The enthalpies of neutralisation for CH3COOH with NaOH and NH4OH with

1.	The enthalpies of neutralisation for CH3COOH with NaOH and NH4OH with HCl are -50.6 and -51.4 kJ eq-1 respectively . Calculate the enthalpy of neutralisation of CH3COOH with NH4OH ?
Answer» Answer: The enthalpy of neutralization will be -44.38 kJ/eq. Explanation: We are GIVEN two neutralization REACTIONS: $CH_3COOH+NaOH\rightarrow CH_3COO^-+Na^++H_2O$ ; $\Delta H_{neutralization}=-50.6kJ/eq$ ...(1) $NH_4OH+HCl\rightarrow NH_4^++Cl^-+H_2O$ ; $\Delta H_{neutralization}=-51.4kJ/eq$ ...(2) Equation 1 is the reaction of strong base and WEAK acid & equation 2 is the reaction of strong acid and weak base. To form the equation in which $CH_3COOH$ reacts with $NH_4OH$ , we write another equation, which will be SUBTRACTED from equation 1 and equation 2. Equation for the reaction of strong acid and strong base: $NaOH+HCl\rightarrow Na^++Cl^-+H_2O$ ; $\Delta H_{neutralization}=-57.62kJ/eq$ ...(3) Subtracting equation 3 from 1 and 2, we get $Eq.1+Eq.2-Eq.3$ $NH_4OH+CH_3COOH\rightarrow CH_3COONH_4+H_2O$ ...(4) So, the calculations for enthalpies will be: $[-50.6-51.4-(-57.62)]kJ/eq=-44.38kJ/eq$ So, the enthalpy of neutralization of equation 4, will be -44.38 kJ/eq.

The enthalpies of neutralisation for CH3COOH with NaOH and NH4OH with HCl are -50.6 and -51.4 kJ eq-1 respectively . Calculate the enthalpy of neutralisation of CH3COOH with NH4OH ?

Answer»

Answer: The enthalpy of neutralization will be -44.38 kJ/eq.

Explanation: We are GIVEN two neutralization REACTIONS:

$CH_3COOH+NaOH\rightarrow CH_3COO^-+Na^++H_2O$ ; $\Delta H_{neutralization}=-50.6kJ/eq$ ...(1)

$NH_4OH+HCl\rightarrow NH_4^++Cl^-+H_2O$ ; $\Delta H_{neutralization}=-51.4kJ/eq$ ...(2)

Equation 1 is the reaction of strong base and WEAK acid & equation 2 is the reaction of strong acid and weak base.

To form the equation in which $CH_3COOH$ reacts with $NH_4OH$ , we write another equation, which will be SUBTRACTED from equation 1 and equation 2.

Equation for the reaction of strong acid and strong base:

$NaOH+HCl\rightarrow Na^++Cl^-+H_2O$ ; $\Delta H_{neutralization}=-57.62kJ/eq$ ...(3)

Subtracting equation 3 from 1 and 2, we get

$Eq.1+Eq.2-Eq.3$

$NH_4OH+CH_3COOH\rightarrow CH_3COONH_4+H_2O$ ...(4)

So, the calculations for enthalpies will be:

$[-50.6-51.4-(-57.62)]kJ/eq=-44.38kJ/eq$

So, the enthalpy of neutralization of equation 4, will be -44.38 kJ/eq.