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

Q: Why is $\Delta U = 0$ for an isothermal process?

For an ideal gas, the internal energy depends solely on its temperature. Since the temperature remains constant during an isothermal process, the change in internal energy ($\Delta U$) must be zero.

Q: What is the difference between $\Delta G$ and $\Delta G^\circ$?

$\Delta G$ is the Gibbs free energy change under any given set of conditions, and it determines the direction of spontaneous change. $\Delta G^\circ$ is the standard Gibbs free energy change, measured when all reactants and products are in their standard states (1 bar pressure, specified temperature). At equilibrium, $\Delta G = 0$, but $\Delta G^\circ$ is a constant related to the equilibrium constant $K$.

Q: Why does the formula for work have a negative sign?

By IUPAC convention, work done *on* the system is positive (energy is added), and work done *by* the system is negative (energy is lost). During expansion ($V_f > V_i$), the system does work on the surroundings, so the work term must be negative.

Q: How do we get the entropy of the surroundings?

The surroundings are considered a massive heat reservoir that absorbs or releases heat reversibly at a constant temperature $T$. The heat gained by the surroundings is the negative of the heat lost by the system ($-\Delta H_{\text{system}}$ at constant pressure). Thus, $\Delta S_{\text{surroundings}} = \frac{q_{\text{surroundings}}}{T} = -\frac{\Delta H_{\text{system}}}{T}$.

Q: Why is option (c) incorrect?

Option (c) misses a crucial negative sign. The correct derivation from $\Delta G^\circ = -RT \ln K$ gives $\ln K = -\frac{\Delta G^\circ}{RT}$. Substituting $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ yields $\ln K = -\frac{(\Delta H^\circ - T\Delta S^\circ)}{RT}$, which is not what option (c) states.

Question

The incorrect expression among the following is

Options:

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

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

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

(D) $K = e^{-\Delta G^\circ / RT}$

Sigma Percentile · Accepted Answer

The beauty of chemical thermodynamics lies in its interconnected equations, each telling a profound story about energy, entropy, and equilibrium. In this problem, we are tasked with identifying the imposter—the incorrect mathematical expression among four fundamental thermodynamic relationships. Let's embark on a journey to verify each one.

### The Entropy of the Universe

Let's begin by evaluating the first expression, which relates the Gibbs free energy of the system to the total entropy of the universe. We know that the total entropy change is the sum of the entropy changes of the system and its surroundings:
$$ \Delta S_{	ext{total}} = \Delta S_{	ext{system}} + \Delta S_{	ext{surroundings}} $$

The surroundings act as a massive thermal reservoir. When the system undergoes a process at constant pressure, the heat it exchanges with the surroundings is equal to its enthalpy change, $\Delta H_{	ext{system}}$. Therefore, the entropy change of the surroundings is:
$$ \Delta S_{	ext{surroundings}} = -\frac{\Delta H_{	ext{system}}}{T} $$

Substituting this back into our total entropy equation, we get:
$$ \Delta S_{	ext{total}} = \Delta S_{	ext{system}} - \frac{\Delta H_{	ext{system}}}{T} $$

Multiplying the entire equation by the absolute temperature $T$ yields:
$$ T\Delta S_{	ext{total}} = T\Delta S_{	ext{system}} - \Delta H_{	ext{system}} $$

Does the right side look familiar? It is exactly the negative of the Gibbs free energy change of the system ($\Delta G_{	ext{system}} = \Delta H_{	ext{system}} - T\Delta S_{	ext{system}}$). Thus:
$$ T\Delta S_{	ext{total}} = -\Delta G_{	ext{system}} \implies \frac{\Delta G_{	ext{system}}}{\Delta S_{	ext{total}}} = -T $$

This confirms that option (a) is a perfectly valid and beautiful expression, showing that for a spontaneous process ($\Delta S_{	ext{total}} > 0$), the Gibbs free energy of the system must decrease ($\Delta G_{	ext{system}} < 0$).

### The Isothermal Journey

Next, let's examine the expression for the reversible work done during an isothermal process. In an isothermal expansion or compression of an ideal gas, the temperature remains constant. Consequently, the change in internal energy is zero ($\Delta U = 0$).

According to the First Law of Thermodynamics ($\Delta U = q + W$), the heat absorbed by the gas is entirely converted into work done by the gas:
$$ q = -W_{	ext{reversible}} $$

The reversible work done by the gas as it expands from an initial volume $V_i$ to a final volume $V_f$ is given by the integral of pressure with respect to volume:
$$ W_{	ext{reversible}} = -\int_{V_i}^{V_f} p \, dV $$

Using the ideal gas law ($p = \frac{nRT}{V}$), we can substitute for pressure:
$$ W_{	ext{reversible}} = -\int_{V_i}^{V_f} \frac{nRT}{V} \, dV = -nRT \ln\left(\frac{V_f}{V_i}ight) $$

This confirms that option (b) is also a correct mathematical statement.

### The Equilibrium State

Now, let's jump to option (d), which deals with the equilibrium constant. The relationship between the standard Gibbs free energy change ($\Delta G^\circ$) and the reaction quotient ($Q$) at any moment is given by the isotherm equation:
$$ \Delta G = \Delta G^\circ + RT \ln Q $$

When the system reaches equilibrium, it can do no more useful work, meaning $\Delta G = 0$. At this precise moment, the reaction quotient $Q$ becomes the equilibrium constant $K$:
$$ 0 = \Delta G^\circ + RT \ln K $$

Rearranging this equation to solve for $\Delta G^\circ$, we get one of the most important equations in chemistry:
$$ \Delta G^\circ = -RT \ln K $$

To isolate $K$, we divide by $-RT$ and take the exponential of both sides:
$$ \ln K = -\frac{\Delta G^\circ}{RT} \implies K = e^{-\frac{\Delta G^\circ}{RT}} $$

This proves that option (d) is absolutely correct.

### The Final Verdict

Finally, we arrive at option (c). We just established that:
$$ \ln K = -\frac{\Delta G^\circ}{RT} $$

We also know the fundamental definition of standard Gibbs free energy:
$$ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ $$

Substituting this definition into our logarithmic equation, we must be incredibly careful with the negative sign:
$$ \ln K = -\frac{(\Delta H^\circ - T\Delta S^\circ)}{RT} = \frac{-\Delta H^\circ + T\Delta S^\circ}{RT} $$

If we look closely at option (c), it states:
$$ \ln K = \frac{\Delta H^\circ - T\Delta S^\circ}{RT} $$

The crucial negative sign in front of the enthalpy term is missing! This subtle yet fatal flaw makes option (c) the **incorrect expression**, and therefore, the correct answer to our problem.

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

Select Answer:

Visualized Solution

$Evaluating Option (a): System and Universe$

$Evaluating Option (b): Isothermal Reversible Work$

$Evaluating Option (d): Equilibrium Constant$

$Evaluating Option (c): The Incorrect Expression$

The Sigma Insight: Entropy and Free Energy

The Entropy of the Universe

The Isothermal Journey

The Equilibrium State

The Final Verdict

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