For the reaction
`C_(6)H_(12)(l)+9O_(2)(g)to6H_(2)O(l)+6CO_(2)(g)` , `DeltaH=-936.9kcal`
Which of the following is true ?

Options

A. `-936.9 = Delta E - (2 xx10^(-3) xx298 xx3) kcal`
B. `+936.9 = Delta E + (2 xx10^(-3) xx298 xx3) kcal`
C. `-936.9 = Delta E - (2 xx10^(-3) xx298 xx2) kcal`
D. `-936.9 = Delta E + (2 xx10^(-3) xx298 xx2) kcal`

Answer

To solve the problem, we need to analyze the given reaction and the relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔE). The reaction is:

\[ C_{6}H_{12}(l) + 9O_{2}(g) \rightarrow 6H_{2}O(l) + 6CO_{2}(g) \]

Given:
- ΔH = -936.9 kcal

### Step 1: Understand the relationship between ΔH and ΔE
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔE) is given by the equation:

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

Where:
- ΔN_g = change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants)
- R = gas constant
- T = temperature in Kelvin

### Step 2: Calculate ΔN_g
To find ΔN_g, we need to count the moles of gaseous products and reactants:

- **Products**: 6 moles of \( CO_{2}(g) \) (6 moles of gas)
- **Reactants**: 9 moles of \( O_{2}(g) \) (9 moles of gas)

Now, calculate ΔN_g:

\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 6 - 9 = -3 \]

### Step 3: Use the gas constant R
The gas constant R in kcal is:

\[ R = 2 \times 10^{-3} \text{ kcal/K/mol} \]

### Step 4: Use the temperature T
Assuming standard temperature, T = 298 K.

### Step 5: Substitute values into the equation
Now, substitute the values into the relationship:

\[ -936.9 \text{ kcal} = \Delta E + (-3) \times (2 \times 10^{-3} \text{ kcal/K/mol}) \times 298 \text{ K} \]

Calculating the right side:

\[ -3 \times (2 \times 10^{-3}) \times 298 = -3 \times 0.002 \times 298 = -1.788 \text{ kcal} \]

### Step 6: Rearranging the equation to find ΔE
Now, we can rearrange the equation to solve for ΔE:

\[ -936.9 = \Delta E - 1.788 \]

Adding 1.788 to both sides:

\[ \Delta E = -936.9 + 1.788 = -935.112 \text{ kcal} \]

### Conclusion
Thus, the change in internal energy (ΔE) for the reaction is approximately -935.112 kcal.