krit.club logo

Chemical Thermodynamics - Enthalpy and Calorimetry

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Enthalpy (HH) is a state function defined as the sum of the internal energy and the product of pressure and volume: H=U+PVH = U + PV.

•

The change in enthalpy (ΔH\Delta H) is equal to the heat exchanged by the system with its surroundings at constant pressure (qP=ΔHq_P = \Delta H).

•

For reactions involving gases, the relationship between enthalpy change and internal energy change is given by ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT, where Δng\Delta n_g is the change in the number of moles of gaseous products and reactants.

•

Exothermic reactions release heat to the surroundings, resulting in ΔH<0\Delta H < 0. Endothermic reactions absorb heat from the surroundings, resulting in ΔH>0\Delta H > 0.

•

Calorimetry is the experimental technique used to measure heat changes. A bomb calorimeter measures ΔU\Delta U at constant volume, while a coffee-cup calorimeter measures ΔH\Delta H at constant pressure.

•

Specific Heat Capacity (cc) is the amount of heat required to raise the temperature of 1 g1 \text{ g} of a substance by 1 K1 \text{ K} or 1∘C1^\circ C. Molar Heat Capacity (CmC_m) refers to the heat required for 1 mole1 \text{ mole} of a substance.

•

Hess's Law of Constant Heat Summation states that the total enthalpy change for a reaction is independent of the path taken, provided the initial and final states are identical.

📐Formulae

H=U+PVH = U + PV

ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT

q=m⋅c⋅ΔTq = m \cdot c \cdot \Delta T

q=C⋅ΔTq = C \cdot \Delta T

ΔrH∘=∑ΔfH∘(products)−∑ΔfH∘(reactants)\Delta_r H^\circ = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants})

Δng=ng(products)−ng(reactants)\Delta n_g = n_{g}(\text{products}) - n_{g}(\text{reactants})

💡Examples

Problem 1:

Calculate the enthalpy change (ΔH\Delta H) for the reaction N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) at 298 K298 \text{ K}, given that the internal energy change (ΔU\Delta U) is −92.22 kJ-92.22 \text{ kJ}. (R=8.314 J K−1 mol−1R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1})

Solution:

  1. Calculate Δng\Delta n_g: Δng=2−(1+3)=−2\Delta n_g = 2 - (1 + 3) = -2.
  2. Use the formula ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT.
  3. Convert ΔU\Delta U to Joules: −92.22×103 J-92.22 \times 10^3 \text{ J}.
  4. ΔH=−92220+(−2×8.314×298)\Delta H = -92220 + (-2 \times 8.314 \times 298).
  5. ΔH=−92220−4955.14=−97175.14 J\Delta H = -92220 - 4955.14 = -97175.14 \text{ J}.
  6. ΔH≈−97.18 kJ\Delta H \approx -97.18 \text{ kJ}.

Explanation:

Since the number of moles of gas decreases during the reaction, work is done on the system by the surroundings, making the enthalpy change more negative than the internal energy change.

Problem 2:

A 0.50 g0.50 \text{ g} sample of benzoic acid is burned in a bomb calorimeter. The temperature of the calorimeter increases by 2.5 K2.5 \text{ K}. If the heat capacity of the calorimeter is 10.0 kJ K−110.0 \text{ kJ K}^{-1}, calculate the heat of combustion at constant volume.

Solution:

  1. Use the formula for heat absorbed by the calorimeter: q=C⋅ΔTq = C \cdot \Delta T.
  2. q=10.0 kJ K−1×2.5 K=25.0 kJq = 10.0 \text{ kJ K}^{-1} \times 2.5 \text{ K} = 25.0 \text{ kJ}.
  3. Since the combustion releases heat, the heat of reaction qVq_V (which is ΔU\Delta U) is −25.0 kJ-25.0 \text{ kJ} for 0.50 g0.50 \text{ g}.

Explanation:

In a bomb calorimeter, the volume is constant, so the measured heat flow corresponds to the change in internal energy (ΔU\Delta U). The negative sign indicates an exothermic process.

Enthalpy and Calorimetry - Revision Notes & Key Formulas | ICSE Class 11 Chemistry