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Using Hess's law we found:
1) By adding reaction 10.2 with the reverse 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 total 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 differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.
Hence, to get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
We need to add reaction 10.2 to the reverse of reaction 10.1
KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)
Canceling the KOH(s) from both sides, we get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:
[tex] \Delta H = \Delta H_{soln} + \Delta H_{neut} [/tex]
The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:
[tex] \Delta H = \Delta H_{neut} - \Delta H_{soln} [/tex]
Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.
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