Confused about delta H for spontaneous reactions (1 Viewer)

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Ok i have over thought this too much, what is the delta H of a spontaneous reaction?

I thought -ve delta H would be spontaneous, but then i realised combustion has a -ve delta H, and you have to put some energy in for combustion (eg. lighting it). So therefore +ve delta H is spontaneous? This still doesn't make sense, that would be gaining energy.

pls help :blink2:
 
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I dont know what your on about -

delta h = negative when exothermic reaction.
delta h = positive when endothermic reaction.
 
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Delta H has nothing to do with a reaction being spontaneous or non-spontaneous. You're confusing it with electrical potentials (if that's the right term), which are found on the data sheet in the HSC exam.
 

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Delta H has nothing to do with a reaction being spontaneous or non-spontaneous. You're confusing it with electrical potentials (if that's the right term), which are found on the data sheet in the HSC exam.
hmm ok. Do we have to know how to find if a reaction is spontaneous or not? Or is that outside syllabus?
 

shved

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hmm ok. Do we have to know how to find if a reaction is spontaneous or not? Or is that outside syllabus?
Yes u have to know how to determine if a reaction would be spontaneous at standard conditions.
It was asked in last year's paper
 

Kurosaki

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Ok i have over thought this too much, what is the delta H of a spontaneous reaction?

I thought -ve delta H would be spontaneous, but then i realised combustion has a -ve delta H, and you have to put some energy in for combustion (eg. lighting it). So therefore +ve delta H is spontaneous? This still doesn't make sense, that would be gaining energy.

pls help :blink2:
What you might want to look into is entropy and Gibbs free energy to see how enthalpy factors into the spontaneity of a process.
 
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someth1ng

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Enthalpy change has nothing to do with spontaneity; what you might want to look into is entropy and Gibbs free energy.
What you just wrote is contradictory.

If you use this formula: ΔG=ΔH-TΔS
If ΔG is negative, it is spontaneous where ΔS is the change in entropy and ΔH is the change in enthalpy.
 

Kurosaki

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What you just wrote is contradictory.

If you use this formula: ΔG=ΔH-TΔS
If ΔG is negative, it is spontaneous where ΔS is the change in entropy and ΔH is the change in enthalpy.
My bad, thanks for the correction =). I really should brush up on first year chem...thanks for reminding me about the formula.
 
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