Murph's Blog of Chem Joy: SCH 4U - Enthalpy & Calorimetry

Properties for Enthalpy

Enthalpy is an extensive property – the magnitude of ΔH ∝ n_reactants. Enthalpy is reported per mole of a substance and given the name molar enthalpy. There are a variety of physical and chemical changes for which molar enthalpy is reported: ΔH_soln, ΔH_c (below), ΔH_vap, ΔH_fr, ΔH_neut, ΔH_f.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) ΔH_c = -802 kJ/mol of CH₄

2CH₄(g) + 4O₂(g) → 2CO₂(g) + 4H₂O(g) ΔH = 2 mol(-802 kJ/mol) = -1604 kJ

Enthalpy change for a reaction is equal in magnitude, but opposite in sign to the enthalpy change for the reverse reaction.

CO₂(g) + 2H₂O(g) → CH₄(g) + 2O₂(g) ΔH = +802 kJ

Enthalpy change for a reaction depends on the state of the reactants and products.

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = -890 kJ

Calorimetry

There are many tables containing molar enthalpy values, but how are they obtained?

A calorimeter is used to carry out these measurements – a simple calorimeter may be constructed from two nested Styrofoam cups with a lid. Calorimetry can be carried out to determine the heats of both physical and chemical changes.

Three assumptions are made during a calorimetry experiment:

the transfer of matter and energy between the system and surroundings does not occur

any heat absorbed or released by the actual calorimeter is insignificant

all dilute aqueous solutions will have the same density (d = 1.00 g/mL) and specific heat capacity (c = 4.184 J/g°C) as pure water

ex. A 50.0 mL sample of 1.0 M hydrochloric acid is mixed with 50.0 mL of 1.0 M sodium hydroxide at 25.0°C in a calorimeter. After fully mixing, the temperature is 31.9°C. Determine the enthalpy of neutralization per mole of hydrochloric acid.

m_NaOH = 50.0 g q = m ΔT c

m_HCl = 50.0 g = (50.0 g + 50.0 g)(31.9°C – 25.0°C)(4.184 J/g°C)

T₁ = 25.0°C = (100.0 g)(6.9°C)(4.184 J/g°C)

T₂ = 31.9°C = 2.9 kJ

c = 4.184 J/g°C

ΔH = ? n = cV ΔH = q/n

= 1.0 M(0.0500 L) = 2.9 kJ/0.050 mol

= 0.050 mol = 58 kJ/mol

∴ the enthalpy of neutralization is -58 kJ/mol

*Notice that the sign on the enthalpy value has been switched from + to -. This is because the data we used to calculate the enthalpy pertained to the surroundings (for instance, the temperature readings were recorded by the thermometer, which is part of the surroundings). So, we switch the sign so that the answer to the question now pertains to the system.

Homework # 1, 3-6 (HINT: For #1, google the formula for ethylene glycol.)