Given `C_(2)H_(2)(g)+H_(2)(g)rarrC_(2)H_(4)(g): DeltaH^(@)=-175 " kJ mol"^(-1)`
`DeltaH_(f(C_(2)H_(4),g))^(@)=50 " kJ mol"^(-1), DeltaH_(f(H_(2)O,l))^(@)=-280 " kJ mol"^(-1), DeltaH_(f(CO_(2)g))^(@)=-390 " kJ mol"^(-1)`
If `DeltaH^(@)` is enthalpy of combustion (in kJ `"mol"^(-1)`) of `C_(2)H_(2)`(g), then calculate the value of `|(DeltaH^(@))/(257)|`

Answer

To solve the problem, we need to calculate the enthalpy of combustion of \( C_2H_2 \) (acetylene) and then find the value of \( \left| \frac{\Delta H^{(@)}}{257} \right| \).

### Step 1: Understand the Reaction
The combustion of \( C_2H_2 \) can be represented by the following reaction:
\[ C_2H_2(g) + O_2(g) \rightarrow 2CO_2(g) + H_2O(l) \]
We need to find the enthalpy change (\( \Delta H^{(@)} \)) for this reaction.

### Step 2: Use Hess's Law
To find the enthalpy of combustion, we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.

### Step 3: Write the Formation Reactions
We know the following:
- The enthalpy of formation of \( C_2H_4(g) \) is \( \Delta H_f(C_2H_4)^{(@)} = 50 \, \text{kJ/mol} \).
- The enthalpy of formation of \( CO_2(g) \) is \( \Delta H_f(CO_2)^{(@)} = -390 \, \text{kJ/mol} \).
- The enthalpy of formation of \( H_2O(l) \) is \( \Delta H_f(H_2O)^{(@)} = -280 \, \text{kJ/mol} \).
- The enthalpy of formation of \( H_2(g) \) is \( 0 \, \text{kJ/mol} \).

### Step 4: Calculate \( \Delta H_f(C_2H_2) \)
From the heat of hydrogenation given:
\[ \Delta H_{hydrogenation} = \Delta H_f(C_2H_4) - \Delta H_f(C_2H_2) - \Delta H_f(H_2) \]
Substituting the known values:
\[ -175 = 50 - \Delta H_f(C_2H_2) - 0 \]
Rearranging gives:
\[ \Delta H_f(C_2H_2) = 50 + 175 = 225 \, \text{kJ/mol} \]

### Step 5: Calculate the Enthalpy of Combustion
Using the formula for the enthalpy of combustion:
\[ \Delta H_{combustion} = [2 \Delta H_f(CO_2) + \Delta H_f(H_2O)] - \Delta H_f(C_2H_2) \]
Substituting the values:
\[ \Delta H_{combustion} = [2(-390) + (-280)] - 225 \]
Calculating:
\[ = [-780 - 280] - 225 \]
\[ = -1060 - 225 = -1285 \, \text{kJ/mol} \]

### Step 6: Calculate \( \left| \frac{\Delta H^{(@)}}{257} \right| \)
Now we need to calculate:
\[ \left| \frac{-1285}{257} \right| \]
Calculating:
\[ = \frac{1285}{257} \approx 5 \]

### Final Answer
Thus, the final value is:
\[ \left| \frac{\Delta H^{(@)}}{257} \right| \approx 5 \]