Equal volumes of `C_(2)H_(2)` and `H_(2)` are combusted under identical conditons. The ratio of evolved in i.e. for `C_(2)H_(2)` and `H_(2)` is:
`H_(2)(g) + 1/2O_(2) (g) to H_(2)O(g), DeltaH=-241.8` kJ
`C_(2)H_(2)(g) + 21/2O_(2)(g) to 2CO_(2)(g) + H_(2)O(g), DeltaH = -1300` kJ

Options

A. `5.37/1`
B. `1/(5.37)`
C. `1/1`
D. None of these

Answer

To solve the problem, we need to find the ratio of heat evolved during the combustion of equal volumes of \( C_2H_2 \) (acetylene) and \( H_2 \) (hydrogen) under identical conditions.

### Step-by-step Solution:

1. **Identify the Combustion Reactions and Their Enthalpy Changes:**
- The combustion reaction for hydrogen:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g), \quad \Delta H = -241.8 \text{ kJ} \]
- The combustion reaction for acetylene:
\[ C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(g), \quad \Delta H = -1300 \text{ kJ} \]

2. **Use Avogadro's Law:**
- According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain the same number of molecules. Therefore, if we take equal volumes of \( C_2H_2 \) and \( H_2 \), they will contain the same number of moles. Let's denote the number of moles of each gas as \( N \).

3. **Calculate the Total Heat Evolved for Each Gas:**
- For \( H_2 \):
- Heat evolved for \( N \) moles of \( H_2 \):
\[ Q_{H_2} = N \times (-241.8) \text{ kJ} \]
- For \( C_2H_2 \):
- Heat evolved for \( N \) moles of \( C_2H_2 \):
\[ Q_{C_2H_2} = N \times (-1300) \text{ kJ} \]

4. **Set Up the Ratio of Heat Evolved:**
- The ratio of heat evolved during the combustion of \( C_2H_2 \) to that of \( H_2 \) is given by:
\[ \text{Ratio} = \frac{Q_{C_2H_2}}{Q_{H_2}} = \frac{N \times (-1300)}{N \times (-241.8)} \]
- The \( N \) cancels out:
\[ \text{Ratio} = \frac{-1300}{-241.8} = \frac{1300}{241.8} \]

5. **Calculate the Numerical Value:**
- Performing the division:
\[ \text{Ratio} = \frac{1300}{241.8} \approx 5.37 \]

6. **Final Result:**
- The ratio of heat evolved for \( C_2H_2 \) to \( H_2 \) is:
\[ C_2H_2 : H_2 = 5.37 : 1 \]