Chapter 9 - Enthalpy

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16 Terms

1
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exothermic

-bonds are made

-energy is released

-temperature increases

-products have less energy

-enthalpy change is negative

<p>-bonds are <strong>made</strong></p><p>-energy is <strong>released</strong></p><p>-temperature <strong>increases</strong></p><p>-products have <strong>less </strong>energy</p><p>-enthalpy change is <strong>negative</strong></p>
2
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endothermic

-bonds are broken

-energy is taken in

-temperature decreases

-products have more energy

-enthalpy change is positive

<p>-bonds are <strong>broken</strong></p><p>-energy is <strong>taken in</strong></p><p>-temperature <strong>decreases</strong></p><p>-products have <strong>more </strong>energy</p><p>-enthalpy change is <strong>positive</strong></p>
3
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average bond enthalpy

the enthalpy change when breaking one mole of covalent bonds in gaseous molecules

breaking - making (reactants - products)

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‘Explain in terms of bond energies why the reaction is exothermic

more energy is released during bond making than is taken in during bond breaking

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specific heat capacity equation

Q = m x c x change in T

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REASONS for differences in calculated + data values

-heat loss to the surroundings

-incomplete combustion

-evaporation of water/alcohol

-non-standard conditions

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standard conditions

pressure = 100kPa

temperature = 298K

concentration = 1 moldm-3

state = physical state

8
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standard enthalpy change of reaction

the energy change in a given reaction in molar quantities in a chemical equation under standard conditions

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standard enthalpy change of formation

the energy change that takes place when one mole of a compound is formed from its elements under standard conditions

10
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standard enthalpy change of combustion

the energy change when one mole of a substance completely reacts with oxygen under standard conditions

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standard enthalpy change of neutralisation

the energy change that takes place when one mole of water is formed from a neutralisation reaction

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why is calculating enthalpy change not always possible?

-slow reaction rate

-high activation energy

-alternative products formed

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Hess’ Law

direct pathway = A

indirect pathway = B + C

A = B + C

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enthalpy change of combustion -EQUATION

-arrows face down

-intermediate = CO2 + H2O

A = B - C

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enthalpy change of formation -EQUATION

-arrows face up

-intermediate = elements

A = C - B

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MULTIPLE equations in Hess’ Law

-match up the equations to main equation using arrows

-then use A=B+C