energetics [P1] PAPER 1+2 Flashcards

1
Q

define the term mean bond enthalpy

A

enthalpy needed in breaking one mole of (covalent) bonds (in gaseous state), averaged (for that type of bond) over a range of compounds

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2
Q

define standard enthalpy of formation

A

The enthalpy change at constant pressure when 1 mole of a product is formed from its constituent elements with all reactants and products in their standard states under standard conditions

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3
Q

define enthalpy change (ΔH)

A

heat energy change at constant pressure

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4
Q

what does standard enthalpy change refer to?

A

standard conditions, ie 100kPa, 298K, 1moldm-3

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5
Q

define standard enthalpy of combustion

A

the enthalpy change at constant pressure when 1 mole of a substance is completely burned in oxygen with all reactants and products in their standard states under standard conditions

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6
Q

how is heat change, q, calculated?

A

q = mcΔT

  • q = heat change (J)
  • m = mass of substance undergoing the temperature change (g)
  • c = specific heat capacity (given)
  • ΔT = temperature change
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7
Q

how is ΔH calculated? (using q)

A
  • q = mcΔT → convert to kJ (divide by 1000)
  • ΔH = q (kJ) /mol
  • mol = the moles of the limiting reagent
  • make sure ΔH is negative if the reaction is exothermic
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8
Q

state Hess’ Law

A

the enthalpy change at constant pressure in a reaction is independent of the route, and depends only on the initial and final states

ΔH1 = ΔH2 + ΔH3
(works like vectors)

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9
Q

the standard enthalpy of formation of an element is ___

A

zero

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10
Q

Hess cycle, given you have enthalpies of formation

A

ΔH = (Σ ΔH products) – (Σ ΔH reactants)

ΔH1 = ΔH3 - ΔH2 [diagram]

⚠️ make sure everything is multiplied according to the numbers in front of each substance in the balanced equation, e.g. you’re given the enthalpy of formation of CO2 → from BALANCED equation - 3CO2 → multiply enthalpy by 3, THEN use it in the above equation

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11
Q

Hess cycle, given you have enthalpies of combustion

A

ΔH = (Σ ΔH reactants) – (Σ ΔH products)

ΔH1 = ΔH2 - ΔH3 [diagram]

⚠️ make sure everything is multiplied according to the numbers in front of each substance in the balanced equation

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12
Q

what is calorimetry used for?

A
  • experimental method for finding enthalpy changes by measuring temperature change over time
  • data can be extrapolated to find ΔT from the graph drawn from obtained results
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13
Q

how to work out ΔH, given you have mean bond enthalpies?

A

ΔH = (break + vaporisation) – (make + vaporisation)
ΔH1 = ΔH2 - ΔH3
• ΔH2 is the sum of all the bond enthalpies (AND enthalpy of vaporisation if a reactant isn’t in a gaseous state) required to BREAK all the bonds of the reactants
• ΔH3 is the sun of all the bond enthalpies required to MAKE all the products

⚠️ make sure everything is multiplied according to the numbers in front of each substance in the balanced equation and that everything must be a gas

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14
Q

Why may the enthalpy change of a reaction calculated using Hess’ Law be similar to the enthalpy change for the same reaction calculated using mean bond enthalpies?

A
  • Same reaction
  • Same reactants and products
  • Same equation
  • Same number and types of bonds broken and formed
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15
Q

Why may the enthalpy change of a reaction calculated using Hess’ Law be different to the enthalpy change for the same reaction calculated using mean bond enthalpies?

A

The mean bond enthalpies differ from the actual bond enthalpies of the substances formed/broken in the reaction

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16
Q

Give one reason why the bond enthalpy calculated for the __ bond in ___ is different from the mean bond enthalpy quoted in a data book.

A

Data book value is derived from a number of different compounds, not just different ___ molecules

17
Q

Suggest one reason, other than the use of mean bond enthalpies, why a value for the enthalpy of combustion of a liquid alkane is different from the value obtained through a calculation.

A
  • Alkane is not gaseous
  • Equation relates to gaseous alkanes
  • It takes energy to convert it into a gas
  • The substances are gaseous in calculations using bond enthalpies
18
Q

State why the value for the standard enthalpy of formation of CO2(g) is the same as the value for the standard enthalpy of combustion of carbon.

A

These two enthalpy changes are for the same reaction

19
Q

State why the heat change calculated from the bomb calorimeter experiment is not an enthalpy change.

A

pressure isn’t constant in a bomb calorimeter