Question

question 2 of 25
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 formula for Gibbs - free energy is $\Delta G=\Delta H - T\Delta S$, where $\Delta G$ is the change in Gibbs - free energy, $\Delta H$ is the change in enthalpy, $T$ is the temperature in Kelvin, and $\Delta S$ is the change in entropy. A reaction is spontaneous when $\Delta G<0$, at equilibrium when $\Delta G = 0$, and non - spontaneous when $\Delta G>0$.

Step2: Calculate the temperature at equilibrium

Set $\Delta G = 0$. Then $0=\Delta H - T\Delta S$. Solving for $T$ gives $T=\frac{\Delta H}{\Delta S}$. Substitute $\Delta H=-890\ kJ/mol$ and $\Delta S=- 0.24\ kJ/(mol\cdot K)$ into the formula: $T=\frac{-890\ kJ/mol}{-0.24\ kJ/(mol\cdot K)}\approx3708\ K$.

Step3: Analyze the spontaneity at different temperatures

Since $\Delta H<0$ and $\Delta S<0$, at low temperatures, the $-T\Delta S$ term is positive but small (because $T$ is small). So, $\Delta G=\Delta H - T\Delta S<0$ at low temperatures. As $T$ increases, the $-T\Delta S$ term becomes more positive. When $T > \frac{\Delta H}{\Delta S}\approx3708\ K$, $\Delta G>0$ and the reaction is non - spontaneous.

Answer:

There is no correct option among A, B, C, D. If we assume there is a small error in the calculation of the equilibrium temperature and consider the closest conceptually correct option, we note that at low temperatures (much lower than 3708 K), the reaction is spontaneous. But among the given options, none are correct based on the accurate calculation. If we had to choose the "best" wrong option, we could say that since at low temperatures the reaction is spontaneous and 371 K is a low temperature compared to 3708 K, we could wrongly assume that the reaction is spontaneous at 371 K (but this is based on an incorrect interpretation of the options and a small error in the equilibrium temperature calculation). However, if we follow the strict thermodynamics and calculations, none of the options are correct.