DF 2 - how much energy

Question 1

which contain more energy - fats or carbohydrates

Answer 1

fats

Question 2

how do fats contain more energy

Answer 2

from burning 1g of a carbohydrate, you get about 17kJ. from burning 1g of a fat you get about 36kJ

Question 3

why do fats contain more energy

Answer 3

for each carbon atom, glucose (carbohydrate) has more oxygen atoms than olive oil (fats) so glucose is much less energy rich

Question 4

is alcohol a fat or carbohydrate

Answer 4

neither

Question 5

what is alcohol used for

Answer 5

ethanol is used in drinks
alcohols are also used as an alternative to petrol for some cars. it burns in the engine releasing energy and releases energy when metabolised in the body.

Question 6

how are enthalpy changes measured

Answer 6

either practically but those that cant be measured directly need to be calculated

Question 7

how can enthalpy changes measured indirectly

Answer 7

using enthalpy cycles

Question 8

give the enthalpy cycle for turning graphite and hydrogen into methane

Answer 8

C(s) +2H₂(g) —-> CH₄
-———————/
CO₂(g) + 2H₂O(l)

Question 9

describe the enthalpy cycle
C(s) +2H₂(g) —-> CH₄
-———————/
CO₂(g) + 2H₂O(l)

Answer 9

this is a direct and indirect way to turn graphite and hydrogen into methane. the enthalpy for the direct route cannot be measured directly.
the indirect route goes via CO₂ and H₂O and involves two anthalpy changes which can both be measured directly

Question 10

when can using enthalpy cycles be used to work out indirect enthalpy changes

Answer 10

most organic compounds burn easily so cycles like this can often be used based on the standard enthalpy change of combustion to work out indirectly the enthalpy change of an organic reaction

Question 11

why can’t enthalpy changes for the direct route be measured directly

Answer 11

could be due to
- a high activation energy
- slow reaction
- more than one reaction taking place

Question 12

state Hess’s law

Answer 12

the enthalpy change for any reaction is independant of the intermediate stages as long as the initial and final conditions are the same for each route

Question 13

explain Hess’s law

Answer 13

the total enthalpy change for the indirect route is the same as the enthalpy change via the direct route. energy cannot be created or destroyed according to the law of conservation of energy. as long as the start or finish points are the same, the enthalpy change will always be the same irrespective of the process to get from start to finish

Question 14

what other names are given to enthalpy cycles

Answer 14

hess’s cycle or thermochemical cycle

Question 15

give the equation to work out the direct route indirectly

Answer 15

ΔH1 = ΔH2 - ΔH3

Question 16

why is -ΔH3

Answer 16

the reaction to which is applies to goes in the opposite direction to the way you want it to go

Question 17

work out the ΔfH in methane 1) using this cycle:
(ΔH2 goes down and ΔH3 goes down)
ΔH1
C(s) +2H₂(g) —-> CH₄
-———————/
ΔH2 ΔH3
CO₂(g) + 2H₂O(l)

2) using these ΔcH values
(C) = -393kJ mol-1
(H₂) = -286kJ mol-1
(CH₄) = -890kJ mol-1

Answer 17

ΔH2 is the sum of ΔHc of 1 mole of carbon and 2 moles of hydrogen
ΔH3 is ΔHc of methane
—> ΔH2 = ΔHc (C) + ΔHc (H₂O)
—> ΔH3 = ΔHc (CH₄)
ΔH1 is ΔHf of methane (what we are trying to find)

ΔH1 = ΔH2 - ΔH3
ΔH1 = ΔHc (C) + 2ΔcH (H₂) - ΔcH (CH₄)
ΔH1 = -393 + 2(-286) - (-890)
ΔH1 = -75 kJmol -1

Question 18

where can we find ΔHf of compounds

Answer 18

ΔHf of many compounds from their elements in their standard conditions are available in data books. some have been measured directly and others indirectly.
its useful to calculate the enthalpy change for a reaction rather than doing experiments yourself

Question 19

work out the ΔH of reaction using this enthalpy cycle (both arrows point down)
route 1
3C(s) + 4H₂ ———> C3H8 (g)
↓ ↓
↓ route 2 ↓
-> 3CO2(g) + 4H2O(g) <-

Answer 19

reactants
sum of
3C –> 3x-394
4H2 –> 4x-286
= -2326

products
C3H8 = -2219
we’re going against the arrow so we have to change the symbol from - to +
+-2219

route 1 = -2326 - (-2219) = -107kJ/mol

Question 20

what is a key point to note when forming an element from an element in ΔHf

Answer 20

if you are forming an element from an element for e.g O2 from O2 there is no chemical change. so this will have a standard enthalpy change of formation of 0 kJmol-1

Question 21

calculate enthalpy change of reaction for the following process
NH3(g) +HCl (g) —–> NH4Cl(s)

ΔfH (NH3) =-46.1kJmol-1
ΔfH (HCl) = -92.3 kJmol-1
ΔfH (NH4Cl) = -315kJmol-1

Answer 21

step 1 - draw Hess’s cycle
(both arrows go up)
ΔH1
NH3 (g) + HCl (g) ——> NH4Cl (s)
↑ ↑
↑ ΔH2 ΔH3 ↑
↑ ↑
1/2 N2(g) +2 H2(g) + 1/2 Cl2(g)

step 2 - identify the reactions involved from the Hess’s cycle
ΔH2 = ΔHf (NH3) + ΔHf (HCl)
ΔH3 = ΔHf (NH4Cl)
ΔH1 = ΔH of reaction of ammonia and hydrogen chloride - the quantity you are trying to find

step 3 - write the equation to calculate the ΔH1 value
ΔH1 = -ΔH2 + ΔH3
ΔH1 = -(ΔHf (NH3) - ΔHf(HCl) +ΔHf(NH4Cl)

step 4 - substitute in the ΔHf values
=- (-46.1) - (-92.3) + (-315)
= -176.6 kJmol-1

Question 22

how to work out the enthalpy change of reaction

Answer 22

sum of ΔHf of products - sum of ΔHf of reactants