In the reaction `CuSO_(4). 3H_(2)O Leftrightarrow CuSO_(4). H_(2)O+2H_(2)O` (vap.), the dissociation pressure is `7 xx 10^(-3)" atm at "25^@C and triangleH^(@)=2700cal`. What will be the dissociation pressure at `127^@C?`

Answer

To solve the problem, we will use the Van't Hoff equation, which relates the change in the equilibrium constant (or dissociation pressure in this case) with temperature and the enthalpy change of the reaction. The equation is given by:

\[ \ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

Where:
- \( K_1 \) is the dissociation pressure at the initial temperature \( T_1 \).
- \( K_2 \) is the dissociation pressure at the final temperature \( T_2 \).
- \( \Delta H \) is the enthalpy change of the reaction.
- \( R \) is the universal gas constant (approximately \( 1.987 \, \text{cal/(mol K)} \)).
- \( T_1 \) and \( T_2 \) are the absolute temperatures in Kelvin.

### Step 1: Convert temperatures to Kelvin
- \( T_1 = 25^\circ C = 298 \, K \)
- \( T_2 = 127^\circ C = 400 \, K \)

### Step 2: Identify given values
- \( K_1 = 7 \times 10^{-3} \, \text{atm} \)
- \( \Delta H = 2700 \, \text{cal} \)
- \( R = 1.987 \, \text{cal/(mol K)} \)

### Step 3: Substitute values into the Van't Hoff equation
We need to find \( K_2 \). Rearranging the Van't Hoff equation gives:

\[ \ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

Substituting the known values:

\[ \ln \left( \frac{K_2}{7 \times 10^{-3}} \right) = -\frac{2700}{1.987} \left( \frac{1}{400} - \frac{1}{298} \right) \]

### Step 4: Calculate \( \frac{1}{T_2} - \frac{1}{T_1} \)
\[ \frac{1}{400} - \frac{1}{298} = \frac{298 - 400}{400 \times 298} = \frac{-102}{119200} \approx -0.000854 \]

### Step 5: Calculate the right-hand side
\[ -\frac{2700}{1.987} \times (-0.000854) \approx 1.020 \]

### Step 6: Solve for \( K_2 \)
Now we have:

\[ \ln \left( \frac{K_2}{7 \times 10^{-3}} \right) \approx 1.020 \]

Exponentiating both sides gives:

\[ \frac{K_2}{7 \times 10^{-3}} \approx e^{1.020} \approx 2.776 \]

Thus,

\[ K_2 \approx 2.776 \times 7 \times 10^{-3} \approx 0.0194 \, \text{atm} \]

### Final Answer
The dissociation pressure at \( 127^\circ C \) is approximately \( 0.0194 \, \text{atm} \).

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