The equilibrium constant of the following isomerisation reaction at 400 K and 298 K are 2.07 and 3.42 respectively. `cis-" butene"overset(k_(1))underset(k_(-1))hArr "trans - butene"`
Which of the following is/are correct?
I. The reaction is exothermic
II. The reaction is endothermic
III. At 400 K 50% of cis - butene and `50%` of trans - butene are present of equilibrium
IV. Both at 298 K and 400 K, `k_(1)=k_(-1)`.

Options

A. I and IV
B. II and IV
C. I and III
D. I only

Answer

To solve the problem, we need to analyze the given information about the isomerization reaction between cis-butene and trans-butene at two different temperatures (400 K and 298 K) and determine the validity of the statements provided.

### Step 1: Understand the Equilibrium Constants
The equilibrium constants (K) for the reaction are given as:
- At 298 K, K1 = 3.42
- At 400 K, K2 = 2.07

### Step 2: Determine the Nature of the Reaction
To determine whether the reaction is exothermic or endothermic, we can use the van 't Hoff equation, which relates the change in the equilibrium constant with temperature to the enthalpy change (ΔH) of the reaction:

\[ \ln \left(\frac{K_2}{K_1}\right) = -\frac{\Delta H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

Where:
- \( K_1 = 3.42 \) (at 298 K)
- \( K_2 = 2.07 \) (at 400 K)
- \( R \) is the gas constant (approximately 8.314 J/mol·K)
- \( T_1 = 298 \) K
- \( T_2 = 400 \) K

### Step 3: Calculate the Natural Logarithm of the Ratio of Equilibrium Constants
Calculate \( \ln \left(\frac{K_2}{K_1}\right) \):

\[ \ln \left(\frac{2.07}{3.42}\right) = \ln(0.605) \approx -0.494 \]

### Step 4: Calculate the Temperature Difference
Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \):

\[ \frac{1}{400} - \frac{1}{298} \approx 0.0025 - 0.003356 \approx -0.000856 \]

### Step 5: Substitute into the van 't Hoff Equation
Substituting into the van 't Hoff equation:

\[ -0.494 = -\frac{\Delta H}{8.314} \times (-0.000856) \]

Rearranging gives:

\[ \Delta H = \frac{0.494 \times 8.314}{0.000856} \approx 4,800 \text{ J/mol} \approx 4.8 \text{ kJ/mol} \]

Since ΔH is positive, the reaction is endothermic.

### Step 6: Analyze the Equilibrium Concentrations
At equilibrium, the relationship between the concentrations of cis-butene and trans-butene can be expressed as:

\[ K = \frac{[\text{trans-butene}]}{[\text{cis-butene}]} = 2.07 \text{ at } 400 K \]

Let \( x \) be the concentration of trans-butene and \( A \) be the initial concentration of cis-butene. Then:

\[ K = \frac{x}{A - x} = 2.07 \]

Cross-multiplying gives:

\[ 2.07(A - x) = x \]

Rearranging gives:

\[ 2.07A - 2.07x = x \implies 2.07A = 3.07x \implies x = \frac{2.07A}{3.07} \]

Calculating the fraction of trans-butene:

\[ \frac{x}{A} = \frac{2.07}{3.07} \approx 0.674 \text{ (or 67.4% trans-butene)} \]

This means that at 400 K, we do not have 50% of each isomer.

### Step 7: Evaluate Each Statement
1. **The reaction is exothermic**: **False** (it is endothermic).
2. **The reaction is endothermic**: **True**.
3. **At 400 K, 50% of cis-butene and 50% of trans-butene are present at equilibrium**: **False** (we found 67% trans-butene).
4. **Both at 298 K and 400 K, \( k_1 = k_{-1} \)**: **False** (the rate constants are not equal as the equilibrium constants differ).

### Conclusion
The correct statement is:
- **II. The reaction is endothermic.**