Question

12g of pure graphite is burnt. Temperature rises from 298 K to 308 K. Heat capacity of the calorimeter is $20.7~kJ~K^{-1}$. Enthalpy change for combustion of 1 mole of graphite $(kJ~mol^{-1})$ is ________.

• A. -2070
• B. -207
• C. +2070
• D. +207
• E. +2.07

Hint:

Combustion always releases heat (negative $\Delta H$).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This problem involves thermochemistry and calorimetry. A bomb calorimeter measures heat changes at constant volume, which corresponds to the change in internal energy, \(\Delta U\). The enthalpy change, \(\Delta H\), is measured at constant pressure. We must first find \(\Delta U\) from the calorimeter data and then convert it to \(\Delta H\) using the relationship between them.
Step 2: Key Formula or Approach
1. Heat absorbed by calorimeter (\(q\)): \(q = C_{cal} \times \Delta T\), where \(C_{cal}\) is the heat capacity of the calorimeter. 2. Internal energy change (\(\Delta U\)): The heat released by the reaction is absorbed by the calorimeter. Since combustion is exothermic, \(\Delta U = -q\). 3. Molar quantities: Convert the calculated \(\Delta U\) to a molar value (kJ/mol). 4. Enthalpy change (\(\Delta H\)): Use the formula \(\Delta H = \Delta U + \Delta n_g RT\), where \(\Delta n_g\) is the change in moles of gas for the reaction.
Step 3: Detailed Explanation
1. Calculate the heat absorbed by the calorimeter (\(q_{cal}\)):
- Heat capacity, \(C_{cal} = 20.7 \text{ kJ K}^{-1}\) - Temperature change, \(\Delta T = 308 \text{ K} - 298 \text{ K} = 10 \text{ K}\) - Heat absorbed, \(q_{cal} = (20.7 \text{ kJ K}^{-1}) \times (10 \text{ K}) = 207 \text{ kJ}\) 2. Calculate the internal energy change for the reaction (\(\Delta U\)):
The heat absorbed by the calorimeter is released by the reaction. Therefore, the internal energy change for the amount of substance reacted is: \[ \Delta U = -q_{cal} = -207 \text{ kJ} \] 3. Calculate moles of graphite reacted:
Graphite is a form of carbon (C). The molar mass of C is 12 g/mol. \[ \text{moles of C} = \frac{\text{mass}}{\text{molar mass}} = \frac{12 \text{ g}}{12 \text{ g/mol}} = 1 \text{ mole} \] Since exactly 1 mole of graphite was burnt, the calculated \(\Delta U\) is the molar internal energy change: \(\Delta U_m = -207 \text{ kJ/mol}\). 4. Calculate the enthalpy change (\(\Delta H\)):
The relationship is \(\Delta H = \Delta U + \Delta n_g RT\). We need to find \(\Delta n_g\) from the balanced chemical equation for the combustion of graphite: \[ C(s, \text{graphite}) + O_2(g) \to CO_2(g) \] \(\Delta n_g\) is the change in moles of gaseous species: \[ \Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) \] \[ \Delta n_g = 1 \text{ (for CO}_2) - 1 \text{ (for O}_2) = 0 \] Since \(\Delta n_g = 0\), the relationship simplifies: \[ \Delta H = \Delta U + (0)RT = \Delta U \] Therefore, the molar enthalpy change is equal to the molar internal energy change. \[ \Delta H_m = \Delta U_m = -207 \text{ kJ/mol} \] Step 4: Final Answer
The enthalpy change for the combustion of 1 mole of graphite is -207 kJ mol\(^{-1}\).