Numerical/Formula

1. TITLE

Converting Bomb Calorimetry $\Delta U$ to $\Delta H$ numerical — The method that works every time

2. PROBLEM TYPE INTRO (50 words)

This is a classic Thermodynamics numerical connecting heat measured at constant volume (bomb calorimeter) to heat at constant pressure (enthalpy). You will need the formula $\Delta H = \Delta U + \Delta n_g RT$. The most common pitfall? Forgetting to convert the gas constant ($R$) from Joules to kiloJoules before adding it to $\Delta U$!

3. MCQ (Full numerical problem)

Question:
The combustion of 1 mole of liquid benzene ($C_6H_6$) occurs in a bomb calorimeter at $298 \text{ K}$. The heat released is measured to be $3263.9 \text{ kJ mol}^{-1}$ at constant volume. What is the enthalpy of combustion ($\Delta H$) for the reaction at the same temperature?
(Given: $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$)

(a) $-3267.6 \text{ kJ mol}^{-1}$
(b) $-3260.2 \text{ kJ mol}^{-1}$
(c) $-6980.2 \text{ kJ mol}^{-1}$
(d) $+3267.6 \text{ kJ mol}^{-1}$

4. STEP-BY-STEP SOLUTION

SOLUTION

Given:
* Heat released at constant volume ($q_v$) = $\Delta U = -3263.9 \text{ kJ mol}^{-1}$ (Note the negative sign because heat is released)
* Temperature ($T$) = $298 \text{ K}$
* Gas constant ($R$) = $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$

To Find:
* Enthalpy of combustion ($\Delta H$) in $\text{kJ mol}^{-1}$

Formula:
* $\Delta H = \Delta U + \Delta n_g RT$
* Explain when to use this formula: Use this relationship to convert the internal energy change ($\Delta U$) measured in a sealed bomb calorimeter (constant volume) to the enthalpy change ($\Delta H$) at constant atmospheric pressure.

Step 1: Write the balanced chemical equation and find $\Delta n_g$
First, write the combustion reaction for liquid benzene:
$C_6H_6(l) + \frac{15}{2} O_2(g) \rightarrow 6 CO_2(g) + 3 H_2O(l)$
Now, calculate $\Delta n_g$ (moles of gaseous products $-$ moles of gaseous reactants).
$\Delta n_g = 6 - 7.5 = -1.5 \text{ mol}$

Step 2: Calculate the $\Delta n_g RT$ term and convert to kiloJoules
$\Delta n_g RT = (-1.5 \text{ mol}) \times (8.314 \text{ J K}^{-1} \text{ mol}^{-1}) \times (298 \text{ K})$
$\Delta n_g RT = -3716.3 \text{ J}$
Convert Joules to kiloJoules (divide by $1000$):
$\Delta n_g RT = -3.716 \text{ kJ}$

Step 3: Calculate final $\Delta H$
$\Delta H = \Delta U + \Delta n_g RT$
$\Delta H = -3263.9 \text{ kJ} + (-3.716 \text{ kJ})$
$\Delta H = -3267.616 \text{ kJ}$

Answer: (a) $-3267.6 \text{ kJ mol}^{-1}$

5. WHERE STUDENTS GO WRONG

  • ❌ Mistake 1: Including liquids or solids in the $\Delta n_g$ calculation. Remember, $\Delta n_g$ specifically tracks ONLY the mole change for gaseous components. Including $C_6H_6(l)$ or $H_2O(l)$ ruins the calculation.
  • ❌ Mistake 2: Forgetting to convert $R$ to kiloJoules. Students often calculate $\Delta n_g RT$ in Joules ($-3716.3 \text{ J}$) and directly subtract it from $\Delta U$ in kiloJoules ($-3263.9$), resulting in Option C ($-6980.2$). Always ensure unit consistency!
  • ❌ Mistake 3: Missing the negative sign for heat released. "Heat released" means the reaction is exothermic, so $\Delta U$ must be negative. Using a positive $\Delta U$ leads to sign errors.

6. QUICK FORMULA REFERENCE

Formulas Used:
* $q_v = \Delta U$ — Use when measuring heat transfer in a closed, constant-volume system like a bomb calorimeter.
* $\Delta H = \Delta U + \Delta n_g RT$ — Use when connecting internal energy to enthalpy for reactions involving gases.

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: The combustion of an organic solid like Benzoic Acid or Naphthalene, where your first critical step is writing the perfectly balanced chemical equation to find the correct $\Delta n_g$."

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter 5, Examples 5.5 and Problem 5.6.

9. CTA

More solved numericals → Chemical Thermodynamics Complete Guide
Practice problems → Mock exams