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Topic: Standard Enthalpies Question  (Read 31575 times)

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Offline Gweedo8

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Standard Enthalpies Question
« on: June 25, 2009, 02:24:17 PM »
Hydrogen sulfide gas is a poisonous gas with the odor of rotten eggs. It occurs in natural gas and is produced during the decay of organic matter, which contains sulfur. The gas burns in oxygen as follow:
2H2S(g)+3O2(g)->2H2O(l)+2SO2(g)
Calculate the standard enthalpy change for this reaction using standard enthalpies of formation.

Okay so here is my work after using the standard enthalpy chart to determine the ∆Hf˚
2H2S(g)=-20.5 kJ/mol
3O2(g)=0 kJ/mol
2H2O(l)=285.8 kJ/mol
2SO2(g)-296.8 kJ/mol
I put them into the reaction ΔH° = Σ(ΔHf°products) - Σ(ΔHf°reactants) which looked like this
ΔH° =(2molH2O×285.8/1)+(2molSO2×(-296.8)/1)- (2molH2S×-20.5/1)+(3molO2×0/1) =19kJ/mol
But the answer is supposed to be -1124.2kJ, can anyone tell me what I did wrong here?
 

Offline sjb

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Re: Standard Enthalpies Question
« Reply #1 on: June 25, 2009, 02:41:23 PM »
Hydrogen sulfide gas is a poisonous gas with the odor of rotten eggs. It occurs in natural gas and is produced during the decay of organic matter, which contains sulfur. The gas burns in oxygen as follow:
2H2S(g)+3O2(g)->2H2O(l)+2SO2(g)
Calculate the standard enthalpy change for this reaction using standard enthalpies of formation.

Okay so here is my work after using the standard enthalpy chart to determine the ∆Hf˚
2H2S(g)=-20.5 kJ/mol
3O2(g)=0 kJ/mol
2H2O(l)=285.8 kJ/mol
2SO2(g)-296.8 kJ/mol
I put them into the reaction ΔH° = Σ(ΔHf°products) - Σ(ΔHf°reactants) which looked like this
ΔH° =(2mol H2O×285.8/1)+(2mol SO2×(-296.8 )/1)- (2mol H2S×-20.5/1)+(3mol O2×0/1) =19kJ/mol
But the answer is supposed to be -1124.2kJ, can anyone tell me what I did wrong here?
 

Right method - are you sure your data is correct?

Offline plankk

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Re: Standard Enthalpies Question
« Reply #2 on: June 25, 2009, 03:03:44 PM »
I'm sure that the enthalpy of formation for H2O(l) is exotermic. Then you will gain the correct answer.

Offline Bryby

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Re: Standard Enthalpies Question
« Reply #3 on: June 27, 2009, 11:03:15 AM »
I see what you did wrong. Remember that you must first add up all the products and all the reactants. Then you must subtract. Here is different way to write it. ΔHf = Σ((-571.6)+(-593.6)) - Σ((-41.2) + (0))
You didnt sum the left side. You must have taken -593.6 and subrtracted -41.2... You should have added up the products first. Hope this helps.

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