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 in the reaction:

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

we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We will break down the reaction into several steps for which we have the enthalpy changes.

### Step-by-Step Solution:

1. **Convert NH₃(g) to NH₃(aq)**:
- The enthalpy change for this step is given as:
\[ \Delta H_1 = -8.4 \, \text{kcal} \]
This means that when gaseous ammonia is converted to aqueous ammonia, it releases 8.4 kcal of heat.

2. **Convert HCl(g) to HCl(aq)**:
- The enthalpy change for this step is given as:
\[ \Delta H_2 = -17.3 \, \text{kcal} \]
This means that when gaseous HCl is converted to aqueous HCl, it releases 17.3 kcal of heat.

3. **React NH₃(aq) with HCl(aq) to form NH₄Cl(aq)**:
- The enthalpy change for this reaction is given as:
\[ \Delta H_3 = -12.5 \, \text{kcal} \]
This means that when aqueous ammonia reacts with aqueous HCl to form aqueous ammonium chloride, it releases 12.5 kcal of heat.

4. **Convert NH₄Cl(s) to NH₄Cl(aq)**:
- The enthalpy change for this step is given as:
\[ \Delta H_4 = +3.9 \, \text{kcal} \]
This means that when solid ammonium chloride is dissolved in water to form aqueous ammonium chloride, it absorbs 3.9 kcal of heat.

5. **Reverse the last step**:
- To find the enthalpy change for the conversion of NH₄Cl(aq) to NH₄Cl(s), we reverse the sign:
\[ \Delta H_4' = -3.9 \, \text{kcal} \]

6. **Add all the enthalpy changes**:
- Now we can sum all the enthalpy changes to find the total heat change for the overall reaction:
\[ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4' \]
\[ \Delta H = (-8.4) + (-17.3) + (-12.5) + (-3.9) \]
\[ \Delta H = -42.1 \, \text{kcal} \]

### Final Answer:
The heat change in the reaction is:
\[ \Delta H = -42.1 \, \text{kcal} \]