Find out the heat evolved in combustion if 112 litre (at 1 atm, 273 K) of water gas (mixture of equal volume of `H_(2)(g)` and CO(g)) is combusted with excess oxygen.
`H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(g), Delta=-241.8 kJ` `CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g), Delta=-283 kJ`

Options

A. `241.8` KJ
B. 283 KJ
C. 1321 KJ
D. 1586 KJ

Explanation

To find the heat evolved in the combustion of 112 liters of water gas (a mixture of equal volumes of H₂ and CO), we can follow these steps:

### Step 1: Determine the volumes of H₂ and CO
Since water gas is a mixture of equal volumes of H₂ and CO, we can divide the total volume by 2.

\[ \text{Volume of } H_2 = \text{Volume of } CO = \frac{112 \text{ L}}{2} = 56 \text{ L} \]

### Step 2: Calculate the number of moles of H₂ and CO
Using the ideal gas law, we can find the number of moles at standard temperature and pressure (STP: 273 K and 1 atm). The molar volume of an ideal gas at STP is 22.4 L/mol.

\[ \text{Number of moles of } H_2 = \frac{56 \text{ L}}{22.4 \text{ L/mol}} = 2.5 \text{ moles} \]

\[ \text{Number of moles of } CO = \frac{56 \text{ L}}{22.4 \text{ L/mol}} = 2.5 \text{ moles} \]

### Step 3: Calculate the heat evolved from the combustion of H₂
The enthalpy change for the combustion of H₂ is given as ΔH = -241.8 kJ per mole.

\[ \text{Heat from } H_2 = 2.5 \text{ moles} \times (-241.8 \text{ kJ/mol}) = -604.5 \text{ kJ} \]

### Step 4: Calculate the heat evolved from the combustion of CO
The enthalpy change for the combustion of CO is given as ΔH = -283 kJ per mole.

\[ \text{Heat from } CO = 2.5 \text{ moles} \times (-283 \text{ kJ/mol}) = -707.5 \text{ kJ} \]

### Step 5: Calculate the total heat evolved
Now, we can sum the heat evolved from both reactions:

\[ \text{Total heat evolved} = -604.5 \text{ kJ} + (-707.5 \text{ kJ}) = -1312 \text{ kJ} \]

### Final Result
The total heat evolved in the combustion of 112 liters of water gas is approximately **-1312 kJ**.

---