The value of equilibrium constant for the reaction `[N_2O_5 (g) implies 2NO_2(g) + 1/2 O_2(g)]` is 0.5. The equilibruim constant for the reaction `[4NO_2(g) + O_2(g) implies 2N_2O_5(g)]` is

Options

A. 0.4
B. 0.5
C. 2
D. 4

Answer

To find the equilibrium constant for the reaction

\[ 4NO_2(g) + O_2(g) \rightleftharpoons 2N_2O_5(g) \]

given that the equilibrium constant for the reaction

\[ N_2O_5(g) \rightleftharpoons 2NO_2(g) + \frac{1}{2} O_2(g) \]

is \( K = 0.5 \), we can follow these steps:

### Step 1: Write the given reaction and its equilibrium constant
The given reaction is:

\[ N_2O_5(g) \rightleftharpoons 2NO_2(g) + \frac{1}{2} O_2(g) \]

with \( K = 0.5 \).

### Step 2: Reverse the reaction
To find the equilibrium constant for the reverse reaction, we need to reverse the given reaction:

\[ 2NO_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons N_2O_5(g) \]

When a reaction is reversed, the equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant:

\[ K' = \frac{1}{K} = \frac{1}{0.5} = 2 \]

### Step 3: Multiply the reaction by an integer
Now, we need to multiply the reversed reaction by 2 to match the target reaction:

\[ 4NO_2(g) + O_2(g) \rightleftharpoons 2N_2O_5(g) \]

When we multiply the entire reaction by 2, the new equilibrium constant is given by:

\[ K'' = K'^{n} \]

where \( n \) is the factor by which the reaction is multiplied. In this case, \( n = 2 \):

\[ K'' = (K')^2 = (2)^2 = 4 \]

### Final Answer
Thus, the equilibrium constant for the reaction

\[ 4NO_2(g) + O_2(g) \rightleftharpoons 2N_2O_5(g) \]

is \( K = 4 \).

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