For the reaction, A(g) - B(g), the value of the equilibrium constant at 300 K and 1 atm is equal to

Q: Why do we use $K_p$ instead of $K_c$?

Since the reaction involves gases and the standard state is defined at 1 atm pressure, the equilibrium constant in terms of partial pressures ($K_p$) is the appropriate one to use for standard Gibbs free energy calculations.

Q: Why didn't we substitute the value of $R$?

The question specifically asks for the answer in the form of $-xR$. Therefore, substituting the numerical value of $R$ would add unnecessary calculations and we would just have to divide by it later.

Q: What does the negative sign in $\Delta G^\circ$ indicate?

A negative $\Delta G^\circ$ indicates that the forward reaction is thermodynamically favorable and spontaneous under standard conditions.

Q: How did $\ln(100)$ become $2 \ln(10)$?

This is a standard property of logarithms: $\ln(a^b) = b \ln(a)$. Since $100 = 10^2$, $\ln(100) = \ln(10^2) = 2 \ln(10)$.

Q: Is $\Delta_r G$ the same as $\Delta G^\circ$?

The question asks for $\Delta_r G$ at 1 atm, which corresponds to standard conditions. Therefore, in this context, $\Delta_r G$ is exactly the standard Gibbs free energy change, $\Delta G^\circ$.

Question

For the reaction, $A(g) \longrightarrow B(g)$, the value of the equilibrium constant at $300	ext{ K}$ and $1	ext{ atm}$ is equal to $100.0$. The value of $\Delta_r G$ for the reaction at $300	ext{ K}$ and $1	ext{ atm}$ in $	ext{J mol}^{-1}$ is $-xR$, where $x$ is ......... . (Rounded off to the nearest integer)
($R = 8.31	ext{ J mol}^{-1}	ext{ K}^{-1}$ and $\ln 10 = 2.3$)

Sigma Percentile · Accepted Answer

Have you ever wondered why some chemical reactions happen spontaneously while others refuse to budge without a massive push of energy? The secret lies in a powerful thermodynamic concept known as **Gibbs Free Energy**.

In this problem, we are exploring a simple gaseous reaction where reactant $A$ transforms into product $B$. We are given the equilibrium constant and asked to find the standard Gibbs free energy change, $\Delta_r G^\circ$. Let's embark on this thermodynamic journey and decode the math behind the spontaneity!

### Analyzing the Setup

Imagine a closed vessel at a comfortable room temperature of $300 	ext{ K}$ and a standard pressure of $1 	ext{ atm}$. Inside, gas $A$ is converting into gas $B$.

The problem tells us that the equilibrium constant, $K_p$, is $100.0$. What does this number physically mean? An equilibrium constant greater than $1$ indicates that at equilibrium, the concentration (or partial pressure) of the products heavily outweighs the reactants. The reaction naturally "wants" to move forward.

In thermodynamic terms, this forward drive means the products exist at a lower, more stable energy state than the reactants. Therefore, we expect the change in Gibbs free energy to be negative.

### The Master Equation

To quantify this energy change, we use the fundamental bridge between thermodynamics and chemical equilibrium:

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

This elegant equation tells us exactly how much "free" energy is released (or absorbed) when reactants convert to products under standard conditions. 
- **$R$** is the universal gas constant.
- **$T$** is the absolute temperature in Kelvin.
- **$K_p$** is the equilibrium constant.

Notice the negative sign! It mathematically ensures that a large equilibrium constant ($K_p > 1$, so $\ln K_p > 0$) results in a negative $\Delta G^\circ$, confirming our intuition about spontaneity.

### Executing the Calculation

Let's substitute our known values into the master equation. We have $T = 300 	ext{ K}$ and $K_p = 100$. The problem cleverly asks for the answer in terms of $R$, so we don't need to plug in $8.31 	ext{ J mol}^{-1} 	ext{ K}^{-1}$ just yet.

$$ \Delta G^\circ = -R (300) \ln(100) $$

Now, we face the natural logarithm of $100$. We can simplify this using a classic logarithm property. Since $100$ is $10^2$, we can bring the exponent to the front:

$$ \ln(100) = \ln(10^2) = 2 \ln(10) $$

The problem generously provides the value of $\ln(10)$ as $2.3$. Let's plug that in:

$$ 2 \ln(10) = 2 	imes 2.3 = 4.6 $$

Now, we bring it all together:

$$ \Delta G^\circ = -R 	imes 300 	imes 4.6 $$

Multiplying $300$ by $4.6$ gives us $1380$. Therefore, our standard Gibbs free energy change is:

$$ \Delta G^\circ = -1380 R $$

### The Final Answer

The question states that the value of $\Delta_r G$ is $-xR$. By directly comparing our calculated expression with the given format, the mystery is solved:

$$ -xR = -1380 R $$

$$ \mathbf{x = 1380} $$

And there we have it! By understanding the deep connection between the equilibrium state and thermodynamic stability, we smoothly navigated the math to arrive at the correct integer. Always remember, the equilibrium constant isn't just a number; it's a direct reflection of the energy landscape of the molecules!

Enter Numerical Value:

Visualized Solution

$A (g) ⇌ B (g)$

$Δ G^{\circ} Formula$

$Substitution$

$Logarithm Property$

$Value of ln (100)$

$Calculating Δ G^{\circ}$

$Finding x$

$Concept Check$

The Sigma Insight: Entropy and Free Energy

Analyzing the Setup

The Master Equation

Executing the Calculation

The Final Answer

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(g)⇌B(g)

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Value of ln(100)

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

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

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