ThermochemistryHard

Question

Consider the following equations.

CH3CH(OH)CHClCOOH + 2KOH → + KCl + 2H2O; ΔH = −14.7 kcal

CH3CH(OH)CHClCOOK + KOH → + KCl + H2O; ΔH = −2.7 kcal

ΔH for the neutralization of HCl and NaOH is −13.75 kcal/eq.

The enthalpy of ionization of CH3CH(OH)CHClCOOH is

Options

A.+1.75 kcal/mol
B.+3.5 kcal/mol
C.+0.875 kcal/mol
D.−1.75 kcal/mol

Solution

$\Delta H_{\text{required}} = \left\lbrack ( - 14.7) - ( - 2.7) \right\rbrack - \lbrack - 13.75\rbrack = + 1.75\text{ kcal/mol}$

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