Question 12

[kzteam5226]

How do we know that CO2 is the product of combustion and CO? I know O2 is always a reactant however I did not know why the product includes C02 and not C0? Is that always the case for combustion reactions?

[mcat_premed3832]

When the MCAT says "combustion" with no other qualifications then indeed one assumes complete combustion. Any hydrocarbon undergoing complete combustion produces carbon dioxide. Carbon monoxide, famous for being a dangerously lethal colorless and odorless gas, is the product of incomplete combustion.

[nicole.mag7972]

Isn't deltaH = Products - reactants? So shouldn't the equation be -1,096- (602+ 394) = -2,092?

[admin]

The Explanation is fine. It considers the bond dissociation energies alone exempting the enthalpies of formation as these are directly not applicable to the reaction. Moreover, the delta H° for the reaction is equal to delta H for the bonds being broken – delta H for the bonds being formed and it is not the same as using the enthalpies of formation values which would be as you stated. This is a combustion or a decomposition reaction. Thus, heat is given off when bonds are formed and heat is consumed when bonds are broken. Thus, the delta H° is a measure of the energy of the bond-making and bond-breaking processes that occurs as reactants turn to products.

Hence, as mentioned for this question the following is applicable:

Delta H° = (energy of bonds being broken) – (energy of bonds being formed).

As the delta H° is negative, this means that the bonds being formed in the reaction are stronger than the bonds that are broken, more energy will be released in the bond-forming process than will be consumed in the bond-breaking process.

[jsfkt78927]

Although not a typical mcat question (ie too calculation intensive) good and tricky. Bring the heat GS, i like it. Many things tripped me up here,
1. didnt realize we were doing bond enthalpies rather than heat of formation
2. wasting time on a non-typical calculation intensive question frustrated me
3. if i knew it was bond enthalpies, i probably would have picked -1900 rather than dividing by two

[jsfkt78927]

also, for the admin, realize that the 'discuss this question' is not properly hyperlinked. clicking on this one for instance brings you to question 17 thread.

[pinoalejan6739]

There is a mistake in the passage or in the question because the DeltaH of formation for MgO in the passage is given for gas while in the question it is asked as a solid.

[sabireen]

I am at a lost to why in the world their balanced equation is stated that way.
We are combusting ethyne: C2H2, so I wrote by equation based on that and the fact that I had to do SOMETHING with those hydrogens:
C2H2 + 3O2 -> 2 CO2 + H2O

then I calculated dissociation heat for 1 cc triple bond, 2 ch bonds and 3 oo double bonds as well as formation heat for 4 co double bonds, and two oh single bond. I got an answer near -760, which is closest to the correct answer. Is my logic ok, because I cannot follow their explanation.

[mohsinalam5969]

@kzteam5226
Not in all cases, the combustion of a hydrocarbon yields CO2 and H2O

Instead, the correct statement would be

"the products of complete combustion of a hydrocarbon are carbon dioxide and water".

It means that any hydrocarbon when reacted with excess of oxygen or is made to undergo complete combustion, it results in CO2 and H2O.

Hope you got what I am saying. Good Luck

[mohsinalam5969]

@sabireen
Hess' law states that the energy change for any chemical or physical process is independent of the pathway or number of steps required to complete the process provided that the final and initial reaction conditions are the same. In other words, an energy change is path independent, only the initial and final states being of importance. This path independence is true for all state functions.
Hess' law allows the enthalpy change (ΔH) for a reaction to be calculated even when it cannot be measured directly. This is accomplished by performing basic algebraic operations based on the chemical equation of reactions using previously determined values for the enthalpies of formation.
this can be calculated only by the help of balanced chemical equation and your approach is wrong because in some cases you are calculating the heat of dissociation of only 1 C C (triple bond) and then 2 C H bonds..... you should have to follow the explanation in order to get the correct answer...thanks