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.

Explanation

To calculate \( K_p \) for the given reaction at two different temperatures, we will follow these steps:

### Given Reaction:
\[ \text{SnO}_2(s) + 2\text{H}_2(g) \rightleftharpoons \text{Sn}(l) + 2\text{H}_2\text{O}(g) \]

### Step 1: Understand the Equilibrium Composition
At equilibrium, we know the percentages of \( H_2 \) and \( H_2O \) at two different temperatures.

1. **At 900 K:**
- \( H_2 \) = 45%
- \( H_2O \) = 100% - 45% = 55%

2. **At 1100 K:**
- \( H_2 \) = 24%
- \( H_2O \) = 100% - 24% = 76%

### Step 2: Calculate \( K_p \) at 900 K
Using the equilibrium expression for \( K_p \):
\[ K_p = \frac{(P_{H_2O})^2}{(P_{H_2})^2} \]

Where \( P_{H_2O} \) and \( P_{H_2} \) are the partial pressures of water vapor and hydrogen gas, respectively.

1. **At 900 K:**
- \( P_{H_2O} = 55 \) (in percentage)
- \( P_{H_2} = 45 \)

Now substituting these values into the \( K_p \) expression:
\[ K_p = \frac{(55)^2}{(45)^2} \]
\[ K_p = \frac{3025}{2025} \approx 1.493 \]

### Step 3: Calculate \( K_p \) at 1100 K
Using the same equilibrium expression:
1. **At 1100 K:**
- \( P_{H_2O} = 76 \)
- \( P_{H_2} = 24 \)

Now substituting these values into the \( K_p \) expression:
\[ K_p = \frac{(76)^2}{(24)^2} \]
\[ K_p = \frac{5776}{576} \approx 10.67 \]

### Final Results:
- \( K_p \) at 900 K = 1.493
- \( K_p \) at 1100 K = 10.67