For the reaction, `SnO_(2)(s)+2H_(2)(g) hArr Sn(l)+2H_(2)O(g)` the equilibrium mixture of steam and hydrogen contained `45%` and `24% H_(2)` at `900 K` and `1100 K` respectively. Calculate `K_(p)` at both the temperature. Generally should it be higher or lower temperatures for better reduction of `SnO_(2)`?

Explanation

To solve the problem, we need to calculate the equilibrium constant \( K_p \) for the reaction at two different temperatures, \( 900 \, K \) and \( 1100 \, K \). The reaction is:

\[ \text{SnO}_2(s) + 2 \text{H}_2(g) \rightleftharpoons \text{Sn}(l) + 2 \text{H}_2O(g) \]

### Step 1: Determine the partial pressures at 900 K
At \( 900 \, K \), the equilibrium mixture contains \( 45\% \) \( H_2 \) and \( 55\% \) \( H_2O \) (since \( 100\% - 45\% = 55\% \)).

Assuming a total pressure \( P \):
- Partial pressure of \( H_2 \) = \( 0.45P \)
- Partial pressure of \( H_2O \) = \( 0.55P \)

### Step 2: Write the expression for \( K_p \) at 900 K
The expression for \( K_p \) is given by:

\[ K_p = \frac{(P_{H_2O})^2}{(P_{H_2})^2} \]

Substituting the partial pressures:

\[ K_p = \frac{(0.55P)^2}{(0.45P)^2} \]

### Step 3: Simplify the expression for \( K_p \) at 900 K
\[ K_p = \frac{0.55^2}{0.45^2} = \frac{0.3025}{0.2025} \approx 1.493 \]

### Step 4: Determine the partial pressures at 1100 K
At \( 1100 \, K \), the equilibrium mixture contains \( 24\% \) \( H_2 \) and \( 76\% \) \( H_2O \) (since \( 100\% - 24\% = 76\% \)).

Assuming a total pressure \( P \):
- Partial pressure of \( H_2 \) = \( 0.24P \)
- Partial pressure of \( H_2O \) = \( 0.76P \)

### Step 5: Write the expression for \( K_p \) at 1100 K
Using the same expression for \( K_p \):

\[ K_p = \frac{(P_{H_2O})^2}{(P_{H_2})^2} \]

Substituting the partial pressures:

\[ K_p = \frac{(0.76P)^2}{(0.24P)^2} \]

### Step 6: Simplify the expression for \( K_p \) at 1100 K
\[ K_p = \frac{0.76^2}{0.24^2} = \frac{0.5776}{0.0576} \approx 10.03 \]

### Conclusion
We have calculated \( K_p \) at both temperatures:
- At \( 900 \, K \), \( K_p \approx 1.493 \)
- At \( 1100 \, K \), \( K_p \approx 10.03 \)

Since \( K_p \) increases with temperature, it indicates that higher temperatures favor the formation of products. Therefore, for better reduction of \( SnO_2 \), higher temperatures should be preferred.