The incorrect expression among the following is
(a) \Delta G_System / \Delta S_Total = -T (at consta

Q: What does the negative sign in the work formula $W = -nRT \ln(V_f/V_i)$ signify?

The negative sign ensures that the work follows the IUPAC sign convention. If a gas expands ($V_f > V_i$), the natural log term is positive, making the overall work negative. This makes sense because the system is doing work on the surroundings, losing energy. Conversely, if the gas is compressed ($V_f < V_i$), the natural log term is negative, making the overall work positive, meaning work is done ON the system.

Q: Why do we use $\Delta G^\circ$ instead of $\Delta G$ in the equilibrium constant equation?

$\Delta G$ represents the free energy change at any arbitrary moment during a reaction, and it equals zero at equilibrium. $\Delta G^\circ$, however, is the standard free energy change—the difference in free energy between pure products and pure reactants in their standard states. The equation $\Delta G^\circ = -RT \ln K$ relates this fixed standard difference to the position of equilibrium.

Q: Is the equation $\Delta G_{\text{System}} / \Delta S_{\text{Total}} = -T$ valid for all processes?

This specific derivation assumes constant temperature and constant pressure. Under these conditions, the heat transfer is equal to the enthalpy change, allowing us to substitute $-\Delta H_{\text{sys}}/T$ for the entropy of the surroundings. So, it is valid specifically for isothermal and isobaric processes.

Q: How does the missing negative sign in option (b) physically affect the result?

The missing negative sign completely flips the temperature dependence of the equilibrium constant. With the correct formula, an exothermic reaction (negative $\Delta H^\circ$) results in a larger $K$ at lower temperatures. The incorrect formula in option (b) would imply the exact opposite, violating Le Chatelier's principle.

Question

The incorrect expression among the following is

Options:

(A) $\frac{\Delta G_{	ext{System}}}{\Delta S_{	ext{Total}}} = -T$ (at constant $p$)

(B) $\ln k = \frac{\Delta H^\circ - T\Delta S^\circ}{RT}$

(C) $k = e^{-\frac{\Delta G^\circ}{RT}}$

(D) For isothermal process, $W_{	ext{reversible}} = -nRT \ln \frac{V_f}{V_i}$

Sigma Percentile · Accepted Answer

This question is a fantastic test of your foundational knowledge in Chemical Thermodynamics. It doesn't ask you to solve a complex numerical problem; instead, it challenges you to verify the mathematical integrity of four fundamental equations. Let's embark on a journey to dissect each option and uncover the imposter.

### Analyzing the Entropy of the Universe

Let's start with option (a). The Second Law of Thermodynamics tells us that for any spontaneous process, the total entropy of the universe must increase. The total entropy change is the sum of the entropy change of the system and the surroundings:

$$\Delta S_{	ext{Total}} = \Delta S_{	ext{System}} + \Delta S_{	ext{Surr}}$$

Now, how do we quantify the entropy change of the surroundings? At constant pressure, the heat exchanged by the system is equal to its enthalpy change, $\Delta H_{	ext{System}}$. If the system releases heat, the surroundings absorb it. Therefore, the entropy change of the surroundings is:

$$\Delta S_{	ext{Surr}} = -\frac{\Delta H_{	ext{System}}}{T}$$

Substituting this back into our total entropy equation:

$$\Delta S_{	ext{Total}} = \Delta S_{	ext{System}} - \frac{\Delta H_{	ext{System}}}{T}$$

To make this equation more elegant, let's multiply the entire expression by the absolute temperature, $T$:

$$T\Delta S_{	ext{Total}} = T\Delta S_{	ext{System}} - \Delta H_{	ext{System}}$$

If we factor out a negative sign on the right side, we get $-(\Delta H_{	ext{System}} - T\Delta S_{	ext{System}})$. Does that term inside the parenthesis look familiar? Yes, it is the definition of the Gibbs Free Energy change of the system, $\Delta G_{	ext{System}}$! Therefore:

$$T\Delta S_{	ext{Total}} = -\Delta G_{	ext{System}}$$

Rearranging this gives us $\frac{\Delta G_{	ext{System}}}{\Delta S_{	ext{Total}}} = -T$. Option (a) is a perfectly valid and profound thermodynamic relationship.

### The Equilibrium Constant and Free Energy

Next, let's evaluate option (c). One of the most critical equations linking thermodynamics to chemical equilibrium is the relationship between the standard Gibbs free energy change, $\Delta G^\circ$, and the equilibrium constant, $K$:

$$\Delta G^\circ = -RT \ln K$$

If we want to solve for $K$, we first isolate the natural logarithm:

$$\ln K = -\frac{\Delta G^\circ}{RT}$$

To remove the natural logarithm, we take the exponential of both sides:

$$K = e^{-\frac{\Delta G^\circ}{RT}}$$

This confirms that option (c) is mathematically sound.

### Reversible Isothermal Work

Option (d) presents the formula for the work done during a reversible isothermal expansion of an ideal gas. The fundamental definition of pressure-volume work is:

$$W_{	ext{reversible}} = -\int_{V_i}^{V_f} p \, dV$$

For an ideal gas, $p = \frac{nRT}{V}$. Substituting this into the integral:

$$W_{	ext{reversible}} = -\int_{V_i}^{V_f} \frac{nRT}{V} \, dV$$

Since the process is isothermal, temperature $T$ is constant, allowing us to pull $nRT$ out of the integral:

$$W_{	ext{reversible}} = -nRT \int_{V_i}^{V_f} \frac{1}{V} \, dV$$

Integrating $\frac{1}{V}$ yields the natural logarithm, resulting in:

$$W_{	ext{reversible}} = -nRT \ln \left(\frac{V_f}{V_i}ight)$$

Option (d) is absolutely correct.

### The Imposter Revealed

Finally, let's scrutinize option (b). We know two fundamental expressions for the standard Gibbs free energy change:

1. $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$
2. $\Delta G^\circ = -RT \ln K$

By equating these two expressions, we can relate the equilibrium constant directly to enthalpy and entropy:

$$-RT \ln K = \Delta H^\circ - T\Delta S^\circ$$

To isolate $\ln K$, we must divide both sides by $-RT$. **This is where the trap lies!**

$$\ln K = \frac{\Delta H^\circ - T\Delta S^\circ}{-RT}$$

Bringing the negative sign to the numerator, we get:

$$\ln K = -\frac{\Delta H^\circ - T\Delta S^\circ}{RT} = \frac{-\Delta H^\circ + T\Delta S^\circ}{RT}$$

Now, look closely at option (b):

$$\ln k = \frac{\Delta H^\circ - T\Delta S^\circ}{RT}$$

It completely misses the negative sign! This missing sign would invert the relationship, suggesting that an exothermic reaction (negative $\Delta H^\circ$) would decrease the equilibrium constant, which contradicts Le Chatelier's principle. Because of this crucial missing negative sign, **option (b) is the incorrect expression**.

Animated Solution for Chemistry - Chemical Thermodynamics: The incorrect expression among the following is

Select Answer:

Visualized Solution

$Checking Option (a)$

$Checking Option (c)$

$Checking Option (d)$

$Checking Option (b)$

$Conclusion$

The Sigma Insight: Entropy and Free Energy

Analyzing the Entropy of the Universe

The Equilibrium Constant and Free Energy

Reversible Isothermal Work

The Imposter Revealed

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