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Chemistry: Post your doubts here!

mahnam

mahnam

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Can someone please help me with this:
Given CO(g) + 1/2 O2(g) → CO2(g) ΔHo = –283 kJ mol–1 H2(g) + 1/2 O2(g) → H2O(l) ΔHo = –286 kJ mol–1H2O(g) → H2O(l) ΔHo = –44 kJ mol–1what is the change in enthalpy, ΔHo, for the following reaction?

CO2(g) + H2(g) → CO(g) + H2O(g)
A –525 kJ mol–1 B –41 kJ mol–1 C +41 kJ mol–1 D +525 kJ mol–1
I am getting B but apparently the correct ans is C.
I don't know where i went wrong so if someone can tell i'd be very grateful.
 
AbbbbY

AbbbbY

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mahnam said:
Can someone please help me with this:
Given CO(g) + 1/2 O2(g) → CO2(g) ΔHo = –283 kJ mol–1 H2(g) + 1/2 O2(g) → H2O(l) ΔHo = –286 kJ mol–1H2O(g) → H2O(l) ΔHo = –44 kJ mol–1what is the change in enthalpy, ΔHo, for the following reaction?

CO2(g) + H2(g) → CO(g) + H2O(g)
A –525 kJ mol–1 B –41 kJ mol–1 C +41 kJ mol–1 D +525 kJ mol–1
I am getting B but apparently the correct ans is C.
I don't know where i went wrong so if someone can tell i'd be very grateful.
Click to expand...

There's a long and correct way and there's a shortcut way.

You could use Hess's law and go through the hassle, but since it's an MCQ there has to be a shortcut. If there's anything I've learnt doing P1's over the years it's this.

Break down the reactions into two parts.

Firstly, CO2 -> CO
Secondly, H2 -> H2O

You've got the value for CO-> CO2 so CO2->CO will be negative of the former.

+283-286.

Now comes the interesting part. H2->H2O will produce H2O in the liquid state. You require H2O in the gaseous state.
You've the value for H2O(g)->H2O(l). The value for H2O(l) -> H2O(g) will be negative of the former

so you're left with +283-286+44
 
1357911

1357911

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S

sitooon

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Can someone tell me how to draw the diagram for 1 (c)
and eg, of ONE method to prepare the concentrations ? in (d)
 

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