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101_1433-crop.jpghi if you could help me in this one it would be really helpful.
Original post by Farhan7
101_1433-crop.jpghi if you could help me in this one it would be really helpful.


First of all, rearrange the second equation so that the REACTANTS are the PRODUCTS, and vice versa:

KCl (s) + H2O (l) => HCl (g) + KOH (s). By switching the equation around, ΔH becomes POSITIVE (+) 204 kJ.

Now, multiply each component in this equation by TWO, as well as ΔH:

2 KCl (S) + 2 H2O (l) => 2 HCl (g) + 2 KOH (s); ΔH = 204 kJ x 2 = + 408 kJ

The first equation looks like this:

H2SO4(l) + 2KOH(s) → 2K2SO4(s) + 2(H2O)(l)
ΔH = -342 kJ

and the second looks like this:

2 KCl (s) + 2 H2O (l) => 2 HCl (g) + 2 KOH (s)
ΔH = + 408 kJ

We need to make sure that these two reactions have the same products and reactants as the original equation:

H2SO4(l) + 2 KCl(s) → 2HCl(g) + K2SO4(s)

Thus, this is why I rearranged the second equation, so that the reactant KCl would be on the LEFT side, and the product HCl would be on the RIGHT side! Now that we have like terms (2 KOH and 2 H2O), we can cancel these out, since they are at OPPOSITE sides of the equation. You now should have:

H2SO4(l) → 2K2SO4(s)
ΔH = -342 kJ

and

2 KCl (S) => 2 HCl (g)
ΔH = + 408 kJ

According to your ORIGINAL equation, you have the products and reactants arranged correctly, so now, just add the products and reactants together, as well as ΔH:

H2SO4(l) + 2 KCl(s) → 2HCl(g) + K2SO4(s); ΔH = -342 kJ + 408 kJ = 66 kJ

This is known as HESS'S LAW, where you add up the reactants on one side and products on the other, as well as the ΔH's for BOTH reactions!

Hope this helps!
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Farhan7
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thanks mate. It helps and i understood everything you said. But now i am having difficulty in drawing the HESS cycle! Could you please help me out on that?
Have a read:
http://www.chemguide.co.uk/physical/energetics/sums.html
Original post by Farhan7
thanks mate. It helps and i understood everything you said. But now i am having difficulty in drawing the HESS cycle! Could you please help me out on that?


There is a chemisrty whatsapp group in which we go through a load of questions. If you want to be part reply to this message or private message me


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