The equilibrium constant of the ester formation of propionic acid with ethyl alcohol is 7-36 at 50°C. Calculate the weight of ethyl propionate, in grams, existing in an equilibrium mixture when 0-5 mole of propionic acid is heated with 05 mole of ethyl alcohol at 50°C.

Answer

To solve the problem of calculating the weight of ethyl propionate at equilibrium when 0.5 moles of propionic acid is heated with 0.5 moles of ethyl alcohol, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction for the formation of ethyl propionate from propionic acid and ethyl alcohol can be written as:
\[ \text{C}_3\text{H}_6\text{O}_2 + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{C}_5\text{H}_{10}\text{O}_2 + \text{H}_2\text{O} \]
Where:
- C₃H₆O₂ is propionic acid,
- C₂H₅OH is ethyl alcohol,
- C₅H₁₀O₂ is ethyl propionate,
- H₂O is water.

### Step 2: Set up the initial concentrations
Initially, we have:
- Propionic acid: 0.5 moles
- Ethyl alcohol: 0.5 moles
- Ethyl propionate: 0 moles
- Water: 0 moles

### Step 3: Define changes at equilibrium
Let \( x \) be the amount of ethyl propionate formed at equilibrium. Then, at equilibrium:
- Propionic acid = \( 0.5 - x \)
- Ethyl alcohol = \( 0.5 - x \)
- Ethyl propionate = \( x \)
- Water = \( x \)

### Step 4: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{C}_5\text{H}_{10}\text{O}_2][\text{H}_2\text{O}]}{[\text{C}_3\text{H}_6\text{O}_2][\text{C}_2\text{H}_5\text{OH}]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{x \cdot x}{(0.5 - x)(0.5 - x)} \]
Given \( K_c = 7.36 \), we can write:
\[ 7.36 = \frac{x^2}{(0.5 - x)^2} \]

### Step 5: Solve for \( x \)
Taking the square root of both sides:
\[ \sqrt{7.36} = \frac{x}{0.5 - x} \]
Calculating \( \sqrt{7.36} \) gives approximately \( 2.71 \):
\[ 2.71 = \frac{x}{0.5 - x} \]
Cross-multiplying gives:
\[ 2.71(0.5 - x) = x \]
\[ 1.355 - 2.71x = x \]
Combining like terms:
\[ 1.355 = 3.71x \]
\[ x = \frac{1.355}{3.71} \approx 0.3653 \text{ moles} \]

### Step 6: Calculate the mass of ethyl propionate
The molar mass of ethyl propionate (C₅H₁₀O₂) can be calculated as follows:
- C: 12.01 g/mol × 5 = 60.05 g/mol
- H: 1.008 g/mol × 10 = 10.08 g/mol
- O: 16.00 g/mol × 2 = 32.00 g/mol
Total molar mass = 60.05 + 10.08 + 32.00 = 102.13 g/mol

Now, calculate the mass of ethyl propionate:
\[ \text{Mass} = \text{moles} \times \text{molar mass} = 0.3653 \text{ moles} \times 102.13 \text{ g/mol} \approx 37.26 \text{ grams} \]

### Final Answer
The weight of ethyl propionate existing in the equilibrium mixture is approximately **37.26 grams**.

---