Chemistry - Thermochemistry Question with Solution | TestHub
Consider the equations:
CH3CH(OH)CHClCOOH + 2KOH → P + KCl + 2H2O; ΔH = -14.7 kcal
CH3CH(OH)CHClCOOK + KOH → P + KCl + H2O; ΔH = -0.522 kcal
where P is the epoxide CH3CH(O)CHCOOK shown below.
ΔH for the neutralization of HCl and NaOH is -13.75 kcal/eq. The enthalpy of ionization of CH3CH(OH)CHClCOOH is
Options:
Answer:
Solution:
Subtract the second equation from the first: CH₃CH(OH)CHClCOOH + KOH → CH₃CH(OH)CHClCOOK + H₂O.
The change in enthalpy, ΔH, for this reaction is calculated as:
ΔH = -14.7 - (-0.522) = -14.178 kcal.
This reaction represents the neutralization of a weak acid by a strong base. The enthalpy of neutralization (ΔHneut) can be expressed as the sum of the enthalpy of ionization (ΔHion) and the enthalpy change for the reaction of H⁺ and OH⁻ ions, which is -13.75 kcal:
ΔHneut = ΔHion + (-13.75)
Therefore, we can calculate the enthalpy of ionization:
ΔHion = -14.178 + 13.75 = -0.428 ≈ -0.429 kcal/mol.
This corresponds to option (d).
