Equilibrium constant for the following equilibrium is given at `0^(@)C` . `Na_(2)HPO_(4)` . `12H_(2)O(s) hArrNa_(2)HPO_(4)` . `7H_(2)O(s) + 5H_(2)O(g)` `K_(p) = 31.25xx10^(-13)`. At equilibrium what will be partial pressure of water vapour:

Answer

To solve the problem, we need to find the partial pressure of water vapor at equilibrium for the given reaction:

\[ \text{Na}_2\text{HPO}_4 \cdot 12\text{H}_2\text{O (s)} \rightleftharpoons \text{Na}_2\text{HPO}_4 \cdot 7\text{H}_2\text{O (s)} + 5\text{H}_2\text{O (g)} \]

Given:
- Equilibrium constant \( K_p = 31.25 \times 10^{-13} \)

### Step 1: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the gaseous products. Since solids do not appear in the equilibrium expression, we only consider the gaseous component:

\[ K_p = \frac{(P_{\text{H}_2\text{O}})^5}{1} = (P_{\text{H}_2\text{O}})^5 \]

### Step 2: Substitute the value of \( K_p \)

Now, we substitute the given value of \( K_p \) into the equation:

\[ (P_{\text{H}_2\text{O}})^5 = 31.25 \times 10^{-13} \]

### Step 3: Solve for \( P_{\text{H}_2\text{O}} \)

To find \( P_{\text{H}_2\text{O}} \), we take the fifth root of both sides:

\[ P_{\text{H}_2\text{O}} = (31.25 \times 10^{-13})^{1/5} \]

### Step 4: Simplify the expression

We can simplify the expression further. First, we can express \( 31.25 \) as \( 3125 \times 10^{-15} \):

\[ P_{\text{H}_2\text{O}} = (3125 \times 10^{-15})^{1/5} \]

### Step 5: Calculate the fifth root

Calculating the fifth root:

\[ P_{\text{H}_2\text{O}} = 3125^{1/5} \times (10^{-15})^{1/5} \]

Calculating \( 3125^{1/5} \):

\[ 3125 = 5^5 \implies 3125^{1/5} = 5 \]

And \( (10^{-15})^{1/5} = 10^{-3} \).

### Step 6: Combine the results

Thus, we have:

\[ P_{\text{H}_2\text{O}} = 5 \times 10^{-3} \text{ atm} \]

### Final Answer

The partial pressure of water vapor at equilibrium is:

\[ P_{\text{H}_2\text{O}} = 0.005 \text{ atm} \text{ or } 5 \times 10^{-3} \text{ atm} \]

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