The gas phase reaction 2A(g) A2(g) at 400 K has ΔG° = +25.2 kJ mol^-1. The equilibrium constant

Q: Why did we convert $\Delta G^\circ$ to Joules?

Because the universal gas constant $R$ is given in $\text{J K}^{-1} \text{mol}^{-1}$. To maintain dimensional consistency in the equation $\Delta G^\circ = -RT \ln K_p$, both energy terms must be in the same unit.

Q: How did we calculate the antilog of -3.3 without a calculator?

We used the laws of exponents. We split $10^{-3.3}$ into $10^{-3} \times 10^{-0.3}$. The problem explicitly provided the value of $10^{-0.3}$ (which is $\text{antilog}(-0.3)$) as $0.501$.

Q: Why is $\Delta n_g = -1$?

$\Delta n_g$ is the difference between the sum of stoichiometric coefficients of gaseous products and gaseous reactants. Here, we have $1$ mole of $A_2(g)$ and $2$ moles of $A(g)$, so $1 - 2 = -1$.

Q: Why did the value of $R$ change when calculating $K_c$?

The units of $K_p$ (in bar) and $K_c$ (in mol/L) dictate the units of $R$ required for the conversion. We needed $R$ in $\text{L bar K}^{-1} \text{mol}^{-1}$, which is $0.083$, instead of $8.3 \text{ J K}^{-1} \text{mol}^{-1}$.

Q: Why did we round 1.66 to 2?

The question explicitly instructed to 'Round off to the nearest integer'. Since $1.66$ is closer to $2$ than to $1$, the final answer is $2$.

Question

The gas phase reaction
$2A(g) ightleftharpoons A_2(g)$
at $400	ext{ K}$ has $\Delta G^\circ = +25.2	ext{ kJ mol}^{-1}$.
The equilibrium constant $K_C$ for this reaction is ......... $	imes 10^{-2}$.
(Round off to the nearest integer).
[Use : $R = 8.3	ext{ J mol}^{-1}	ext{ K}^{-1}$, $\ln 10 = 2.3$]
$\log_{10} 2 = 0.30$, $1	ext{ atm} = 1	ext{ bar}$]
$[	ext{antilog}(-0.3) = 0.501]$

Sigma Percentile · Accepted Answer

### The Thermodynamic Setup

Imagine you are in a laboratory, observing a gas-phase reaction where two moles of gas **A** are trying to combine to form a single mole of **A_2**. The universe, however, has its own plans. We are given that the standard Gibbs free energy change, $\Delta G^\circ$, is $+25.2 	ext{ kJ mol}^{-1}$.

What does this positive value physically mean? It tells us that under standard conditions, the reaction is **non-spontaneous** in the forward direction. The universe favors the reactants, and the equilibrium position will lie far to the left.

### The Master Equation

To find exactly where this equilibrium lies, we need to calculate the equilibrium constant. The bridge between thermodynamics and chemical equilibrium is given by the master equation:

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

Since the problem provides the value of $\ln 10 = 2.3$, it is a massive hint to convert our natural logarithm into a base-10 logarithm. The equation transforms into:

$$ \Delta G^\circ = -2.303 RT \log_{10} K_p $$

For simplicity, we will use $2.3$ as instructed.

### Substituting the Values

Here is where many students fall into a classic trap: **unit mismatch**. The gas constant $R$ is given as $8.3 	ext{ J K}^{-1} 	ext{mol}^{-1}$, which is in Joules. However, our $\Delta G^\circ$ is in kilo-Joules. We must convert it!

$$ 25200 = -2.3 	imes 8.3 	imes 400 	imes \log_{10} K_p $$

Let's isolate the unknown term. We divide the energy by the product of our constants:

$$ \log_{10} K_p = \frac{-25200}{2.3 	imes 8.3 	imes 400} $$

The denominator beautifully simplifies to $7636$. Dividing $-25200$ by $7636$ gives us approximately $-3.3$.

### The Antilog Magic

We now have $\log_{10} K_p = -3.3$. To extract $K_p$, we must take the antilogarithm. This might look intimidating without a calculator, but let's break it down using the properties of exponents:

$$ K_p = 10^{-3.3} = 10^{-3} 	imes 10^{-0.3} $$

The problem generously provides the value of $	ext{antilog}(-0.3)$ as $0.501$. Substituting this back in:

$$ K_p = 0.501 	imes 10^{-3} = 5.01 	imes 10^{-4} 	ext{ bar}^{-1} $$

### The Kp to Kc Bridge

Take a breath. We have found $K_p$, but the question specifically asks for $K_c$. How do we cross this bridge? We use the fundamental relation:

$$ K_p = K_c (RT)^{\Delta n_g} $$

First, let's determine $\Delta n_g$, the change in the number of moles of gas. We have $1$ mole of product and $2$ moles of reactants.

$$ \Delta n_g = 1 - 2 = -1 $$

Substituting this into our relation gives $K_p = K_c (RT)^{-1}$, which rearranges to:

$$ K_c = K_p (RT) $$

### The Unit Trap

This is the most critical moment of the problem. If you blindly plug in $R = 8.3$ again, your answer will be completely wrong! Why? Because $K_c$ is defined in terms of concentration ($	ext{mol L}^{-1}$), while $K_p$ is in terms of pressure ($	ext{bar}$).

To make the units perfectly align, we need $R$ in $	ext{L bar K}^{-1} 	ext{mol}^{-1}$. We know that $1 	ext{ J} = 10^{-2} 	ext{ L bar}$. Therefore, our new $R$ value is:

$$ R = 8.3 	imes 10^{-2} = 0.083 	ext{ L bar K}^{-1} 	ext{mol}^{-1} $$

### Final Calculation

Let's bring it all together. We substitute our carefully chosen values into the rearranged equation:

$$ K_c = 5.01 	imes 10^{-4} 	imes (0.083 	imes 400) $$

$$ K_c = 5.01 	imes 10^{-4} 	imes 33.2 $$

$$ K_c = 166.3 	imes 10^{-4} = 1.66 	imes 10^{-2} $$

The question asks for the value multiplied by $10^{-2}$, rounded off to the nearest integer. Our value is $1.66$, and rounding it to the nearest integer gives us our final, glorious answer: **2**.

Enter Numerical Value:

Visualized Solution

\text{The Reaction Setup}

\text{Gibbs Free Energy \& Equilibrium}

\text{Substituting the Values}

\text{Calculating } \log K_p

\text{Finding } K_p

\text{Relating } K_p \text{ and } K_c

\text{The Unit Catch for R}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Law of Mass Action

The Thermodynamic Setup

The Master Equation

Substituting the Values

The Antilog Magic

The Kp to Kc Bridge

The Unit Trap

Final Calculation

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