Find the heat change in the reaction:
`NH_3(g)+HCI(g) to NH_4CI(s)`
from the following data
`NH_(3)(g)+aq to NH_(3)(aq) triangleH=-8.4KCal`
`HCl(g)+aq to HCl (aq) triangleH=-17.3KCal`
`NH_(3)(aq)+HCl(aq) to NH_(4)Cl(aq) triangleH=-12.5KCal`
`NH_(4)Cl(s) +aq to NH_(4)Cl(aq) triangleH=+3.9Kcal`

Options

A. `-42.1`
B. `-34.3`
C. `+34.3`
D. `42.1`

Answer

To find the heat change for the reaction:

\[ \text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s) \]

we will use the provided thermodynamic data from the four reactions given. We need to manipulate these reactions to arrive at the desired equation.

### Step 1: Write down the given reactions and their enthalpy changes.

1. \( \text{NH}_3(g) + \text{H}_2O \rightarrow \text{NH}_3(aq) \) \(\Delta H_1 = -8.4 \, \text{kcal}\)
2. \( \text{HCl}(g) + \text{H}_2O \rightarrow \text{HCl}(aq) \) \(\Delta H_2 = -17.3 \, \text{kcal}\)
3. \( \text{NH}_3(aq) + \text{HCl}(aq) \rightarrow \text{NH}_4\text{Cl}(aq) \) \(\Delta H_3 = -12.5 \, \text{kcal}\)
4. \( \text{NH}_4\text{Cl}(aq) + \text{H}_2O \rightarrow \text{NH}_4\text{Cl}(s) \) \(\Delta H_4 = +3.9 \, \text{kcal}\)

### Step 2: Identify the necessary transformations.

- We need \( \text{NH}_3(g) \) and \( \text{HCl}(g) \) on the left side.
- We need \( \text{NH}_4\text{Cl}(s) \) on the right side.

### Step 3: Write the reactions in the correct order.

1. Keep the first reaction as it is:
\[ \text{NH}_3(g) + \text{H}_2O \rightarrow \text{NH}_3(aq) \quad \Delta H_1 = -8.4 \, \text{kcal} \]

2. Keep the second reaction as it is:
\[ \text{HCl}(g) + \text{H}_2O \rightarrow \text{HCl}(aq) \quad \Delta H_2 = -17.3 \, \text{kcal} \]

3. Keep the third reaction as it is:
\[ \text{NH}_3(aq) + \text{HCl}(aq) \rightarrow \text{NH}_4\text{Cl}(aq) \quad \Delta H_3 = -12.5 \, \text{kcal} \]

4. Reverse the fourth reaction:
\[ \text{NH}_4\text{Cl}(aq) \rightarrow \text{NH}_4\text{Cl}(s) + \text{H}_2O \quad \Delta H_4 = -3.9 \, \text{kcal} \]

### Step 4: Combine the reactions.

Now, we add these reactions together:

- The \( \text{NH}_3(aq) \) and \( \text{HCl}(aq) \) will cancel out.
- The \( \text{NH}_4\text{Cl}(aq) \) will also cancel out.

This gives us:

\[ \text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s) \]

### Step 5: Calculate the overall enthalpy change.

Now we sum the enthalpy changes:

\[ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 \]
\[ \Delta H = (-8.4) + (-17.3) + (-12.5) + (-3.9) \]
\[ \Delta H = -42.1 \, \text{kcal} \]

### Final Answer:

The heat change for the reaction \( \text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s) \) is:

\[ \Delta H = -42.1 \, \text{kcal} \]

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