flyingpanda786
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C(s) + O2(g) → CO2(g) DH° = –396kJ mol–1
Pb(s) + ½O2(g) → PbO(s) DH° = –210kJ mol–1
PbO(s) + CO(g) → Pb(s) + CO2(g) DH° = –74kJ mol–1

What is the value of DH°, in kJ mol–1, for the
following reaction?

C(s) + ½O2(g) → CO(g)

A –260
B –112
C +112
D +260

I have tried for really long and the answer that i get isn't even an option, hahaha. Could you please explain also???
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NDVA
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(Original post by flyingpanda786)
C(s) + O2(g) → CO2(g) DH° = –396kJ mol–1
Pb(s) + ½O2(g) → PbO(s) DH° = –210kJ mol–1
PbO(s) + CO(g) → Pb(s) + CO2(g) DH° = –74kJ mol–1

What is the value of DH°, in kJ mol–1, for the
following reaction?

C(s) + ½O2(g) → CO(g)

A –260
B –112
C +112
D +260

I have tried for really long and the answer that i get isn't even an option, hahaha. Could you please explain also???
How about replacing that reaction with the reaction Pb(s)+C(s)+O_2(g) \rightarrow PbO(s)+CO(g)
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flyingpanda786
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(Original post by NDVA)
How about replacing that reaction with the reaction Pb(s)+C(s)+O_2(g) \rightarrow PbO(s)+CO(g)
hmmm still not sure i keep getting -532kJ mol-1
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NDVA
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(Original post by flyingpanda786)
hmmm still not sure i keep getting -532kJ mol-1
I think you just made a mistake somewhere about the signs

Let the required \Delta H be x.

Pb(s)+C(s)+O_2(g) \rightarrow PbO(s)+CO(g) has \Delta H = -322kJmol^{-1}

But Pb(s)+\frac{1}{2} O_2 \rightarrow PbO has \Delta H = -210kJmol^{-1}

Hence  (-210)+x=(-322) so x=-112kJmol^{-1}, which is option B
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charco
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(Original post by flyingpanda786)

equation 1: C(s) + O2(g) → CO2(g) DH° = –396kJ mol–1
equation 2: Pb(s) + ½O2(g) → PbO(s) DH° = –210kJ mol–1
equation 3: PbO(s) + CO(g) → Pb(s) + CO2(g) DH° = –74kJ mol–1

What is the value of DH°, in kJ mol–1, for the
following reaction?

C(s) + ½O2(g) → CO(g)

A –260
B –112
C +112
D +260

I have tried for really long and the answer that i get isn't even an option, hahaha. Could you please explain also???
Because of Hess' law any mathematical operation may be carried out on an equation provided you do the same to the energy change.

The idea here is that you construct the required equation by manipulation of the given equations. My advice is to do so one step at a time.

You need C(s) and this appears in equation 1

C(s) + O2(g) → CO2(g)

But you don't need CO2(g) so subtract equation 3 from 1
C(s) + O2(g) → CO2(g)
PbO(s) + CO(g) → Pb(s) + CO2(g)
---------------------------------------------- subtract
C(s) + O2(g) - PbO(s) - CO(g) → CO2(g) - Pb(s) - CO2(g)

cancelling down

C(s) + O2(g) - PbO(s) - CO(g) → -Pb(s)

rearranging to make all terms positive gives

C(s) + O2(g) + Pb(s) → PbO(s) + CO(g)

Now you want rid of the PbO(s) term so subtract equation 2

C(s) + O2(g) + Pb(s) → PbO(s) + CO(g)
Pb(s) + ½O2(g) → PbO(s)
------------------------------------- subtract
C(s) + ½O2(g) → CO(g)

Which is your target equation. Simply follow each step using the energy to get the answer ...
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username986184
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If it's tricky then draw a piccy
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flyingpanda786
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(Original post by TeachChemistry)
If it's tricky then draw a piccy
wow!! that must have taken you a while haha. That was really nice of you thank you for your help and time budd, really appreciate it!!!
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