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Kaplan Subject Test#1---G CHEM

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tsitneD

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I need help with the fourth question on gchem subject test 1

everything in KJ/mol.

#4) refer to this

H20 (g) ---> H20(l) delta H = -44

C(s) + O2 (g) ----> CO2(g) delta H = -394

H2(g) + 1/2 O2 (g) ----> H2O (l) delta H = -286

C2H5OH (l) + 3O2 (g) ------> 2CO2 (g) + 3 H2O (l) delta H= -1367

Question) calculate enthalpy change for the reaction?

2 C(s) + 2H2 (g) + H2O (l) -----> C2H5OH (l)

A) -226 B) +7 C) +109 D) +344 E) +687


I apply hess's law and i dont take H2 a diatomic gas into account but still with all the correct reversing and all i dont get the right answer which is B
 
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faerielynx

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i need help with the fourth question on gchem subject test 1

everything in kj/mol.

#4) refer to this

h20 (g) ---> h20(l) delta h = -44

c(s) + o2 (g) ----> co2(g) delta h = -394

h2(g) + 1/2 o2 (g) ----> h2o (l) delta h = -286

c2h5oh (l) + 3o2 (g) ------> 2co2 (g) + 3 h2o (l) delta h= -1367

question) calculate enthalpy change for the reaction?

2 c(s) + 2h2 (g) + h2o (l) -----> c2h5oh (l)

a) -226 b) +7 c) +109 d) +344 e) +687


i apply hess's law and i dont take h2 a diatomic gas into account but still with all the correct reversing and all i dont get the right answer which is b

To calculate thermodynamic values for a reaction or process we can apply Hess's Law, which states that the total change
in a thermodynamic state function (such as enthalpy here) can be found as the sum of the changes for a series of steps with the
same net result. Such a series of steps and net result for this particular example, with the _H value for each step, are shown
below:
2 C(​
s) + 2 O2(g) → 2 CO2(g) 2(–394)
H
2(g) + O2(g) → 2 H2O(l) 2(–286)
2 CO
2(g) + 3 H2O(l) → C2H5OH(l)+ 3 O2(g) –(–1367)
________________________________________________________________________________
2 C(
s) + 2 H2(g) + H2O(l) → C2H5OH(l) ΔH= 2(–394) + 2(–286) – (–1367)

Adding the values to arrive at the net
ΔH, the result is +7 kJ, choice B.
 
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tsitneD

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i thought according hess law for this i add the enthalpies?

and i was getting it wrong because i switched H2O enthalpy sign. And correct me we never take diatomic gasses into account right?
 

tsitneD

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what i dont get it is that when H2O (l) are on different side so wouldn't we switch or reverse the enthalpy?

so it should be

2(-394) + 2(286) - (-1367)

but i dont know how come it is 2(-286).
 

UCfan

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what i dont get it is that when H2O (l) are on different side so wouldn't we switch or reverse the enthalpy?

so it should be

2(-394) + 2(286) - (-1367)

but i dont know how come it is 2(-286).

faerielynx's solution is right. You need to multiply second and third reaction by two and reverse the last reaction. If you look at the overall reaction 3 moles of H2O (l) in the reactant side and 2 moles of H2O (l) in the product side. If you reverse the third reaction as you suggested, you cannot get the overall reaction. So the answer needs to be 7 kJ/mol.
2 C(s) + 2 O2(g) → 2 CO2(g) 2(–394)
2 H
2(g) + O2(g) → 2 H2O(l) 2(–286)
2 CO
2(g) + 3 H2O(l) → C2H5OH(l)+ 3 O2(g) –(–1367)
__________________________________________________ ______________________________
2 C(
s) + 2 H2(g) + H2O(l) → C2H5OH(l) ΔH= 2(–394) + 2(–286) – (–1367)
 

tsitneD

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faerielynx's solution is right. You need to multiply second and third reaction by two and reverse the last reaction. If you look at the overall reaction 3 moles of H2O (l) in the reactant side and 2 moles of H2O (l) in the product side. If you reverse the third reaction as you suggested, you cannot get the overall reaction. So the answer needs to be 7 kJ/mol.
2 C(s) + 2 O2(g) → 2 CO2(g) 2(–394)
2 H
2(g) + O2(g) → 2 H2O(l) 2(–286)
2 CO
2(g) + 3 H2O(l) → C2H5OH(l)+ 3 O2(g) –(–1367)
__________________________________________________ ______________________________
2 C(
s) + 2 H2(g) + H2O(l) → C2H5OH(l) ΔH= 2(–394) + 2(–286) – (–1367)


thanks.
so we dont switch the sgin because overall we have ended up with one mole H2O on reactant side so sign remains the same because in our reaction of interest H2O is on reactant side. Correct? let me know if im still wrong.

sorry i get everything about right except these three step reactions for enthalpy.
 
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