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Chemistry AS > 2.1 Energetics > Flashcards

2.1 Energetics Flashcards

0
Q

State Hess’s law.

A

The enthalpy change of a reaction is independent of the route taken.

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1
Q

What is the enthalpy of formation?

A

The enthalpy change when one mole of a substance is formed from their constituent elements under standard conditions, all products and reactants in their standard states.

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2
Q

What is the enthalpy of combustion?

A

The enthalpy change when one mole of a substance is burned completely in oxygen, under standard conditions, products and reactant in their standard states.

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3
Q

Which way do the arrows point when drawing a Hess cycle with ‘enthalpy of formation’ data?

A

Upwards.

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4
Q

Which way do the arrows point when drawing a Hess cycle with ‘enthalpy of combustion’ data?

A

Downwards

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5
Q

What equation could you use to calculate the enthalpy change when given ‘formation’ data?

A

ΔH = ∑ΔHf (products) - ∑ΔHf (reactants)

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6
Q

A)Use the standard enthalpy of formation data from the table and the equation for the extraction of manganese to calculate a value for the standard enthalpy change of this extraction.
Mn2O3(s) + 3CO(g) –> 2Mn(s) + 3CO2(g)

/_\Hf: Mn2O3(s)=-971
CO(g)=-111
Mn(s)=0
CO2(g)= -394

B) State why the standard enthalpy of formation of Mn(s) is zero.

A

A) 122 kJmol-1 (1 mark for correct Hess cycle/equation), second mark for correct calculation, third mark for correct answer)

B) Because it’s an element / By definition

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7
Q

What equation could you use when working out the enthalpy change of a reaction using ‘combustion’ data?

A

ΔH = ∑ΔHc (reactants) - ∑ΔHc (products)

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8
Q

Define ‘mean bond enthalpy’

A

The enthalpy change to break a covalent bond, averaged over different molecules.

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9
Q

When burning a known mass of fuel in a calorimeter and measuring the temperature change, what equation can you use to calculate energy released?

A

Q=MC/_\T

Q is energy in Joules
M is the mass of the substance(don’t use the mass of fuel! It’s normally the mass of water that’s being heated up)

C is the specific heat capacity (usually 4.18 Jg-1K-1)

/_\T is the change in temperature (you don’t need to convert from degrees Celsius to Kelvin for this calculation as its the temperature CHANGE)

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10
Q

CH3OH(l) + 1.5O2(g) –> CO2(g) + 2H2O(l)
0.32g of methanol was burned, the temperature of the 200cm3 of water rose by 4.0 K. Calculate the enthalpy change in kJ mol-1. (Specific heat capacity is 4.18 J g-1 K-1).

A

Q=MC/_\T

200 x 4.18 x 4 = 3344J

3344/1000=3.344KJ

Moles methanol 0.32/32= 0.01 mol

3.344/0.01=334.4

Enthalpy change= -334.4KJmol-1 (it’s an exothermic reaction, meaning heat is being released to the surroundings, so the answer is negative)

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11
Q

When measuring enthalpy changes in a solution (such as neutralisation reactions), why are polystyrene beakers commonly used?

A
  • They are good insulators, so it reduces heat loss through the sides of the beakers.
  • they have a low heat capacity, so little heat is absorbed by them.
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12
Q

50cm3 of 1.0 moldm-3 HCl and 50cm3 of 1.0 moldm-3 NaOH solution were mixed in an expanded polystyrene beaker. The temperature rose by 6.6 K.(specific heat capacity 4.18 Jg-1mol-1)
Calculate the enthalpy change in KJmol-1. Give your final answer to 3 s.f.

A

Q=MC/_\T

100 x 4.18 x 6.6 = 2758.8 J

2758.8/1000=2.7588KJ

50/1000= 0.05dm3

Moles=0.05 x 1.0= 0.05 mol

  1. 7588/0.05=55.176
    - 55.2 KJmol-1
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13
Q

0.50g Zinc was added to 25.0 cm3 of 0.20moldm-3 copper Sulfate solution. The temperature rose by 10 K. Calculate the enthalpy change, the specific heat capacity is 4.2.

