The standard state Gibbs's energy change for the isomerisation reaction `cis-2-pentene to trans-2-pentene` is `-3.67 kJ mol^(-1)` at `400 K`. If more `trans-2-pentene` is added to the reaction vessel, then:

Options

A. (A) more cis-2-pentene is formed
B. (B) equilibrium remains unaffected
C. (C) additional trans-2-pentene is formed
D. (D) equilibrium is shifted in forward direction

Explanation

To solve the problem regarding the isomerization reaction of cis-2-pentene to trans-2-pentene, we will follow these steps:

### Step 1: Understand the Reaction and Given Data
The isomerization reaction is:
\[ \text{cis-2-pentene} \rightleftharpoons \text{trans-2-pentene} \]

We are given that the standard Gibbs free energy change (\( \Delta G^\circ \)) for this reaction at 400 K is:
\[ \Delta G^\circ = -3.67 \, \text{kJ/mol} \]

### Step 2: Relate Gibbs Free Energy to Equilibrium Constant
The relationship between the Gibbs free energy change and the equilibrium constant (\( K_c \)) is given by:
\[ \Delta G^\circ = -RT \ln K_c \]
Where:
- \( R \) is the universal gas constant (8.314 J/mol·K)
- \( T \) is the temperature in Kelvin

### Step 3: Calculate the Equilibrium Constant \( K_c \)
We can rearrange the equation to find \( K_c \):
\[ K_c = e^{-\Delta G^\circ / RT} \]

First, convert \( \Delta G^\circ \) from kJ to J:
\[ \Delta G^\circ = -3.67 \, \text{kJ/mol} = -3670 \, \text{J/mol} \]

Now substituting the values:
\[ K_c = e^{3670 / (8.314 \times 400)} \]

Calculating the exponent:
\[ K_c = e^{(3670 / 3325.6)} \approx e^{1.103} \approx 3.01 \]

### Step 4: Analyze the Effect of Adding More trans-2-pentene
According to Le Chatelier's principle, if we add more trans-2-pentene to the system, the equilibrium will shift to counteract this change. Since we are increasing the concentration of the product (trans-2-pentene), the equilibrium will shift to the left to produce more reactant (cis-2-pentene).

### Step 5: Conclusion
Thus, when more trans-2-pentene is added, the equilibrium will shift towards the formation of cis-2-pentene. Therefore, the correct answer is that more cis-2-pentene is formed.

### Final Answer
More cis-2-pentene is formed.

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