Heat of combustion of ethanol at constant pressure and at temperature TK is found to be `-q J" mol"^(-1)`. Hence, heat of combustion (in J `mol^(-1)`) of ethanol at the same temperature and at constant volume will be :

Options

A. `RT-q`
B. `-(q+RT)`
C. `q-RT`
D. `q+RT`

Answer

To find the heat of combustion of ethanol at constant volume (denoted as \( Q_v \)) given the heat of combustion at constant pressure (denoted as \( Q_p \)), we can use the relationship between the two:

### Step 1: Understand the relationship between \( Q_p \) and \( Q_v \)
The heat of combustion at constant pressure is related to the heat of combustion at constant volume through the equation:
\[ Q_p = Q_v + \Delta N GRT \]
where:
- \( Q_p \) is the heat of combustion at constant pressure,
- \( Q_v \) is the heat of combustion at constant volume,
- \( \Delta N \) is the change in the number of moles of gas,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Write the combustion reaction of ethanol
The combustion reaction of ethanol (\( C_2H_5OH \)) can be written as:
\[ C_2H_5OH (l) + 3 O_2 (g) \rightarrow 2 CO_2 (g) + 3 H_2O (l) \]

### Step 3: Calculate \( \Delta N \)
To find \( \Delta N \), we need to determine the number of moles of gaseous products and reactants:
- Moles of gaseous products: \( 2 \, (CO_2) \)
- Moles of gaseous reactants: \( 3 \, (O_2) \)

Thus, the change in the number of moles of gas (\( \Delta N \)) is:
\[ \Delta N = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 2 - 3 = -1 \]

### Step 4: Substitute values into the equation
Now, substituting \( \Delta N \) into the equation:
\[ Q_p = Q_v + (-1)RT \]
Given that \( Q_p = -Q \), we can write:
\[ -Q = Q_v - RT \]

### Step 5: Solve for \( Q_v \)
Rearranging the equation to solve for \( Q_v \):
\[ Q_v = -Q + RT \]

### Final Answer
Thus, the heat of combustion of ethanol at constant volume is:
\[ Q_v = -Q + RT \quad \text{(in J mol}^{-1}\text{)} \]

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