The two reactions illustrated in the diagrams below often occur when a foreign substance enters the body.
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Using Hess's law we found:
1) By <em>adding </em>reaction 10.2 with the <em>reverse </em>of reaction 10.1 we get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)
2) The ΔHsoln must be subtracted from ΔHneut to get the <em>total </em>change in enthalpy (ΔH).
The reactions of dissolution (10.1) and neutralization (10.2) are:
KOH(s) → KOH(aq) ΔHsoln (10.1)
KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)
1) According to Hess's law, the total change in enthalpy of a reaction resulting from <u>differents changes</u> in various <em>reactions </em>can be calculated as the <u>sum</u> of all the <em>enthalpies</em> of all those <em>reactions</em>.
Hence, to get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
We need to <em>add </em>reaction 10.2 to the <u>reverse</u> of reaction 10.1
KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)
<u>Canceling</u> the KOH(s) from both sides, we get <em>reaction 10.3</em>:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
2) The change in enthalpy for <em>reaction 10.3</em> can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:
The enthalpy of <em>reaction 10.1 </em>(ΔHsoln) changed its sign when we reversed reaction 10.1, so:
Therefore, the ΔHsoln must be <u>subtracted</u> from ΔHneut to get the total change in enthalpy ΔH.
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