Energy Diagrams
The activation energy of a reaction is measured as the difference between these two points on an energy diagram.
The activation complex and the reactants.
This type of process releases heat into the surroundings.
Exothermic.
Hess’s Law allows you to calculate ΔH for a reaction by doing what with multiple known equations.
Adding their ΔH values after combining the equations.
Bond breaking is always this type of process.
Endothermic.
The standard enthalpy of formation of any element in its standard state is this value.
0 kJ/mol.
A reaction has reactants at 50 kJ and products at 10 kJ. Calculate ΔH.
–40 kJ (exothermic)
A flask feels hot to the touch, what direction is heat flowing in (specify in terms of system and surrounding)?
From the system into the surroundings.
When reversing a reaction in Hess’s Law, this must be done to ΔH.
Change its sign.
Bond enthalpy values represent this physical quantity.
Energy required to break one mole of a bond in the gas phase.
When calculating ΔH° for a reaction, you must ensure that the chemical equation is ___.
Balanced
A catalyst lowers the activation energy of a reaction. Which parts of the energy diagram remain unchanged?
The energies of reactants, products, and ΔH remain unchanged.
Bond formation is always this type of thermodynamic process.
Exothermic.
When multiplying a reaction by a coefficient, what happens to ΔH?
Multiply ΔH by the same factor.
A reaction is exothermic when the energy released from forming bonds is ______ than the energy required to break bonds.
Greater
A reaction with a highly negative enthalpy of formation indicates that the process is ___ (in terms of heat)?
Exothermic
A reaction has Eₐ = 120 kJ and ΔH = +15 kJ. Describe the shape of the energy diagram.
Large uphill activation barrier with products slightly above reactants.
When ammonium nitrate dissolves in water and the beaker becomes cold, the dissolution is this type of process.
Endothermic.
Given:
C + O₂ → CO₂ ΔH = –393 kJ
CO + ½O₂ → CO₂ ΔH = –283 kJ
Calculate ΔH for: C + ½O₂ → CO.
-110 kJ.
Explain why O=O has a higher bond enthalpy than O–O.
Double bonds are stronger and require more energy to break.
What are the standard conditions?
1 mole, 1 atm, 25°C
A reaction has reactants at 30 kJ, an activation complex at 140 kJ, and products at 80 kJ. Calculate both Eₐ and ΔH.
Eₐ = 110 kJ; ΔH = +50 kJ (endothermic)
This term describes the total heat content of a system at constant pressure.
Enthalpy (ΔH).
Given:
N₂ + O₂ → 2NO ΔH = +180 kJ
2NO + O₂ → 2NO₂ ΔH = –114 kJ
Find ΔH for: N₂ + 2O₂ → 2NO₂.
+66 kJ.
Estimate ΔH for the reaction:
H₂ + Cl₂ → 2HCl
Bond enthalpies:
H–H = 436 kJ/mol
Cl–Cl = 243 kJ/mol
H–Cl = 431 kJ/mol
ΔH = (436 + 243) – 2(431) = –183 kJ.
Using the following standard enthalpies of formation:
CO(g) = –110.5 kJ/mol
CO₂(g) = –393.5 kJ/mol
H₂O(g) = –241.8 kJ/mol
C₂H₅OH(l) = –277.0 kJ/mol
Calculate ΔH° for the combustion of ethanol:
C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(g)
ΔH∘=[2(−393.5)+3(−241.8)]−[−277.0]
ΔH∘=[−787.0−725.4]+277.0
ΔH∘=−1235.4 kJ