When the following reaction was carried out in bomb calorimeter, `DeltaU` is found to be `-740.0kJ/mol" of NH"_(2)CN(S)" at 300 K "NH_(2)CN_((s))+(3)/(2)O_(2(g))rarrN_(2(g))+CO_(2(g))+H_(2)O_((l))`
Calculate `DeltaH_("300 K")` for the reaction.

Options

A. `-738.75kJ`
B. `+738.75kJ`
C. `-824.75kJ`
D. `-919.57kJ.`

Answer

To calculate \( \Delta H \) for the given reaction, we can use the relationship between \( \Delta H \) and \( \Delta U \):

\[ \Delta H = \Delta U + \Delta N_g RT \]

Where:
- \( \Delta H \) is the change in enthalpy,
- \( \Delta U \) is the change in internal energy,
- \( \Delta N_g \) is the change in the number of moles of gas,
- \( R \) is the universal gas constant (8.314 J/(K·mol)),
- \( T \) is the temperature in Kelvin.

### Step 1: Identify \( \Delta U \)
From the problem statement, we know:
\[ \Delta U = -740.0 \, \text{kJ/mol} \]

### Step 2: Calculate \( \Delta N_g \)
The reaction is:
\[ \text{NH}_2\text{CN}_{(s)} + \frac{3}{2} \text{O}_{2(g)} \rightarrow \text{N}_{2(g)} + \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)} \]

To find \( \Delta N_g \), we need to count the moles of gaseous products and reactants:
- Moles of gas products: \( 1 \, (\text{N}_2) + 1 \, (\text{CO}_2) = 2 \)
- Moles of gas reactants: \( \frac{3}{2} \, (\text{O}_2) = 1.5 \)

Thus,
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} = 2 - 1.5 = 0.5 \]

### Step 3: Calculate \( \Delta H \)
Now we can substitute the values into the equation:

1. Convert \( \Delta U \) from kJ to J:
\[ \Delta U = -740.0 \, \text{kJ/mol} = -740000 \, \text{J/mol} \]

2. Calculate \( \Delta N_g RT \):
\[ \Delta N_g RT = 0.5 \times 8.314 \, \text{J/(K·mol)} \times 300 \, \text{K} \]
\[ = 0.5 \times 8.314 \times 300 = 1249.1 \, \text{J/mol} \]

3. Now substitute back into the equation for \( \Delta H \):
\[ \Delta H = -740000 \, \text{J/mol} + 1249.1 \, \text{J/mol} \]
\[ = -738750.9 \, \text{J/mol} = -738.75 \, \text{kJ/mol} \]

### Final Answer
\[ \Delta H_{(300 \, K)} = -738.75 \, \text{kJ/mol} \]

### Conclusion
The correct answer is \( -738.75 \, \text{kJ/mol} \).