Advanced Higher Chemistry Help!!!

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??? :smile:

Reply 1

NDVA

8

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??? :smile:

How about replacing that reaction with the reaction

Pb(s)+C(s)+O_2(g) \rightarrow PbO(s)+CO(g)

(edited 8 years ago)

Reply 2

flyingpanda786

OP

5

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

Reply 3

NDVA

8

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 :smile:

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 :smile:

(edited 8 years ago)

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??? :smile:

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 ...