As level chemistry

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Darkwings_Lok
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#1
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#1
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





Is the equation 2(CH3CONH2)+6O2 -> 4CO2 + 5H2O +N2
And whats to do next?
Please help, don't paste me a link to website.
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ganesteaches
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#2
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So this is a Hess's Law question you need to draw the Hess cycle and establish which route you are taking. As they have provided enthalpy of formation which way do you inital arrows go and also what do you have in the bottom compartment of the cycle i.e.

reactants -> products

______?here? (sorry had to do the lines here cause it's meant to be under the arrow)

Once you establish the way the arrows go from the enthalpy of formation and from that establish a route should be straight forward. Hopefully that gives you a hint.
Last edited by ganesteaches; 5 months ago
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Aleksander Krol
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#3
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(Original post by Darkwings_Lok)
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





Is the equation 2(CH3CONH2)+6O2 -> 4CO2 + 5H2O +N2
And whats to do next?
Please help, don't paste me a link to website.
We can calculate the Enthalpy Change of a Reaction by subtracting the Enthalpy Change of Formation of the Total Products from the Enthalpy Change of Formation of the Total Reactants.
This is only applicable when they've given you the values of the Enthalpy Change of Formation for the Products and the Reactants in the question.
Also note that, the Enthalpy Change of Formation for all elements is zero. e.g.: Enthalpy Change for the formation of N2, O2, Fe etc. is zero.

Balanced equation for the combustion of Ethanamide is:

1C2H5NO + O2 ----> 2CO2 + 2.5 H2O + 0.5N2

According to the definition of the Enthalpy Change of Combustion, when 1 mole of a substance is burnt in sufficient amount of oxygen under standard conditions.
Therefore, Ethanamide has to be 1 mole over here.

The Reactants in this question are, O2 and Ethanamide.
The Total Enthalpy Change for the Formation of O2 will be zero.
Reason: Because the Enthalpy Change of Formation for all elements is zero.
The Total Enthalpy Change for the Formation of Ethanamide will be -320 KJmol^-1

So, the Total Enthalpy Change of Formation of the Reactants are (-320) + 0 = - 320 KJmol^-1

The products are, 2 CO2, 0.5 N2 and 2.5 H2O
So, according to the definition of the Enthalpy Change of Formation, when 1 mole of a compound is formed from its constituent elements in their standard state, under standard conditions.
In the balanced equation, we have 2 CO2 + 0.5 N2 + 2.5 H2O
not 1 CO2 + 1 N2 + 1 H2O
So, the data they have given in the question, are when 1 mole of CO2 and H2O are formed.
Also, they have not given you the Enthalpy Change of Formation for O2 and N2 , because you must know they're zero.
Therefore the Total Enthalpy Change for the Formation of the products 2CO2 + 0.5 N2 + H20 would be (2 x - 394) + 0 + (2.5 x - 286) = -1503 KJmol^-1

Since now when you finally have the Enthalpy Change of Formation of the Total Products and Reactants, you can now directly subtract the Enthalpy Change of Formation of the Total Reactant from the Enthalpy Change of Formation of the Total Products: (-1503) - (-320) = -1183 KJmol^-1
This would be your final answer.

You can also solve this by drawing the Hess' Cycle, but this can be used to save time.
That depends on you.
Last edited by Aleksander Krol; 5 months ago
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Darkwings_Lok
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#4
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Great, thanks
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