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  • Find the enthalpy of ionization of acetic acid Given : (i) <img alt="HCl (aq) + NaOH(aq) \rightarrow H_2O(l) + NaCl \;\;\;\; \Delta H_1 = -52.7 kJ" src="https://learn.careers360.com/lat

Find the enthalpy of ionization of acetic acid
Given : (i) HCl (aq) + NaOH(aq) \rightarrow H_2O(l) + NaCl \;\;\;\; \Delta H_1 = -52.7 kJ
             (ii) \\CH_3COOH (aq) + NaOH(aq) \rightarrow CH_3COONa + H_2O (l)\;\;\;\; \Delta H_2 = -54.4 kJ

Option: 1

+1.7 kJ


Option: 2

-0.7 kJ


Option: 3

-1.7 kJ


Option: 4

Can’t be calculated


Answers (1)

best_answer

We have to find the enthalpy of ionization of acetic acid, which means we have to find the enthalpy of reaction :

CH_3COOH(aq) \rightarrow CH_3COO^- + H^+ \;\;\;\; \Delta H_3 = ?

For solving this question, let's eliminate the spectator ions, namely Cl and Na ions.
Now the reactions will be written as:

(i) H^+ + OH^- \rightarrow H_2O(l) \;\;\;\; \Delta H_1 = -52.7kJ

(ii) CH_3COOH(aq) + OH^-\rightarrow CH_3COO^- + H_2O(l) \;\;\;\; \Delta H_2 = -54.4 kJ

Subtracting reaction (i) from reaction (ii) will give us the desired reaction
 

\Delta H_3 = \Delta H_2 - \Delta H_1 \\\Rightarrow\Delta H_3 = -54.4 - (-52.7) \\\Rightarrow \Delta H_3 = -1.7

Hence, option number (3) is correct
 

Posted by

Irshad Anwar

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