The enthalpies of neutralization for ch3cooh with naoh and nh4oh with hcl are -50.6 and -51.4 Kj respectively. Calculate the enthalpy of neutralization of ch3cooh with nh4oh.

To find the enthalpy of neutralization of acetic acid (CH₃COOH) with ammonium hydroxide (NH₄OH), we can utilize Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction. In this case, we can break down the neutralization reactions into steps and use the given enthalpy values to find the desired enthalpy of neutralization.

Understanding the Reactions

First, let's summarize the reactions and their enthalpies:

• Neutralization of acetic acid with sodium hydroxide (NaOH):
• CH₃COOH + NaOH → CH₃COONa + H₂O (ΔH = -50.6 kJ)
• Neutralization of ammonium hydroxide with hydrochloric acid (HCl):
• NH₄OH + HCl → NH₄Cl + H₂O (ΔH = -51.4 kJ)

Setting Up the Calculation

We want to find the enthalpy change for the reaction:

• CH₃COOH + NH₄OH → CH₃COONH₄ + H₂O

To do this, we can rearrange the reactions we have. We can express the reaction of acetic acid with ammonium hydroxide in terms of the other two reactions.

Using Hess's Law

We can manipulate the reactions as follows:

1. Start with the reaction of acetic acid with NaOH:
• CH₃COOH + NaOH → CH₃COONa + H₂O (ΔH = -50.6 kJ)
2. Next, we need to consider the reaction of NH₄OH with HCl, but we will reverse it to get NH₄OH:
• NH₄Cl + H₂O → NH₄OH + HCl (ΔH = +51.4 kJ)

Now, we can combine these two reactions:

• CH₃COOH + NaOH + NH₄Cl + H₂O → CH₃COONa + H₂O + NH₄OH + HCl

Notice that the water molecules cancel out, and we are left with:

• CH₃COOH + NH₄OH → CH₃COONH₄ + HCl

Calculating the Enthalpy Change

Now we can calculate the overall enthalpy change for the desired reaction:

ΔH = ΔH (CH₃COOH + NaOH) + ΔH (NH₄Cl + H₂O → NH₄OH + HCl)

Substituting the values:

ΔH = (-50.6 kJ) + (+51.4 kJ) = 0.8 kJ

Final Result

Thus, the enthalpy of neutralization of acetic acid with ammonium hydroxide is approximately +0.8 kJ. This indicates that the reaction is slightly endothermic, meaning it absorbs a small amount of heat from the surroundings.