What is the heat of reaction for the following reaction?
`CH_(4)(g)+NH_(3)(g) to 3H_(2) (g)+HCN(g)`
Use the following thermodynamic data in kJ/mole
`N_(2)(G)+3H_(2)(g) to 2NH_(3) (g), triangle_(r)H^(@)=-91.8`
`C(s)+2H_(2)(g) to CH_(3) (g), triangle_(r)H^(@)=+74.9`
`H_(2)(g)+2C(s)+N_(2)(g) to 2HCN(g), triangle_(r)H^(@)=+261`

Options

A. 299.3kJ
B. 256kJ
C. `-149.5kJ`
D. `101.5kJ`

Answer

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 Hess's Law and the provided thermodynamic data.

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

1. **Reaction 1:**
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
\[ \Delta H_r = -91.8 \, \text{kJ} \]

2. **Reaction 2:**
\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]
\[ \Delta H_r = +74.9 \, \text{kJ} \]

3. **Reaction 3:**
\[ H_2(g) + 2C(s) + N_2(g) \rightarrow 2HCN(g) \]
\[ \Delta H_r = +261 \, \text{kJ} \]

### Step 2: Manipulate the reactions to derive the target reaction.

To find the heat of reaction for the target reaction, we need to express it in terms of the given reactions.

1. **Reverse Reaction 2** (to produce \( CH_4 \)):
\[ CH_4(g) \rightarrow C(s) + 2H_2(g) \]
\[ \Delta H_r = -74.9 \, \text{kJ} \]

2. **Use half of Reaction 1** (to produce \( NH_3 \)):
\[ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g) \]
\[ \Delta H_r = \frac{-91.8}{2} = -45.9 \, \text{kJ} \]

3. **Use half of Reaction 3** (to produce \( HCN \)):
\[ H_2(g) + C(s) \rightarrow HCN(g) \]
\[ \Delta H_r = \frac{261}{2} = +130.5 \, \text{kJ} \]

### Step 3: Combine the manipulated reactions.

Now, we can add these modified reactions together:

1. From Reaction 2 (reversed):
\[ CH_4(g) \rightarrow C(s) + 2H_2(g) \]
\[ \Delta H_r = -74.9 \, \text{kJ} \]

2. From Reaction 1 (half):
\[ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g) \]
\[ \Delta H_r = -45.9 \, \text{kJ} \]

3. From Reaction 3 (half):
\[ H_2(g) + C(s) \rightarrow HCN(g) \]
\[ \Delta H_r = +130.5 \, \text{kJ} \]

### Step 4: Add the enthalpy changes.

Now we can sum the enthalpy changes:
\[ \Delta H_{reaction} = (-74.9) + (-45.9) + (+130.5) \]

Calculating this gives:
\[ \Delta H_{reaction} = -74.9 - 45.9 + 130.5 = 9.7 \, \text{kJ} \]

### Final Answer:
The heat of reaction for the given reaction is:
\[ \Delta H_{reaction} = 9.7 \, \text{kJ} \]

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