The equilibrium constant `(K_(c))` for the reaction
`2HCl(g)hArrH_(2)(g)+Cl_(2)(g)`
is `4xx10^(-34)`at `25^(@)C` .what is the equilibrium constant `K_p` for the reaction ?

Answer

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

\[ 2 \text{HCl}(g) \rightleftharpoons \text{H}_2(g) + \text{Cl}_2(g) \]

given that \( K_c = 4 \times 10^{-34} \) at \( 25^\circ C \), we can use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c (RT)^{\Delta n_g} \]

### Step 1: Determine \( \Delta n_g \)

First, we need to calculate \( \Delta n_g \), which is the change in the number of moles of gas during the reaction. This is given by the formula:

\[ \Delta n_g = n_p - n_r \]

where \( n_p \) is the number of moles of gaseous products and \( n_r \) is the number of moles of gaseous reactants.

- From the balanced equation:
- Products: \( \text{H}_2(g) + \text{Cl}_2(g) \) → 2 moles
- Reactants: \( 2 \text{HCl}(g) \) → 2 moles

Calculating \( \Delta n_g \):

\[ \Delta n_g = (1 + 1) - 2 = 2 - 2 = 0 \]

### Step 2: Substitute values into the equation

Now that we have \( \Delta n_g = 0 \), we can substitute the values into the equation for \( K_p \):

\[ K_p = K_c (RT)^{\Delta n_g} \]

Since \( \Delta n_g = 0 \):

\[ K_p = K_c (RT)^0 \]

### Step 3: Simplify the equation

We know that any number raised to the power of 0 is 1:

\[ (RT)^0 = 1 \]

Thus, we have:

\[ K_p = K_c \cdot 1 = K_c \]

### Step 4: Final calculation

Now we can substitute the value of \( K_c \):

\[ K_p = 4 \times 10^{-34} \]

### Conclusion

Therefore, the equilibrium constant \( K_p \) for the reaction is:

\[ \boxed{4 \times 10^{-34}} \]