Given the following heats of combustion.
CH3OH(l) + 3/2 O2(g) -> CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ
C(graphite) + O2(g) ->CO2(g) ΔH°rxn = -393.5 kJ
H2(g) + 1/2 O2(g) -> H2O(l) ΔH°rxn = -285.8 kJ
Calculate the enthalpy of formation of methanol (CH3OH) from its elements.
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C(graphite) + 2 H2(g) + 1/2 O2(g)
-> CH3OH(l)
I don't even know where to begin on this problem
This is an exercise in the Born-Haber cycle. Look that up in your text or use the Internet (Google). You add the separate equations you have (you can multiply them or reverse them if needed) to get the final equation for the formation from the elements. If you multiply them, multiply the heats of formation. If you reverse an equation, change the sign of the heat of formation but keep the number the same. Just taking a quick look, it appears equation 2 stays as is, equation 1 is reversed, and equation 3 stays as is but you will need to check me out on that. I tried to do it in my head and sometimes these things don't work very well with that many numbers and symbols flying around.
Addendum:
This is an example of a Hess's Law problem. Your textbook should have some solved examples. Here is one solved example online:
http://answers.yahoo.com/question/index?qid=20080514171625AAsqOoC
And I need to correct my first response. After reading GK's response, I realize I called this the Born-Haber cycle. That is similar but this is an example of Hess's Law.
good thanks
well!!thanks
To calculate the enthalpy of formation of methanol (CH3OH) from its elements, we need to use the given thermochemical equations and apply the Hess's Law.
Hess's Law states that the overall enthalpy change of a reaction is independent of the pathway taken. This means that if we can express the desired reaction as a sum of known reactions, we can use their enthalpy changes to calculate the enthalpy change of the desired reaction.
In this case, we want to find the enthalpy of formation of methanol (CH3OH) from its elements (C(graphite), H2(g), and O2(g)). We have the following thermochemical equations:
1. CH3OH(l) + 3/2 O2(g) -> CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ
2. C(graphite) + O2(g) -> CO2(g) ΔH°rxn = -393.5 kJ
3. H2(g) + 1/2 O2(g) -> H2O(l) ΔH°rxn = -285.8 kJ
We need to manipulate these equations to obtain the desired reaction: C(graphite) + 2 H2(g) + 1/2 O2(g) -> CH3OH(l).
Step 1: Reverse equation 2
C(graphite) + CO2(g) -> O2(g) ΔH°rxn = +393.5 kJ
Step 2: Multiply equation 3 by 2
2 H2(g) + O2(g) -> 2 H2O(l) ΔH°rxn = -571.6 kJ
Step 3: Add equation 1, equation 2 (reversed), and equation 3 (multiplied by 2)
C(graphite) + CO2(g) + 2 H2(g) + O2(g) + 2 H2(g) + O2(g) -> CO2(g) + 2 H2O(l) + 2 H2O(l) + CH3OH(l)
C(graphite) + 4 H2(g) + 3 O2(g) -> 2 CO2(g) + 4 H2O(l) + CH3OH(l)
Step 4: Calculate the overall enthalpy change of the reaction
ΔH°rxn(overall) = ΔH°rxn(1) + ΔH°rxn(2) + ΔH°rxn(3)
ΔH°rxn(overall) = -726.4 kJ + 393.5 kJ - 571.6 kJ
ΔH°rxn(overall) = -904.5 kJ
So, the enthalpy of formation of methanol (CH3OH) from its elements is -904.5 kJ.
REWRITE the three reactions and their ∆H's
(a) Reverse the 1st reaction and change the algebraic sign of its ∆H
(b) Leave the 2nd reaction and its ∆H unchanged.
(c) Double the 3rd reaction (multiply all its coefficients by 2). Double its ∆H, also.
(d) ADD the three rewritten reactions and do the cancellations of like terms. If you did it correctly, the simplified reaction should be:
C(graphite) + 2 H2(g) + 1/2 O2(g) -> CH3OH(l)
(e) COMBINE the ∆H's of the three rewritten reactions algebraically to get the ∆H of the formation reaction of CH3OH.
Your answer should be between -200kJ and -300kJ