For the reaction `PCl_(3)(g)+Cl_(2)(g)rarrPCl_(5)(g),K_(c)` is 26 at `250^(@)C`. `K_(p)` at the same temperature is `(R=8.314JK^(-1)mol^(-1))`

Options

A. `4.6xx10^(3)`
B. `5.7xx10^(3)`
C. `6.0xx10^(-3)`
D. `8.3xx10^(-3)`

Answer

To find the value of \( K_p \) for the reaction \( PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g) \) given that \( K_c = 26 \) at \( 250^\circ C \), we can follow these steps:

### Step 1: Convert Temperature to Kelvin
The temperature is given in Celsius and needs to be converted to Kelvin. The conversion formula is:
\[ T(K) = T(°C) + 273 \]
Substituting the given temperature:
\[ T = 250 + 273 = 523 \, K \]

### Step 2: Determine \( \Delta N_g \)
To find \( \Delta N_g \), we need to calculate the change in the number of moles of gas from reactants to products.

- On the reactant side, we have:
- 1 mole of \( PCl_3 \)
- 1 mole of \( Cl_2 \)

Total moles of reactants = \( 1 + 1 = 2 \)

- On the product side, we have:
- 1 mole of \( PCl_5 \)

Total moles of products = \( 1 \)

Now, we can calculate \( \Delta N_g \):
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} = 1 - 2 = -1 \]

### Step 3: Use the Relationship Between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c \cdot R^T \cdot (RT)^{\Delta N_g} \]
Since \( \Delta N_g = -1 \), we can simplify the equation to:
\[ K_p = \frac{K_c}{R^T} \]

### Step 4: Substitute the Values
Now we can substitute the known values into the equation:
- \( K_c = 26 \)
- \( R = 8.314 \, J \, K^{-1} \, mol^{-1} \)
- \( T = 523 \, K \)

Calculating \( R \cdot T \):
\[ R \cdot T = 8.314 \, J \, K^{-1} \, mol^{-1} \cdot 523 \, K = 4343.282 \, J \, mol^{-1} \]

Now substituting into the equation for \( K_p \):
\[ K_p = \frac{26}{4343.282} \approx 0.00598 \]

### Step 5: Final Result
Expressing the result in scientific notation:
\[ K_p \approx 5.98 \times 10^{-3} \]

### Conclusion
Thus, the value of \( K_p \) at \( 250^\circ C \) is approximately \( 6 \times 10^{-3} \).

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