The equilibrium constant for the reaction `H_(2)+Br_(2)hArr2HBr` is `67.8` at `300 K`. The equilibrium constant for the dissociation of `HBr` is:

Options

A. `0.0147`
B. `67.80`
C. `33.90`
D. `8.349`

Answer

To find the equilibrium constant for the dissociation of HBr from the given reaction, we can follow these steps:

### Step 1: Write the reaction for the formation of HBr
The formation of HBr from hydrogen and bromine can be represented as:
\[ \text{H}_2 + \text{Br}_2 \rightleftharpoons 2 \text{HBr} \]
The equilibrium constant \( K_f \) for this reaction is given as \( 67.8 \) at \( 300 \, K \).

### Step 2: Write the reaction for the dissociation of HBr
The dissociation of HBr can be represented as:
\[ 2 \text{HBr} \rightleftharpoons \text{H}_2 + \text{Br}_2 \]
This is the reverse reaction of the formation of HBr.

### Step 3: Relate the equilibrium constants
For the reverse reaction, the equilibrium constant \( K_d \) is the reciprocal of the equilibrium constant for the forward reaction. Therefore:
\[ K_d = \frac{1}{K_f} \]
Substituting the given value of \( K_f \):
\[ K_d = \frac{1}{67.8} \]

### Step 4: Calculate the value of \( K_d \)
Now we perform the calculation:
\[ K_d = \frac{1}{67.8} \approx 0.0147 \]

### Conclusion
The equilibrium constant for the dissociation of HBr is approximately \( 0.0147 \).

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