In accordance with Hess's Law, manipulate the following equations and their change in Hrxn to calculate change in Hrxn's for the reaction

HClO4 + NH3 �N NH4ClO4...

HClO4(aq)+ LiOH(aq)�¨ LiClO4(aq)+H2O(l)
CH3COOH(aq)+ NH3(aq) �¨ NH4CH3COO(aq)
CH3COOH(aq)+LiOH(aq)�¨ LiCH3COO(aq)+H2O(l)

To calculate the change in enthalpy (ΔHrxn) for the reaction HClO4 + NH3 → NH4ClO4, we can use Hess's Law, which states that the overall change in enthalpy for a chemical reaction is equal to the sum of the enthalpy changes for the individual steps of the reaction.

Let's assign ΔH values to each of the given equations and manipulate them to obtain the desired reaction:

1. HClO4(aq) + LiOH(aq) → LiClO4(aq) + H2O(l) (ΔH1)

2. CH3COOH(aq) + NH3(aq) → NH4CH3COO(aq) (ΔH2)

3. CH3COOH(aq) + LiOH(aq) → LiCH3COO(aq) + H2O(l) (ΔH3)

Now, we can manipulate these equations to match the desired reaction equation:

1. HClO4(aq) + LiOH(aq) → LiClO4(aq) + H2O(l) (ΔH1)
-1 × HClO4(aq) + 1 × NH4ClO4(aq) → 1 × NH3(aq) + 1 × H2O(l)

2. CH3COOH(aq) + NH3(aq) → NH4CH3COO(aq) (ΔH2)
-1 × HClO4(aq) + 1 × NH4ClO4(aq) → 1 × NH3(aq) + 1 × H2O(l)

3. CH3COOH(aq) + LiOH(aq) → LiCH3COO(aq) + H2O(l) (ΔH3)
-1 × NH3(aq) + 1 × NH4ClO4(aq) → 1 × NH3(aq) + 1 × H2O(l)

Now, let's add the manipulated equations to get the overall desired reaction:

-1 × HClO4(aq) + 1 × NH4ClO4(aq) + (-1 × NH3(aq)) + 1 × NH4ClO4(aq) →
1 × NH3(aq) + 1 × H2O(l) + 1 × NH3(aq) + 1 × H2O(l)

Simplifying the equation:

HClO4(aq) + 2 × NH4ClO4(aq) → 2 × NH3(aq) + 2 × H2O(l)

Next, we calculate the corresponding change in enthalpy (ΔH) for each reaction:

ΔHtotal = ΔH1 + ΔH2 + ΔH3

Finally, plug in the known values of each ΔH:

ΔHtotal = ΔH1 + ΔH2 + ΔH3
ΔHtotal = (ΔH1 of HClO4 + LiOH) + (ΔH2 of CH3COOH + NH3) + (ΔH3 of CH3COOH + LiOH)

Note: You will need to have the ΔH values provided for each of the reactions mentioned above in order to calculate ΔHtotal.

To calculate the change in enthalpy of the reaction HClO4 + NH3 -> NH4ClO4, we can use Hess's Law by manipulating the given equations and their corresponding enthalpy changes.

Step 1: Reverse the second equation:
NH4CH3COO(aq) -> CH3COOH(aq) + NH3(aq)

Step 2: Multiply the second equation by -1:
-1 * (NH4CH3COO(aq) -> CH3COOH(aq) + NH3(aq))

This gives you:
-NH4CH3COO(aq) + CH3COOH(aq) + NH3(aq)

Step 3: Add the three equations together, making sure to multiply the equations and their enthalpy changes as needed to balance the coefficients:

HClO4(aq)+ LiOH(aq) -> LiClO4(aq) + H2O(l)
ΔH1 = ? (given)

-1 * (NH4CH3COO(aq) -> CH3COOH(aq) + NH3(aq))
ΔH2 = -ΔH2 (given)

CH3COOH(aq) + LiOH(aq) -> LiCH3COO(aq) + H2O(l)
ΔH3 = ? (given)

HClO4(aq) + NH3(aq) -> NH4ClO4(aq)
ΔHrxn = ΔH1 + ΔH2 + ΔH3

Step 4: Add the enthalpy changes:
ΔHrxn = ΔH1 - ΔH2 + ΔH3

Remember to consider the sign of the enthalpy change when adding or subtracting. The enthalpy change of the overall reaction can be determined by manipulating and adding the given equations using Hess's Law.