Which is the heat of reaction for the following reaction:
`CH_(4)(g) + NH_(3)(g) rarr 3H_(2)(g) +HCN(g)`
Use the following thermodynamic data in kJ/mol.
`N_(2)(g) + 3H_(2)(g) rarr 2NH_(3) (g) , Delta_(r )H^@ = -91.8`
`C(s) + 2H_(2)(g) rarr CH_(4)(g), Delta_(r )H^(@) = +74.9`
`H_(2)(g) + 2C(s) + N_(2)(g) rarr 2HCN(g), Delta_(r )H^(@) = 261.0`

Options

A. 299.3 kJ
B. 256.0 kJ
C. `-149 .5 kJ`
D. `101.5 kJ`

Explanation

To find the heat of reaction for the given reaction:

**Reaction:**
\[ CH_4(g) + NH_3(g) \rightarrow 3H_2(g) + HCN(g) \]

We will use the provided thermodynamic data:

1. \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g), \Delta_rH^\circ = -91.8 \, \text{kJ} \)
2. \( C(s) + 2H_2(g) \rightarrow CH_4(g), \Delta_rH^\circ = +74.9 \, \text{kJ} \)
3. \( H_2(g) + 2C(s) + N_2(g) \rightarrow 2HCN(g), \Delta_rH^\circ = 261.0 \, \text{kJ} \)

### Step 1: Calculate the heat of formation for each compound involved in the reaction.

**For \( NH_3(g) \):**
- The reaction for the formation of ammonia is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
- To find the heat of formation for 1 mole of \( NH_3 \), divide the reaction by 2:
\[ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g) \]
- The heat of formation of \( NH_3 \):
\[ \Delta_fH^\circ(NH_3) = \frac{-91.8}{2} = -45.9 \, \text{kJ/mol} \]

**For \( HCN(g) \):**
- The reaction for the formation of HCN is:
\[ H_2(g) + 2C(s) + N_2(g) \rightarrow 2HCN(g) \]
- To find the heat of formation for 1 mole of \( HCN \), divide the reaction by 2:
\[ \frac{1}{2}H_2(g) + C(s) + \frac{1}{2}N_2(g) \rightarrow HCN(g) \]
- The heat of formation of \( HCN \):
\[ \Delta_fH^\circ(HCN) = \frac{261.0}{2} = 130.5 \, \text{kJ/mol} \]

**For \( CH_4(g) \):**
- The reaction for the formation of methane is:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]
- The heat of formation of \( CH_4 \):
\[ \Delta_fH^\circ(CH_4) = +74.9 \, \text{kJ/mol} \]

### Step 2: Apply the heat of formation values to the reaction.

The heat of reaction can be calculated using the formula:
\[ \Delta_rH^\circ = \sum \Delta_fH^\circ \text{(products)} - \sum \Delta_fH^\circ \text{(reactants)} \]

**Products:**
- For \( 3H_2(g) \): The heat of formation is \( 0 \) (elements in standard state).
- For \( HCN(g) \): \( 130.5 \, \text{kJ/mol} \)

Total heat of formation for products:
\[ \Delta_fH^\circ(\text{products}) = 0 + 130.5 = 130.5 \, \text{kJ} \]

**Reactants:**
- For \( CH_4(g) \): \( 74.9 \, \text{kJ/mol} \)
- For \( NH_3(g) \): \( -45.9 \, \text{kJ/mol} \)

Total heat of formation for reactants:
\[ \Delta_fH^\circ(\text{reactants}) = 74.9 + (-45.9) = 29.0 \, \text{kJ} \]

### Step 3: Calculate the heat of the reaction.

Now, substituting the values into the equation:
\[ \Delta_rH^\circ = 130.5 - 29.0 = 101.5 \, \text{kJ} \]

### Final Answer:
The heat of reaction for the given reaction is:
\[ \Delta_rH^\circ = 101.5 \, \text{kJ} \]

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