`k_(P) & k_(C)` are ….. for reaction at equilibrium of type `H_(2)(g) + Br_(2)(g) hArr 2HBr(g)`.

Answer

To find the relationship between \( K_p \) and \( K_c \) for the reaction:

\[ H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g) \]

we can follow these steps:

### Step 1: Understand the Definitions
- \( K_p \) is the equilibrium constant in terms of partial pressures of the gases.
- \( K_c \) is the equilibrium constant in terms of concentrations of the gases.

### Step 2: Use the Relationship Between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:

\[ K_p = K_c \cdot R^{\Delta n} \cdot T^{\Delta n} \]

where:
- \( R \) is the universal gas constant.
- \( T \) is the temperature in Kelvin.
- \( \Delta n \) is the change in the number of moles of gas, calculated as the number of moles of products minus the number of moles of reactants.

### Step 3: Calculate \( \Delta n \)
For the given reaction:

- Products: \( 2 \) moles of \( HBr \)
- Reactants: \( 1 \) mole of \( H_2 \) + \( 1 \) mole of \( Br_2 \) = \( 2 \) moles

Now, calculate \( \Delta n \):

\[ \Delta n = \text{(moles of products)} - \text{(moles of reactants)} = 2 - 2 = 0 \]

### Step 4: Substitute \( \Delta n \) into the Equation
Since \( \Delta n = 0 \), we can substitute this into the relationship:

\[ K_p = K_c \cdot R^{0} \]

Since \( R^{0} = 1 \):

\[ K_p = K_c \cdot 1 \]

Thus, we have:

\[ K_p = K_c \]

### Conclusion
For the reaction \( H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g) \), the relationship between \( K_p \) and \( K_c \) is:

\[ K_p = K_c \]