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RioNii RioNii
wrote...
Posts: 163
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A month ago
Given the following two reactions:

(1)Pb(s) + Cl2(g)PbCl2(s)       H(1) = -359.4 kJ

(2)Hg(l) + Cl2(g)HgCl2(s)       H(2) = -224.3 kJ

Calculate the enthalpy change for the following reaction:

(3) Pb(s) + HgCl2(s) PbCl2(s) + Hg(l)     
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bio_manbio_man
wrote...
Educator
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Posts: 31070
A month ago
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Hi RioNii, change equation (2) to:

HgCl2(s) <-> (2)Hg(l) + Cl2(g)       ΔH = +224.3 kJ

Now add them:

ΔH = -359.4 kJ + 224.3 kJ = -135.1 KJ
1

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wrote...
A month ago
Thank you sir!
wrote...
Educator
A month ago
No problem. I have marked it solved Slight Smile
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