A reaction is at equilibrium at `100^(@)C` and the enthalpy change for the reaction is `"42.6 kJ mol"^(-1)`. What will be the value of `DeltaS` in `"J K"^(-1)"mol"^(-1)`?

Options

A. 120
B. 426.2
C. 373.1
D. 114.2

Explanation

To solve the problem, we need to find the value of ΔS (change in entropy) for a reaction at equilibrium at 100°C with an enthalpy change (ΔH) of 42.6 kJ/mol. We will follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Kelvin (T) can be calculated using the formula:
\[ T(K) = T(°C) + 273 \]

Given:
\[ T(°C) = 100 \]

Calculating:
\[ T(K) = 100 + 273 = 373 \, K \]

### Step 2: Convert the enthalpy change from kJ/mol to J/mol
Since we need ΔS in J/K·mol, we must convert ΔH from kJ to J. We know that:
\[ 1 \, \text{kJ} = 1000 \, \text{J} \]

Given:
\[ ΔH = 42.6 \, \text{kJ/mol} \]

Calculating:
\[ ΔH = 42.6 \times 1000 = 42600 \, \text{J/mol} \]

### Step 3: Use the relationship between ΔH, ΔS, and T to find ΔS
At equilibrium, the relationship between ΔH, ΔS, and T is given by:
\[ ΔG = ΔH - TΔS \]

At equilibrium, ΔG = 0, so:
\[ 0 = ΔH - TΔS \]
This rearranges to:
\[ ΔS = \frac{ΔH}{T} \]

### Step 4: Substitute the values of ΔH and T into the equation
Now we can substitute the values we calculated:
\[ ΔS = \frac{42600 \, \text{J/mol}}{373 \, K} \]

Calculating:
\[ ΔS = 114.2 \, \text{J/K·mol} \]

### Final Answer
The value of ΔS is:
\[ ΔS = 114.2 \, \text{J/K·mol} \]

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