`H_(2)(g) + Cl_(2) rarr 2HCl (g), Delta H = -44K.cals 2 Na(s)+2HCl (g)rarr 2NaCl(s)+H_(2)(g), Delta H = -152K.cal`
`Na(s)+(1)/(2)Cl_(2)(g) rarr NaCl(s), Delta H = ?`

Options

A. `+108K.cal`
B. `-196 K.cal`
C. `-98 K.cal`
D. `50 K.cal`

Explanation

To find the enthalpy change (ΔH) for the reaction:

\[ \text{Na(s)} + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{NaCl(s)} \]

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 of the reaction.

### Given Reactions:

1. \( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g), \quad \Delta H = -44 \, \text{kcal} \)
2. \( 2 \text{Na}(s) + 2 \text{HCl}(g) \rightarrow 2 \text{NaCl}(s) + \text{H}_2(g), \quad \Delta H = -152 \, \text{kcal} \)

### Step 1: Manipulate the Reactions

We need to manipulate these reactions to derive the desired reaction.

- From the first reaction, we have:
\[ \text{H}_2 + \text{Cl}_2 \rightarrow 2 \text{HCl} \]

- From the second reaction, we can divide the entire equation by 2 to get:
\[ \text{Na}(s) + \text{HCl}(g) \rightarrow \text{NaCl}(s) + \frac{1}{2} \text{H}_2(g) \]
The ΔH for this reaction will be:
\[ \Delta H = \frac{-152 \, \text{kcal}}{2} = -76 \, \text{kcal} \]

### Step 2: Add the Reactions

Now, we can add the modified second reaction to the first reaction:

1. \( \text{H}_2 + \text{Cl}_2 \rightarrow 2 \text{HCl} \) (ΔH = -44 kcal)
2. \( \text{Na}(s) + \text{HCl}(g) \rightarrow \text{NaCl}(s) + \frac{1}{2} \text{H}_2(g) \) (ΔH = -76 kcal)

To combine these, we need to express the first reaction in terms of HCl:

- Rearranging the first reaction gives:
\[ 2 \text{HCl} \rightarrow \text{H}_2 + \text{Cl}_2 \]
The ΔH for this reaction will be:
\[ \Delta H = +44 \, \text{kcal} \]

### Step 3: Combine the Reactions

Now we can add the two reactions:

1. \( 2 \text{HCl} \rightarrow \text{H}_2 + \text{Cl}_2 \) (ΔH = +44 kcal)
2. \( \text{Na}(s) + \text{HCl}(g) \rightarrow \text{NaCl}(s) + \frac{1}{2} \text{H}_2(g) \) (ΔH = -76 kcal)

Adding these gives:

\[ \text{Na}(s) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{NaCl}(s) \]

### Step 4: Calculate ΔH

Now, we can calculate the total ΔH for the desired reaction:

\[ \Delta H = +44 \, \text{kcal} - 76 \, \text{kcal} = -32 \, \text{kcal} \]

### Final Result

Thus, the enthalpy change for the reaction:

\[ \text{Na}(s) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{NaCl}(s) \]

is:

\[ \Delta H = -32 \, \text{kcal} \]