The variation of equilibrium constant with temperature is given below:
Temperature Equilibrium const

Q: Why do we use 2.303 in the formulas?

The formulas naturally use the natural logarithm ($\ln$ or base $e$). To convert it to the common logarithm ($\log$ or base 10), which is easier to calculate with numbers like 10 and 100, we multiply by 2.303.

Q: Why must temperature be in Kelvin?

Thermodynamic equations are derived based on the absolute temperature scale. Using Celsius will give completely incorrect results because it doesn't start at absolute zero.

Q: Is $\Delta H^\circ$ assumed to be constant with temperature?

Yes, over small temperature ranges, $\Delta H^\circ$ is generally assumed to be constant unless heat capacity data ($\Delta C_p$) is provided. The Van't Hoff equation relies on this assumption.

Q: Why did we get a positive $\Delta H^\circ$?

A positive $\Delta H^\circ$ indicates an endothermic reaction. Notice that as temperature increased from $25^\circ\text{C}$ to $100^\circ\text{C}$, the equilibrium constant increased from 10 to 100. According to Le Chatelier's principle, this happens for endothermic reactions.

Q: Why is $\Delta G^\circ$ negative?

A negative $\Delta G^\circ$ indicates that the reaction is spontaneous in the forward direction under standard conditions. Since $K > 1$ for both temperatures, the products are favored, which aligns with a negative $\Delta G^\circ$.

Question

The variation of equilibrium constant with temperature is given below:
$\begin{array}{cc} 	ext{Temperature} & 	ext{Equilibrium constant} \ T_1 = 25^\circ	ext{C} & K_1 = 10 \ T_2 = 100^\circ	ext{C} & K_2 = 100 \end{array}$
The values of $\Delta H^\circ$, $\Delta G^\circ$ at $T_1$ and $\Delta G^\circ$ at $T_2$ (in $	ext{kJ mol}^{-1}$) respectively, are close to [use $R = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$]

Options:

(A) $28.4, -7.14 	ext{ and } -5.71$

(B) $0.64, -7.14 	ext{ and } -5.71$

(C) $28.4, -5.71 	ext{ and } -14.29$

(D) $0.64, -5.71 	ext{ and } -14.29$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
We are given the equilibrium constants for a reaction at two different temperatures. At $25^\circ	ext{C}$, the equilibrium constant $K_1$ is $10$. When the temperature is raised to $100^\circ	ext{C}$, the equilibrium constant $K_2$ increases to $100$.

Our goal is to find the standard enthalpy change, $\Delta H^\circ$, and the standard Gibbs free energy change, $\Delta G^\circ$, at both temperatures. Before we plug anything into formulas, we must convert our temperatures to the absolute Kelvin scale. 
$T_1 = 25 + 273 = 298	ext{ K}$
$T_2 = 100 + 273 = 373	ext{ K}$

### The Master Equation: Van't Hoff
To find the standard enthalpy change, we need a relationship that connects the equilibrium constant with temperature. The **Van't Hoff equation** is the perfect tool for this job:
$$\log \left( \frac{K_2}{K_1} ight) = \frac{\Delta H^\circ}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} ight)$$

Let's substitute our known values into this equation:
$$\log \left( \frac{100}{10} ight) = \frac{\Delta H^\circ}{2.303 	imes 8.314} \left( \frac{1}{298} - \frac{1}{373} ight)$$

The left side simplifies beautifully to $\log(10)$, which is just $1$. On the right side, the product $2.303 	imes 8.314$ is approximately $19.147$. 
$$1 = \frac{\Delta H^\circ}{19.147} \left( \frac{373 - 298}{298 	imes 373} ight)$$
$$1 = \frac{\Delta H^\circ}{19.147} \left( \frac{75}{111154} ight)$$

Solving for $\Delta H^\circ$, we get:
$$\Delta H^\circ = \frac{19.147 	imes 111154}{75} \approx 28377	ext{ J mol}^{-1}$$

Since the options are in kilojoules, we divide by $1000$:
**$\Delta H^\circ \approx 28.4	ext{ kJ mol}^{-1}$**

### Calculating Gibbs Free Energy
Now, we need to find the standard Gibbs free energy change at both temperatures. The fundamental thermodynamic relation we use is:
$$\Delta G^\circ = -RT \ln K$$

Converting the natural logarithm to base-10, we get:
$$\Delta G^\circ = -2.303 RT \log K$$

