For the reaction `H_(2(g))+ I_(2(g)) Leftrightarrow 2HI_((g)) ` at 741K, the value of equlibrium constant, `K_c` is 50. The value of `K_p` under the same conditions will be

Options

A. 0.02
B. 0.2
C. 50
D. 50/RT

Answer

To find the value of \( K_p \) for the reaction

\[ H_{2(g)} + I_{2(g)} \leftrightarrow 2HI_{(g)} \]

at 741 K, given that the equilibrium constant \( K_c \) is 50, we can use the relationship between \( K_p \) and \( K_c \):

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

**Step 1: Identify the values needed for the equation.**

- \( K_c = 50 \)
- \( R \) (universal gas constant) = 0.0821 L·atm/(K·mol)
- \( T = 741 \, K \)

**Step 2: Calculate \( \Delta N_g \).**

\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

For the given reaction:
- Moles of products (2 moles of \( HI \)) = 2
- Moles of reactants (1 mole of \( H_2 \) + 1 mole of \( I_2 \)) = 2

So,

\[ \Delta N_g = 2 - 2 = 0 \]

**Step 3: Substitute the values into the equation.**

Now, substituting the values into the equation for \( K_p \):

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

Since \( \Delta N_g = 0 \), we have:

\[ e^{\Delta N_g} = e^0 = 1 \]

Thus, the equation simplifies to:

\[ K_p = K_c \cdot R^T \cdot 1 \]

**Step 4: Calculate \( R^T \).**

Now we need to calculate \( R^T \):

\[ R^T = 0.0821 \, \text{L·atm/(K·mol)} \cdot 741 \, K \]

Calculating this gives:

\[ R^T = 60.89 \, \text{L·atm/mol} \]

**Step 5: Calculate \( K_p \).**

Now substituting back into the equation:

\[ K_p = 50 \cdot 1 = 50 \]

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

\[ K_p = 50 \]

### Final Answer:
The value of \( K_p \) under the same conditions is \( 50 \).

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