The enthalpy of neutralization of monobasic acid in decinormal solution by a dilute solution of KOH is -12,200 cal. The enthalpy of neutralization of strong acid by strong base is -13,700 cal. Assumin

Question

The enthalpy of neutralization of monobasic acid in decinormal solution by a dilute solution of KOH is -12,200 cal. The enthalpy of neutralization of strong acid by strong base is -13,700 cal. Assuming that the acid is 25% dissociated in decinormal solution, the enthalpy of dissociation of the acid (in kcal/mole) is

TestHub · Accepted Answer

Only the 75% undissociated acid absorbs the dissociation energy. The observed enthalpy change (Δ H_obs) is given by:

Δ H_obs = -13700 + 0.75 Δ H_diss

Given that Δ H_obs = -12200, we can substitute this value into the equation:

-12200 = -13700 + 0.75 Δ H_diss

Now, solve for 0.75 Δ H_diss:

0.75 Δ H_diss = -12200 + 13700

0.75 Δ H_diss = 1500 cal

Finally, calculate Δ H_diss:

Δ H_diss = 1500/0.75 cal

Δ H_diss = 2000 cal/mol = 2 kcal/mol

The answer is 2.

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