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Chemical Thermodynamics - Measurement of ΔU and ΔH: Calorimetry

Grade 11CBSEChemistry

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

🔑Concepts

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Calorimetry is the experimental technique used to measure the heat changes (qq) associated with chemical or physical processes using a device called a calorimeter.

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The measurement of internal energy change (ΔU\Delta U) is carried out at constant volume in a 'Bomb Calorimeter'. Since ΔV=0\Delta V = 0, no work is done (w=0w=0), thus qv=ΔUq_v = \Delta U.

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The measurement of enthalpy change (ΔH\Delta H) is carried out at constant pressure, usually in a 'Coffee-cup Calorimeter' or an open container. In this case, qp=ΔHq_p = \Delta H.

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Specific heat capacity (csc_s) is the amount of heat required to raise the temperature of 11 gram of a substance by 11 unit (Kelvin or Celsius).

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Molar heat capacity (CmC_m) is the amount of heat required to raise the temperature of 11 mole of a substance by 11 unit (Kelvin or Celsius).

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The heat capacity (CC) of the calorimeter is the amount of heat required to raise the temperature of the entire calorimeter assembly by 11 Kelvin.

📐Formulae

q=C⋅ΔTq = C \cdot \Delta T

q=m⋅cs⋅ΔTq = m \cdot c_s \cdot \Delta T

q=n⋅Cm⋅ΔTq = n \cdot C_m \cdot \Delta T

ΔU=qv\Delta U = q_v

ΔH=qp\Delta H = q_p

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

💡Examples

Problem 1:

In a bomb calorimeter, 1.0 g1.0 \text{ g} of Magnesium was burnt in excess of oxygen at 298 K298 \text{ K}. The temperature of the calorimeter was found to rise from 298.15 K298.15 \text{ K} to 302.65 K302.65 \text{ K}. If the heat capacity of the calorimeter system is 20.0 kJ K−120.0 \text{ kJ K}^{-1}, calculate the internal energy change (ΔU\Delta U) for the combustion of 1 mole1 \text{ mole} of Magnesium.

Solution:

  1. Calculate the temperature change (ΔT\Delta T): 302.65−298.154.50\begin{array}{r} 302.65 \\ - 298.15 \\ \hline 4.50 \end{array} ΔT=4.50 K\Delta T = 4.50 \text{ K}

  2. Calculate heat absorbed by the calorimeter (qcalq_{cal}): qcal=C⋅ΔTq_{cal} = C \cdot \Delta T qcal=20.0 kJ K−1×4.50 K=90.0 kJq_{cal} = 20.0 \text{ kJ K}^{-1} \times 4.50 \text{ K} = 90.0 \text{ kJ}

  3. Since heat is released by the reaction, qrxn=−qcal=−90.0 kJq_{rxn} = -q_{cal} = -90.0 \text{ kJ}. This is the heat for 1 g1 \text{ g} of MgMg.

  4. Molar mass of Mg=24 g mol−1Mg = 24 \text{ g mol}^{-1}. Calculate ΔU\Delta U for 1 mole1 \text{ mole}: ΔU=−90.0 kJ1.0 g×24 g mol−1=−2160 kJ mol−1\Delta U = \frac{-90.0 \text{ kJ}}{1.0 \text{ g}} \times 24 \text{ g mol}^{-1} = -2160 \text{ kJ mol}^{-1}

Explanation:

The bomb calorimeter measures heat at constant volume, so the heat evolved is equal to the change in internal energy (ΔU\Delta U). The negative sign indicates an exothermic reaction.