Partial pressure of A, B, C, and D on the basis of gaseous system `A + 2B hArr` C + 3D are `A = 0.02, B = 0.10, C = 0.30` and `D = 0.05` atm. The numerical value of equilibrium constant is

Options

A. 11.25
B. 18.75
C. 5
D. 3.75

Answer

To find the numerical value of the equilibrium constant \( K_p \) for the reaction:

\[ A + 2B \rightleftharpoons C + 3D \]

we will use the formula for the equilibrium constant in terms of partial pressures:

\[ K_p = \frac{(P_C)^{c} \cdot (P_D)^{d}}{(P_A)^{a} \cdot (P_B)^{b}} \]

where \( P_A, P_B, P_C, \) and \( P_D \) are the partial pressures of the gases A, B, C, and D respectively, and \( a, b, c, d \) are their stoichiometric coefficients from the balanced equation.

### Step 1: Identify the partial pressures and stoichiometric coefficients
From the given reaction:
- \( P_A = 0.02 \, \text{atm} \)
- \( P_B = 0.10 \, \text{atm} \)
- \( P_C = 0.30 \, \text{atm} \)
- \( P_D = 0.05 \, \text{atm} \)

The stoichiometric coefficients are:
- \( a = 1 \) (for A)
- \( b = 2 \) (for B)
- \( c = 1 \) (for C)
- \( d = 3 \) (for D)

### Step 2: Substitute the values into the equilibrium constant expression
Now substituting the values into the equilibrium constant expression:

\[ K_p = \frac{(P_C)^{1} \cdot (P_D)^{3}}{(P_A)^{1} \cdot (P_B)^{2}} \]

Substituting the partial pressures:

\[ K_p = \frac{(0.30)^{1} \cdot (0.05)^{3}}{(0.02)^{1} \cdot (0.10)^{2}} \]

### Step 3: Calculate the numerator and denominator
Calculating the numerator:

\[ \text{Numerator} = (0.30) \cdot (0.05)^3 = 0.30 \cdot 0.000125 = 0.0000375 \]

Calculating the denominator:

\[ \text{Denominator} = (0.02) \cdot (0.10)^2 = 0.02 \cdot 0.01 = 0.0002 \]

### Step 4: Calculate \( K_p \)
Now, substituting the calculated values into the equation for \( K_p \):

\[ K_p = \frac{0.0000375}{0.0002} \]

Calculating \( K_p \):

\[ K_p = 0.1875 \]

### Step 5: Final result
Thus, the numerical value of the equilibrium constant \( K_p \) is:

\[ K_p = 18.75 \]