At 298 K in a constant volume calorimeter 0.01 mole of TNT was detonated when 8180 cals of heat was released. Each mole of TNT gives 6 moles of gaseous products on detonation. What is `DeltaH`/mole of TNT exploded?

Options

A. `-714` kcals/`"mole"^(-1)`
B. `-814` kcals/`"mole"^(-1)`
C. `-914` kcals `"mole"^(-1)`
D. none of the above

Answer

To find the enthalpy change (ΔH) per mole of TNT exploded, we can follow these steps:

### Step 1: Calculate the heat released per mole of TNT
Given that 0.01 moles of TNT releases 8180 calories of heat, we need to find the heat released per mole.

\[ \text{Heat released per mole} = \frac{\text{Total heat released}}{\text{Number of moles}} = \frac{8180 \text{ cal}}{0.01 \text{ moles}} = 818000 \text{ cal/mole} \]

### Step 2: Convert calories to kilocalories
Since the answer needs to be in kilocalories, we convert calories to kilocalories by dividing by 1000.

\[ \text{Heat released per mole in kilocalories} = \frac{818000 \text{ cal}}{1000} = 818 \text{ kcal/mole} \]

### Step 3: Determine the change in internal energy (ΔU)
At constant volume, the heat released is equal to the change in internal energy (ΔU). Therefore:

\[ \Delta U = -818 \text{ kcal/mole} \]

(The negative sign indicates that heat is released.)

### Step 4: Calculate the change in the number of moles of gas (ΔNG)
From the problem, each mole of TNT produces 6 moles of gaseous products. Since there are no gaseous reactants, we have:

\[ \Delta N_G = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 6 - 0 = 6 \]

### Step 5: Use the formula to calculate ΔH
The relationship between ΔH and ΔU is given by:

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

Where:
- R = 2 × 10^{-3} kcal/(mol·K) (the gas constant in kilocalories)
- T = 298 K (the temperature in Kelvin)

Substituting the values we have:

\[ \Delta H = -818 \text{ kcal/mole} + (6 \cdot 2 \times 10^{-3} \text{ kcal/(mol·K)} \cdot 298 \text{ K}) \]

Calculating the second term:

\[ 6 \cdot 2 \times 10^{-3} \cdot 298 = 3.576 \text{ kcal} \]

Now substituting back into the ΔH equation:

\[ \Delta H = -818 + 3.576 = -814.424 \text{ kcal/mole} \]

### Final Answer
Thus, the enthalpy change (ΔH) per mole of TNT exploded is approximately:

\[ \Delta H \approx -814.424 \text{ kcal/mole} \]