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

Explanation

To solve the problem, we need to relate the heat of combustion at constant pressure (ΔH) to the heat of combustion at constant volume (ΔU) using the appropriate thermodynamic relationship. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
The heat of combustion of ethanol at constant pressure is given as:
\[ \Delta H = -q \, \text{J mol}^{-1} \]
where ΔH is the change in enthalpy (heat of combustion at constant pressure).

### Step 2: Use the Thermodynamic Relationship
We know the relationship between enthalpy change (ΔH) and internal energy change (ΔU) at constant temperature and pressure:
\[ \Delta H = \Delta U + \Delta N_g RT \]
where:
- ΔN_g = change in the number of moles of gas during the reaction
- R = universal gas constant
- T = temperature in Kelvin

### Step 3: Determine ΔN_g
For the combustion of ethanol (C₂H₅OH), the balanced equation is:
\[ C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O \]
In this reaction:
- Products (gaseous): 2 moles of CO₂
- Reactants (gaseous): 3 moles of O₂

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

### Step 4: Substitute ΔN_g into the Equation
Now substitute ΔN_g into the relationship:
\[ \Delta H = \Delta U + (-1)RT \]
This simplifies to:
\[ \Delta H = \Delta U - RT \]

### Step 5: Solve for ΔU
Rearranging the equation gives us:
\[ \Delta U = \Delta H + RT \]
Substituting ΔH = -q:
\[ \Delta U = -q + RT \]

### Final Answer
Thus, the heat of combustion of ethanol at constant volume is:
\[ \Delta U = RT - q \, \text{J mol}^{-1} \]