta G=-890 - T\times(-0.24)=-890 + 0.24T$.

Step3: Find the temperature for equilibrium

Set $\Delta G = 0$:
\[

$$\begin{align*} 0&=-890+0.24T\\ 0.24T&=890\\ T&=\frac{890}{0.24}\approx3708\ K \end{align*}$$

\]
When $T < 3708\ K$, $\Delta G=-890 + 0.24T<0$ (since the negative $\Delta H$ term dominates at lower temperatures). When $T>3708\ K$, $\Delta G=-890 + 0.24T>0$.

Answer:

A. It is always spontaneous.