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 reaction given the \(\Delta U\), 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/(mol·K)),
- \(T\) is the temperature in Kelvin.

### Step 1: Identify \(\Delta U\)
From the problem, we know that:
\[ \Delta U = -740.0 \, \text{kJ/mol} = -740,000 \, \text{J/mol} \]

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

- Reactants: \(\frac{3}{2} \text{O}_2\) (1.5 moles of gas)
- Products: \(\text{N}_2\) (1 mole of gas) + \(\text{CO}_2\) (1 mole of gas) = 2 moles of gas

Now, calculate \(\Delta n_g\):
\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 2 - 1.5 = 0.5 \]

### Step 3: Substitute values into the equation
Now we can substitute the values into the equation for \(\Delta H\):
\[ \Delta H = \Delta U + \Delta n_g RT \]
\[ \Delta H = -740,000 \, \text{J/mol} + (0.5) \times (8.314 \, \text{J/(mol·K)}) \times (300 \, \text{K}) \]

### Step 4: Calculate the second term
Calculate \(0.5 \times 8.314 \times 300\):
\[ 0.5 \times 8.314 \times 300 = 1247.1 \, \text{J/mol} \]

### Step 5: Combine the terms
Now, convert \(1247.1 \, \text{J/mol}\) to kJ:
\[ 1247.1 \, \text{J/mol} = 1.2471 \, \text{kJ/mol} \]

Now substitute back into the equation:
\[ \Delta H = -740.0 \, \text{kJ/mol} + 1.2471 \, \text{kJ/mol} \]
\[ \Delta H = -738.7529 \, \text{kJ/mol} \]

### Final Answer
\[ \Delta H \approx -738.75 \, \text{kJ/mol} \]