The reaction of cyanamide, `NH_(2)CN(s)`, with dioxygen was carried out in a bomb calorimeter, and `DeltaU` was found to be `-742.7 kJ mol^(-1)` at `298 K`. Calculate enthalpy change for the reaction at `298 K`.
`NH_(2)CN(g)+3/2O_(2)(g) rarr N_(2)(g)+CO_(2)(g)+H_(2)O(l)`

Explanation

To calculate the enthalpy change (ΔH) for the reaction of cyanamide (NH₂CN) with dioxygen (O₂) at 298 K, we can use the relationship between internal energy change (ΔU) and enthalpy change (ΔH). The formula we will use is:

\[ \Delta H = \Delta U + \Delta N_G \cdot R \cdot T \]

Where:
- ΔU is the internal energy change (given as -742.7 kJ/mol)
- ΔN_G is the change in the number of moles of gas
- R is the universal gas constant (8.314 J/mol·K, or 0.008314 kJ/mol·K)
- T is the temperature in Kelvin (given as 298 K)

### Step 1: Calculate ΔN_G
ΔN_G is calculated as the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.

- **Products**:
- N₂(g) = 1 mole
- CO₂(g) = 1 mole
- H₂O(l) = 0 moles (liquid does not count)

Total gaseous products = 1 + 1 = 2 moles

- **Reactants**:
- NH₂CN(g) = 1 mole
- O₂(g) = 3/2 moles

Total gaseous reactants = 1 + 3/2 = 5/2 moles

Now, we can calculate ΔN_G:
\[ \Delta N_G = \text{(moles of products)} - \text{(moles of reactants)} = 2 - \frac{5}{2} = 2 - 2.5 = -0.5 \]

### Step 2: Substitute values into the ΔH equation
Now we can substitute the values into the ΔH equation:

\[ \Delta H = \Delta U + \Delta N_G \cdot R \cdot T \]
\[ \Delta H = -742.7 \, \text{kJ/mol} + (-0.5) \cdot (0.008314 \, \text{kJ/mol·K}) \cdot (298 \, \text{K}) \]

### Step 3: Calculate the second term
Calculating the second term:
\[ -0.5 \cdot 0.008314 \cdot 298 = -1.239 \, \text{kJ/mol} \]

### Step 4: Final calculation of ΔH
Now, substituting this back into the ΔH equation:
\[ \Delta H = -742.7 - 1.239 = -743.939 \, \text{kJ/mol} \]

### Conclusion
Thus, the enthalpy change for the reaction at 298 K is approximately:
\[ \Delta H \approx -743.9 \, \text{kJ/mol} \]