# Using standard enthalpies of formation

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Thread starter 5 months ago
#1
Hi can someone explain to me how to do this question. Much appreciated.
Calculate the enthalpy of combustion of gaseous diborane and gaseous benzene given that they burn according to the following equations:
B2H6(g) + 3O2(g) = B2O3(s) + 3H2O(g), C6H6(g) + 7.5O2(g) = 6CO2(g) + 3H2O(g)
B2H6(g) = +31.4, B2O3(s) = -1270, C6H6(g) = +83.9, CO2(g) = -393, H2O(g) = -242

These are the answers - diborane: -2027.4 kJmol-1 benzene: -3167.9 kJmol-1 but I am unsure how you get to them.
Last edited by Kenz8473202; 5 months ago
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5 months ago
#2
(Original post by Kenz8473202)
Hi can someone explain to me how to do this question. Much appreciated.
Calculate the enthalpy of combustion of gaseous diborane and gaseous benzene given that they burn according to the following equations:
B2H6(g) + 3O2(g) = B2O3(s) + 3H2O(g), C6H6(g) + 7.5O2(g) = 6CO2(g) + 3H2O(g)
B2H6(g) = +31.4, B2O3(s) = -1270, C6H6(g) = +83.9, CO2(g) = -393, H2O(g) = -242

These are the answers - diborane: -2027.4 kJmol-1 benzene: -3167.9 kJmol-1 but I am unsure how you get to them.
Are you given some sort of a data booklet along which gives you vales for the bond energy of Oxygen or the Enthalpy change of formation for O2?
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5 months ago
#3
(Original post by Kenz8473202)
Hi can someone explain to me how to do this question. Much appreciated.
Calculate the enthalpy of combustion of gaseous diborane and gaseous benzene given that they burn according to the following equations:
B2H6(g) + 3O2(g) = B2O3(s) + 3H2O(g), C6H6(g) + 7.5O2(g) = 6CO2(g) + 3H2O(g)
B2H6(g) = +31.4, B2O3(s) = -1270, C6H6(g) = +83.9, CO2(g) = -393, H2O(g) = -242

These are the answers - diborane: -2027.4 kJmol-1 benzene: -3167.9 kJmol-1 but I am unsure how you get to them.
Each of the two questions (for diborane and benzene) have the same approach.

If you've heard of Hess Law you will know that you can draw a Hess Cycle. Have you done this before?
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Thread starter 5 months ago
#4
(Original post by qwert7890)
Each of the two questions (for diborane and benzene) have the same approach.

If you've heard of Hess Law you will know that you can draw a Hess Cycle. Have you done this before?
Yes I have learnt about the Hess Cycle but I am a bit confused - there are 3 different equations so would I need have each equation as delta H 1, delta H 2 and so on
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Thread starter 5 months ago
#5
(Original post by qwert7890)
Are you given some sort of a data booklet along which gives you vales for the bond energy of Oxygen or the Enthalpy change of formation for O2?
No. This is all that is given
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5 months ago
#6
(Original post by Kenz8473202)
Yes I have learnt about the Hess Cycle but I am a bit confused - there are 3 different equations so would I need have each equation as delta H 1, delta H 2 and so on
The diborane question is different and the benzene question is different.
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Thread starter 5 months ago
#7
(Original post by qwert7890)
The diborane question is different and the benzene question is different.
Do both the reactants to the diborane and benzene equations produce 6CO2 and 3H2O
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5 months ago
#8
(Original post by Kenz8473202)
Do both the reactants to the diborane and benzene equations produce 6CO2 and 3H2O
No.

The first reaction is:
1. B2H6(g) + 3O2(g) = B2O3(s) + 3H2O(g)

The second reaction is:
2. C6H6(g) + 7.5O2(g) = 6CO2(g) + 3H2O(g)
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5 months ago
#9
Look at this (the Hess cycle for the diborane reaction): https://pasteboard.co/JMwVZQQ.png

Ignore the balancing of the box in the lower middle.

Basically you are given the enthalpy change of formation, so you can indirectly from these reactants and products from their elements. It's difficult to put to words..
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Thread starter 5 months ago
#10
(Original post by qwert7890)
Look at this (the Hess cycle for the diborane reaction): https://pasteboard.co/JMwVZQQ.png

Ignore the balancing of the box in the lower middle.

Basically you are given the enthalpy change of formation, so you can indirectly from these reactants and products from their elements. It's difficult to put to words..
Thank you, I managed to get -2027.4 for diborane
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5 months ago
#11
(Original post by Kenz8473202)
Thank you, I managed to get -2027.4 for diborane
Were you able to understand why? Perhaps try working out the answer for benzene
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Thread starter 5 months ago
#12
(Original post by qwert7890)
Were you able to understand why? Perhaps try working out the answer for benzene
Yes, don't worry. I managed to work out the Hess Cycle and from there calculate the standard enthalpy. Here it is:
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