At `298K`, if `/_\G_(f)^(@)` of `HCL_((g))` is `1.72kJmol^(-1)`, then calculate `K_(p)` for the following reversible reaction: `2HCl_((g))hArrH_(2(g))+Cl_(2(g))`
(use: at `298K:2.303RT=5700Jmol^(-1)` and `log2=0.30`)

Explanation

To calculate \( K_p \) for the reaction \( 2 \text{HCl}_{(g)} \rightleftharpoons \text{H}_2_{(g)} + \text{Cl}_2_{(g)} \) at \( 298 \, \text{K} \), we can follow these steps:

### Step 1: Determine \( \Delta G^\circ \) for the reaction
Given that \( \Delta G_f^\circ \) for \( \text{HCl}_{(g)} \) is \( 1.72 \, \text{kJ/mol} \), we need to find \( \Delta G^\circ \) for the reaction.

Since the reaction involves 2 moles of \( \text{HCl} \), we can calculate:
\[ \Delta G^\circ = -2 \times \Delta G_f^\circ (\text{HCl}) = -2 \times 1.72 \, \text{kJ/mol} = -3.44 \, \text{kJ/mol} \]

### Step 2: Convert \( \Delta G^\circ \) to Joules
To use the equation involving \( R \) and \( T \), we convert \( \Delta G^\circ \) to Joules:
\[ \Delta G^\circ = -3.44 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -3440 \, \text{J/mol} \]

### Step 3: Use the relationship between \( \Delta G^\circ \) and \( K_p \)
The relationship is given by:
\[ \Delta G^\circ = -2.303 \times R \times T \times \log K_p \]
We know:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \)
- \( 2.303RT = 5700 \, \text{J/mol} \) (as provided)

Substituting the values:
\[ -3440 \, \text{J/mol} = -5700 \, \log K_p \]

### Step 4: Solve for \( \log K_p \)
Rearranging the equation gives:
\[ \log K_p = \frac{-3440}{-5700} = \frac{3440}{5700} \]
Calculating this:
\[ \log K_p \approx 0.6035 \]

### Step 5: Convert \( \log K_p \) to \( K_p \)
Using the property of logarithms:
\[ K_p = 10^{\log K_p} = 10^{0.6035} \]
Since \( \log 2 = 0.3010 \), we can express \( 10^{0.6035} \) as:
\[ K_p = 10^{0.6} \approx 4 \quad (\text{since } 10^{0.6} \approx 4) \]

### Final Answer
Thus, the value of \( K_p \) for the reaction is:
\[ K_p = 4 \]

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