Chemistry A Level AQA Energetics Question

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MadyanC1
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I have no idea how to work this out, the answer to 6 is D and the answer to 7 is B, can someone please explain how?

6) Use the information below to answer this question.
C(s) + O2(g) → CO2(g) ∆H = −393.5 kJ mol−1
H2(g) + O2(g) → H2O(l) ∆H = −285.8 kJ mol−1
3C(s) + 4H2(g) → C3H8(g) ∆H = −104.0 kJ mol−1
4C(s) + 5H2(g) → C4H10(g) ∆H = −125.2 kJ mol−1
The value in kJ mol−1 of the enthalpy of thermal dissociation when butane forms propane, hydrogen and carbon is
A −26.3
B −17.5
C +17.5
D +21.2

7) When ethanamide (CH3CONH2) burns in oxygen the carbon is converted into carbon dioxide, the hydrogen is converted into water and the nitrogen forms nitrogen gas.

Substance ethanamide carbon dioxide water
Enthalpy of formation ( ) / kJ mol−1 −320 −394 −286
Using the data above, which one of the following is a correct value for the enthalpy of combustion of ethanamide?
A −1823 kJ mol−1
B −1183 kJ mol−1
C −1000 kJ mol−1
D −360 kJ mo1−1
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charco
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(Original post by MadyanC1)
I have no idea how to work this out, the answer to 6 is D and the answer to 7 is B, can someone please explain how?

6) Use the information below to answer this question.
C(s) + O2(g) → CO2(g) ∆H = −393.5 kJ mol−1
H2(g) + O2(g) → H2O(l) ∆H = −285.8 kJ mol−1
3C(s) + 4H2(g) → C3H8(g) ∆H = −104.0 kJ mol−1
4C(s) + 5H2(g) → C4H10(g) ∆H = −125.2 kJ mol−1
The value in kJ mol−1 of the enthalpy of thermal dissociation when butane forms propane, hydrogen and carbon is
A −26.3
B −17.5
C +17.5
D +21.2
6) Use the information below to answer this question.
1 ==> C(s) + O2(g) → CO2(g) ∆H = −393.5 kJ mol−1
2 ==> H2(g) + O2(g) → H2O(l) ∆H = −285.8 kJ mol−1
3 ==> 3C(s) + 4H2(g) → C3H8(g) ∆H = −104.0 kJ mol−1
4 ==> 4C(s) + 5H2(g) → C4H10(g) ∆H = −125.2 kJ mol−1
The value in kJ mol−1 of the enthalpy of thermal dissociation when butane forms propane, hydrogen and carbon is
A −26.3
B −17.5
C +17.5
D +21.2

STEP 1: Write out the target equation (butane forms propane, hydrogen and carbon):

C4H10 ==> C3H8 + C + H2

STEP 2: Construct this equation from the ones given

reverse equation 4 (change sign of energy)

==> C4H10(g) → 4C(s) + 5H2(g) ∆H = +125.2 kJ mol−1

Add equation 3

3C(s) + 4H2(g) → C3H8(g) ∆H = −104.0 kJ mol−1
C4H10(g) → 4C(s) + 5H2(g) ∆H = +125.2 kJ mol−1
------------------------------------------------------------------------- add and cancel out common terms
C4H10(g) → C3H8(g) + C(s) + H2(g) ∆H = −104.0 +125.2 = 21.2 kJ mol−1
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charco
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(Original post by MadyanC1)
I have no idea how to work this out, the answer to 6 is D and the answer to 7 is B, can someone please explain how?


7) When ethanamide (CH3CONH2) burns in oxygen the carbon is converted into carbon dioxide, the hydrogen is converted into water and the nitrogen forms nitrogen gas.

Substance ethanamide carbon dioxide water
Enthalpy of formation ( ) / kJ mol−1 −320 −394 −286
Using the data above, which one of the following is a correct value for the enthalpy of combustion of ethanamide?
A −1823 kJ mol−1
B −1183 kJ mol−1
C −1000 kJ mol−1
D −360 kJ mo1−1
Write out the equation that you wish to find.

The reactants are converted into the products, so you lose the enthalpy of formation of the reactants (negative enthalpy of formation values) and you gain the enthalpy of formation of the products (no change to the enthalpy values)

Hence:

Formation enthalpy of products - Formation enthalpy of reactants.

Check this out:

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MadyanC1
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Thanks a lot for the help!!
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