A schematic plot of ln K_eq versus inverse of temperature for a reaction is shown below. The reactio

Q: Why do we use $\ln K_{eq}$ instead of $\log_{10} K_{eq}$ in the graph?

The standard thermodynamic derivation of the van't Hoff equation naturally results in the natural logarithm ($\ln$) because it originates from the integration of $1/T$. If we use base 10 logarithm, we must include the conversion factor $2.303$ in the denominator, making the slope $-\Delta H^\circ / (2.303 R)$.

Q: What exactly does a positive slope indicate in this specific graph?

A positive slope means that as $1/T$ increases (which actually means the temperature $T$ is decreasing), the value of $\ln K_{eq}$ increases. Mathematically, it implies $-\Delta H^\circ / R > 0$, which forces $\Delta H^\circ$ to be negative.

Q: Can we determine the exact numerical value of $\Delta H^\circ$ from this graph?

Absolutely! By calculating the exact slope using the given coordinates $(1.5 \times 10^{-3}, 2.0)$ and $(2.0 \times 10^{-3}, 6.0)$, we can find the numerical value of $-\Delta H^\circ / R$. Multiplying that slope by $-R$ will give you the exact enthalpy change.

Q: Why is the reaction exothermic if $\Delta H^\circ$ is negative?

By thermodynamic convention, $\Delta H^\circ$ represents the heat absorbed by the system. If it is negative, it means the system has a net loss of heat to the surroundings. Releasing heat is the fundamental definition of an exothermic reaction.

Q: Does the y-intercept of this graph have any physical significance?

Yes, it does! If you extrapolate the line to where it crosses the y-axis (where $1/T = 0$), the intercept corresponds to $\Delta S^\circ / R$. This allows chemists to determine the standard entropy change of the reaction purely from experimental equilibrium data.

Question

A schematic plot of $\ln K_{eq}$ versus inverse of temperature for a reaction is shown below

The reaction must be

Options:

(A) highly spontaneous at ordinary temperature

(B) one with negligible enthalpy change

(C) endothermic

(D) exothermic

Sigma Percentile · Accepted Answer

### Decoding the Thermodynamics of Equilibrium: The van't Hoff Plot

Imagine you are trying to predict how a chemical reaction will behave when you turn up the heat. Will it push forward to create more products, or will it retreat back to reactants? The secret to answering this lies in the beautiful intersection of thermodynamics and chemical equilibrium, perfectly captured by the **van't Hoff equation**.

In this problem, we are presented with a graph plotting the natural logarithm of the equilibrium constant, $\ln K_{eq}$, against the inverse of temperature, $\frac{1}{T}$. At first glance, it's just a straight line heading upwards. But to a trained eye, this line is a treasure map revealing the very nature of the reaction's energy flow.

### The Master Equation

To decode the graph, we must first recall the fundamental thermodynamic relationship:

$$\Delta G^\circ = -RT \ln K_{eq}$$

We also know that the standard Gibbs free energy change is related to enthalpy and entropy:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

By equating these two expressions and rearranging them to solve for $\ln K_{eq}$, we arrive at the van't Hoff equation:

$$\ln K_{eq} = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$$

### Analyzing the Setup

Now, let's look at this equation through the lens of coordinate geometry. It perfectly mirrors the equation of a straight line, $y = mx + c$.

If we set our y-axis to be $\ln K_{eq}$ and our x-axis to be $\frac{1}{T}$, the equation maps out as follows:
- **y-variable:** $\ln K_{eq}$
- **x-variable:** $\frac{1}{T}$
- **Slope ($m$):** $-\frac{\Delta H^\circ}{R}$
- **y-intercept ($c$):** $\frac{\Delta S^\circ}{R}$

### Final Calculation and Conclusion

Let's turn our attention back to the graph provided in the question. As we move from left to right (meaning $\frac{1}{T}$ is increasing), the value of $\ln K_{eq}$ is also increasing. This upward trajectory means the line has a **positive slope** ($m > 0$).

Mathematically, this translates to:

$$-\frac{\Delta H^\circ}{R} > 0$$

Since the universal gas constant $R$ is a strictly positive value ($8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$), the only way for the entire fraction to be positive is if the numerator itself is negative. Therefore, the standard enthalpy change, $\Delta H^\circ$, must be less than zero:

$$\Delta H^\circ < 0$$

In the language of chemistry, a negative enthalpy change signifies that the system is releasing heat into its surroundings. This is the textbook definition of an **exothermic reaction**.

So, just by glancing at the upward slope of a van't Hoff plot, we can confidently declare that the reaction is exothermic. If the line had been sloping downwards, it would have been a dead giveaway for an endothermic process. Always let the math tell the physical story!

Animated Solution for Chemistry - Chemical Thermodynamics: A schematic plot of $ln K_{e q}$ versus inverse of temperature for a reaction is shown below The reaction must be

Select Answer:

Visualized Solution

$Analyzing the Graph$

$The van’t Hoff Equation$

$Comparing with Straight Line Equation$

$Determining the Sign of the Slope$

$Conclusion on Enthalpy Change$

$The Way Forward$

The Sigma Insight: Entropy and Free Energy

Decoding the Thermodynamics of Equilibrium

The Master Equation

Analyzing the Setup

Final Calculation and Conclusion

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Animated Solution for Chemistry - Chemical Thermodynamics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A schematic plot of lnKeq​ versus inverse of temperature for a reaction is shown below The reaction must be