Let's calculate this for our first temperature, $T_1 = 298	ext{ K}$:
$$\Delta G^\circ_{T_1} = -2.303 	imes 8.314 	imes 298 	imes \log(10)$$
$$\Delta G^\circ_{T_1} = -19.147 	imes 298 	imes 1 \approx -5706	ext{ J mol}^{-1}$$
**$\Delta G^\circ_{T_1} = -5.71	ext{ kJ mol}^{-1}$**

Next, we calculate it for the second temperature, $T_2 = 373	ext{ K}$:
$$\Delta G^\circ_{T_2} = -2.303 	imes 8.314 	imes 373 	imes \log(100)$$
$$\Delta G^\circ_{T_2} = -19.147 	imes 373 	imes 2 \approx -14283	ext{ J mol}^{-1}$$
**$\Delta G^\circ_{T_2} = -14.29	ext{ kJ mol}^{-1}$**

### Final Conclusion
We have successfully calculated all three required thermodynamic parameters. Comparing these results with the given options, we can confidently conclude that option (c) is the correct answer.

Select Answer:

Visualized Solution

$Given Data$

$Van’t Hoff Equation$

$Substituting Values$

$Calculating Δ H^{\circ}$

$Gibbs Free Energy Equation$

$Calculating Δ G^{\circ} at T_{1}$

$Calculating Δ G^{\circ} at T_{2}$

$Final Answer$

The Sigma Insight: Entropy and Free Energy

Analyzing the Setup We are given the equilibrium constants for a reaction at two different temperatures

The Master Equation

Calculating Gibbs Free Energy Now, we need to find the standard Gibbs free energy change at both temperatures

Final Conclusion We have successfully calculated all three required thermodynamic parameters

Similar Questions

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Comprehension Passage

The value of entropy change, $S_{β} - S_{α}$ (in $J mol^{- 1} K^{- 1}$ ), at 300 K is ______. [Use : $ln 2 = 0.69$ Given : $S_{β} - S_{α} = 0$ at 0 K]

The value of enthalpy change, $H_{β} - H_{α}$ (in $J mol^{- 1}$ ), at 300 K is _______.

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Data

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Van’t Hoff Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔH∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gibbs Free Energy Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔG∘ at T1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔG∘ at T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

The Sigma Insight: Entropy and Free Energy

Analyzing the Setup We are given the equilibrium constants for a reaction at two different temperatures

The Master Equation

Calculating Gibbs Free Energy Now, we need to find the standard Gibbs free energy change at both temperatures

Final Conclusion We have successfully calculated all three required thermodynamic parameters

Similar Questions

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the equilibrium, 2H2​O⇌H3​O++OH−, the value of ΔG∘ at 298 K is approximately

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of standard enthalpy, ΔH⊖ (in kJ mol−1) for the reaction is________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ΔS⊖ (in J K−1 mol−1) for the given reaction, at 1000 K is________.

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of entropy change, Sβ​−Sα​ (in J mol−1 K−1), at 300 K is ______. [Use : ln2=0.69 Given : Sβ​−Sα​=0 at 0 K]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of enthalpy change, Hβ​−Hα​ (in J mol−1), at 300 K is _______.

$Given Data$

$Van’t Hoff Equation$

$Substituting Values$

$Calculating Δ H^{\circ}$

$Gibbs Free Energy Equation$

$Calculating Δ G^{\circ} at T_{1}$

$Calculating Δ G^{\circ} at T_{2}$

$Final Answer$

For the reaction; $A (l) ⟶ 2 B (g)$ $Δ U = 2.1 kcal, Δ S = 20 cal K^{- 1}$ at $300 K$ . Hence, $Δ G$ in kcal is ......

For the equilibrium, $2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ , the value of $Δ G^{\circ}$ at $298 K$ is approximately

The value of standard enthalpy, $Δ H^{⊖}$ (in $kJ mol^{- 1}$ ) for the reaction is________.

The value of $Δ S^{⊖}$ (in $J K^{- 1} mol^{- 1}$ ) for the given reaction, at $1000 K$ is________.

The value of entropy change, $S_{β} - S_{α}$ (in $J mol^{- 1} K^{- 1}$ ), at 300 K is ______. [Use : $ln 2 = 0.69$ Given : $S_{β} - S_{α} = 0$ at 0 K]

The value of enthalpy change, $H_{β} - H_{α}$ (in $J mol^{- 1}$ ), at 300 K is _______.