One mole of `H_(2)O` and one mole of `CO` are taken in a `10 litre` vessel and heated to `725 K`. At equilibrium, `40 per cent` of water (by mass) reacts with carbon monoxide according to the equation,
`H_(2)O_((g))+CO_((g))hArrH_(2(g))+CO_(2(g))`
Calculate the equilibrium constant for the reaction.

Explanation

To solve the problem step by step, we will follow the reaction and the changes in concentration as water reacts with carbon monoxide.

### Step 1: Determine Initial Moles
We start with:
- 1 mole of \( H_2O \)
- 1 mole of \( CO \)

### Step 2: Calculate Mass of Water Reacting
We are given that 40% of water by mass reacts. The molar mass of water \( H_2O \) is approximately 18 g/mol. Therefore, the mass of 1 mole of water is 18 g.

Calculating 40% of water:
\[ \text{Mass of water reacting} = 0.4 \times 18 \, \text{g} = 7.2 \, \text{g} \]

### Step 3: Convert Mass of Water to Moles
To find the moles of water that reacted, we convert the mass back to moles:
\[ \text{Moles of water reacting} = \frac{7.2 \, \text{g}}{18 \, \text{g/mol}} = 0.4 \, \text{mol} \]

### Step 4: Determine Moles at Equilibrium
At equilibrium, the changes in moles will be:
- Moles of \( H_2O \) remaining: \( 1 - 0.4 = 0.6 \, \text{mol} \)
- Moles of \( CO \) reacting: \( 0.4 \, \text{mol} \) (since the reaction consumes 1:1 ratio)
- Moles of \( H_2 \) formed: \( 0.4 \, \text{mol} \)
- Moles of \( CO_2 \) formed: \( 0.4 \, \text{mol} \)

### Step 5: Calculate Concentrations
The volume of the vessel is 10 L. Therefore, we can calculate the equilibrium concentrations:
- Concentration of \( H_2O \):
\[ [H_2O] = \frac{0.6 \, \text{mol}}{10 \, \text{L}} = 0.06 \, \text{mol/L} \]
- Concentration of \( CO \):
\[ [CO] = \frac{1 - 0.4}{10} = \frac{0.6 \, \text{mol}}{10 \, \text{L}} = 0.06 \, \text{mol/L} \]
- Concentration of \( H_2 \):
\[ [H_2] = \frac{0.4 \, \text{mol}}{10 \, \text{L}} = 0.04 \, \text{mol/L} \]
- Concentration of \( CO_2 \):
\[ [CO_2] = \frac{0.4 \, \text{mol}}{10 \, \text{L}} = 0.04 \, \text{mol/L} \]

### Step 6: Write the Expression for the Equilibrium Constant
The equilibrium constant \( K_c \) for the reaction:
\[ H_2O(g) + CO(g) \rightleftharpoons H_2(g) + CO_2(g) \]
is given by:
\[ K_c = \frac{[H_2][CO_2]}{[H_2O][CO]} \]

### Step 7: Substitute the Concentrations into the Expression
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.04)(0.04)}{(0.06)(0.06)} \]

### Step 8: Calculate \( K_c \)
Calculating the value:
\[ K_c = \frac{0.0016}{0.0036} = 0.4444 \approx 0.44 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction at 725 K is approximately \( 0.44 \).

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