Question

0.4 g of propane burns completely at 300 K in a bomb calorimeter. The temperature of the calorimeter and surrounding water rises by \(0.4^\circ C\). If the heat capacity of the calorimeter and contents is \(24 \, kJ \, K^{-1}\), what is the enthalpy for the reaction? Assume that propane gas shows ideal behaviour.

• A. \(-1006.9 \, kJ\)
• B. \(-1063.5 \, kJ\)
• C. \(-2084.6 \, kJ\)
• D. \(-1902.8 \, kJ\)

Hint:

In a bomb calorimeter, heat measured is \(\Delta U\), because the process occurs at constant volume. To find \(\Delta H\), use \(\Delta H = \Delta U + \Delta n_gRT\).

Answer 1

The Correct Option is B

Step 1: Calculate heat released in the bomb calorimeter.
In a bomb calorimeter, the reaction takes place at constant volume. Therefore, heat released is equal to change in internal energy for the given amount of propane.
\[ q_v = -C \Delta T \] Given:
\[ C = 24 \, kJ \, K^{-1}, \quad \Delta T = 0.4 \, K \] \[ q_v = -24 \times 0.4 = -9.6 \, kJ \] So, \(0.4 \, g\) propane releases \(9.6 \, kJ\) heat.

Step 2: Calculate moles of propane burnt.
Molar mass of propane \((C_3H_8)\) is:
\[ 3(12) + 8(1) = 44 \, g \, mol^{-1} \] \[ \text{Moles of propane} = \frac{0.4}{44} \] \[ \text{Moles of propane} = 0.00909 \, mol \]

Step 3: Calculate internal energy change per mole.
For \(0.00909 \, mol\), heat released is \(-9.6 \, kJ\).
Therefore, for \(1 \, mol\):
\[ \Delta U = \frac{-9.6}{0.00909} \] \[ \Delta U = -1056 \, kJ \, mol^{-1} \]

Step 4: Write the combustion reaction of propane.
The complete combustion reaction of propane is:
\[ C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l) \] Now, calculate change in moles of gaseous species:
\[ \Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} \] \[ \Delta n_g = 3 - (1+5) \] \[ \Delta n_g = 3 - 6 = -3 \]

Step 5: Convert internal energy change into enthalpy change.
For ideal gases:
\[ \Delta H = \Delta U + \Delta n_g RT \] Substituting the values:
\[ \Delta H = -1056 + (-3)(8.314 \times 10^{-3})(300) \] \[ \Delta H = -1056 - 7.4826 \] \[ \Delta H = -1063.48 \, kJ \, mol^{-1} \]

Step 6: Choose the correct option.
The calculated value is approximately:
\[ \Delta H = -1063.5 \, kJ \] Therefore, the correct option is (B).
Final Answer:
The enthalpy change for the reaction is:
\[ \boxed{-1063.5 \, kJ} \]