Given the reaction btw .3 gases represented `X_(2),Y_(2),Z_(2) ` to give the compound `XYZ_(g)`
`X_(2(g))+Y_(2(g))+Z_(2(g)) hArr 2XYZ_(g)`
At equilibrium `conc^(n)`
of `X_(2)=3M,Y_(2)=6M,Z_(2)`=9M
`XYZ=6M`
If the reaction take place in a sealed vessel at `527^(@)C`, then the
value of `K_(C)` will be :-

Options

A. `(27)/(2)`
B. `(2)/(9)`
C. 36
D. 24

Explanation

To find the equilibrium constant \( K_c \) for the reaction

\[ X_2(g) + Y_2(g) + Z_2(g) \rightleftharpoons 2XYZ(g) \]

given the equilibrium concentrations:

- \([X_2] = 3 \, M\)
- \([Y_2] = 6 \, M\)
- \([Z_2] = 9 \, M\)
- \([XYZ] = 6 \, M\)

we can follow these steps:

### Step 1: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) is defined as:

\[ K_c = \frac{[products]}{[reactants]} \]

For our reaction, this becomes:

\[ K_c = \frac{[XYZ]^2}{[X_2][Y_2][Z_2]} \]

### Step 2: Substitute the equilibrium concentrations into the expression

Now we substitute the given equilibrium concentrations into the expression:

\[ K_c = \frac{(6)^2}{(3)(6)(9)} \]

### Step 3: Calculate the numerator and denominator

Calculating the numerator:

\[ (6)^2 = 36 \]

Calculating the denominator:

\[ (3)(6)(9) = 162 \]

### Step 4: Divide the numerator by the denominator

Now we can calculate \( K_c \):

\[ K_c = \frac{36}{162} \]

### Step 5: Simplify the fraction

To simplify \( \frac{36}{162} \):

\[ K_c = \frac{36 \div 18}{162 \div 18} = \frac{2}{9} \]

### Final Answer

Thus, the value of \( K_c \) is:

\[ K_c = \frac{2}{9} \]

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