A

Q=MC/_\T

25x4.2x10= 1050J

1050/1000= 1.05 KJ

Moles copper sulfate solution= 0.025 x 0.20= 0.005 mol
Moles zinc = 0.50/65.4 = 0.00765 mol

In the reaction 0.005 moles of copper sulfate is reacting with 0.00765 moles of Zinc, and so there will be an excess of Zinc afterwards (0.00265 moles). Therefore the moles of copper sulfate is used to calculate the enthalpy change because all of it reacts and so will provide a more accurate value for the enthalpy change.

  1. 05/0.005 = 210
    - 210 KJ mol-1
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14
Q

Why might the value of the enthalpy change you calculate from an experiment be lower than the accepted value?

A
  • heat may have been lost to the surroundings, or transferred to the beaker.
  • incomplete reaction, or if burning a fuel it may have been due to incomplete combustion
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15
Q

Calculate the enthalpy of combustion of the following reaction using the data given:

CH3CHO(l) + H2(g) –> CH3CH2OH(l)

/_\Hc:
CH3CHO = -1167KJmol-1
H2= -286 KJmol-1
CH3CH2OH = -1367KJmol-1

A

Hess cycle with arrows pointing downwards and 4CO2 and 3H20 as the combustion products.

(-1167) + (-286) + -(-1367)= -86KJmol-1

16
Q

Calculate the standard enthalpy change for the following reaction using the data given:

SO2(g) + 2H2S(g) —> 3S(s) + 2H2O(l)

/_\Hc:
S=-297kJmol-1

/_\Hf:
H2O=-286KJmol-1
H2s=-20kJmol-1

A

Hess cycle with arrows pointing upwards and 3S(s) O2(g) and 2H2(g) on the bottom. Enthalpy of formation data filled in and must multiply by 2 to get 2H2S and by 2 to get 2H20. The enthalpy of combustion of S is said to be -297KJmol-1, (S + O2 –> SO2). So the enthalpy of formation of SO2 must be the same as the enthalpy change of a reaction is independent of the route taken.

-(-297) + -(-40) + (-572)=-235 KJmol-1

17
Q

When burning a known mass of fuel in a calorimeter and measuring the temperature change, what equation can you use to calculate energy released?

A

Q=MC/_\T

Q is energy in Joules
M is the mass of the substance(don’t use the mass of fuel! It’s normally the mass of water that’s being heated up)

C is the specific heat capacity (usually 4.18 Jg-1K-1)

/_\T is the change in temperature (you don’t need to convert from degrees Celsius to Kelvin for this calculation as its the temperature CHANGE)

18
Q

CH3OH(l) + 1.5O2(g) –> CO2(g) + 2H2O(l)
0.32g of methanol was burned, the temperature of the 200cm3 of water rose by 4.0 K. Calculate the enthalpy change in kJ mol-1. (Specific heat capacity is 4.18 J g-1 K-1).

A

Q=MC/_\T

200 x 4.18 x 4 = 3344J

3344/1000=3.344KJ

Moles methanol 0.32/32= 0.01 mol

3.344/0.01=334.4

Enthalpy change= -334.4KJmol-1 (it’s an exothermic reaction, meaning heat is being released to the surroundings, so the answer is negative)

19
Q

When measuring enthalpy changes in a solution (such as neutralisation reactions), why are polystyrene beakers commonly used?

A
  • They are good insulators, so it reduces heat loss through the sides of the beakers.
  • they have a low heat capacity, so little heat is absorbed by them.
20
Q

When burning a known mass of fuel in a calorimeter and measuring the temperature change, what equation can you use to calculate energy released?

A

Q=MC/_\T

Q is energy in Joules
M is the mass of the substance(don’t use the mass of fuel! It’s normally the mass of water that’s being heated up)

C is the specific heat capacity (usually 4.18 Jg-1K-1)

/_\T is the change in temperature (you don’t need to convert from degrees Celsius to Kelvin for this calculation as its the temperature CHANGE)

Chemistry AS (17 decks)

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