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:**
\[ \text{CH}_4(g) + \text{NH}_3(g) \rightarrow 3\text{H}_2(g) + \text{HCN}(g) \]

We will use the provided thermodynamic data for three reactions and manipulate them to derive the desired reaction.

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

1. \[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g), \quad \Delta H = -91.8 \, \text{kJ} \]
2. \[ \text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g), \quad \Delta H = +74.9 \, \text{kJ} \]
3. \[ \text{H}_2(g) + 2\text{C}(s) + \text{N}_2(g) \rightarrow 2\text{HCN}(g), \quad \Delta H = +261 \, \text{kJ} \]

### Step 2: Reverse the second reaction to get CH4 on the left side.

Reversing the second reaction:
\[ \text{CH}_4(g) \rightarrow \text{C}(s) + 2\text{H}_2(g), \quad \Delta H = -74.9 \, \text{kJ} \]

### Step 3: Reverse the first reaction to get NH3 on the left side.

Reversing the first reaction:
\[ 2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g), \quad \Delta H = +91.8 \, \text{kJ} \]

### Step 4: Write the third reaction as it is.

The third reaction remains unchanged:
\[ \text{H}_2(g) + 2\text{C}(s) + \text{N}_2(g) \rightarrow 2\text{HCN}(g), \quad \Delta H = +261 \, \text{kJ} \]

### Step 5: Adjust the third reaction to obtain 1 mole of HCN.

To obtain 1 mole of HCN, we divide the entire third reaction by 2:
\[ \frac{1}{2} \text{H}_2(g) + \text{C}(s) + \frac{1}{2} \text{N}_2(g) \rightarrow \text{HCN}(g), \quad \Delta H = \frac{261}{2} = 130.5 \, \text{kJ} \]

### Step 6: Combine the modified reactions.

Now we will combine the modified reactions:

1. From the reversed second reaction:
\[ \text{CH}_4(g) \rightarrow \text{C}(s) + 2\text{H}_2(g), \quad \Delta H = -74.9 \, \text{kJ} \]

2. From the reversed first reaction:
\[ 2\text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g), \quad \Delta H = +91.8 \, \text{kJ} \]

3. From the adjusted third reaction:
\[ \frac{1}{2} \text{H}_2(g) + \text{C}(s) + \frac{1}{2} \text{N}_2(g) \rightarrow \text{HCN}(g), \quad \Delta H = +130.5 \, \text{kJ} \]

### Step 7: Calculate the total enthalpy change.

Now we sum the enthalpy changes:
- For the first reaction: \(-74.9 \, \text{kJ}\)
- For the second reaction (halved): \( \frac{91.8}{2} = 45.9 \, \text{kJ}\)
- For the third reaction (halved): \(130.5 \, \text{kJ}\)

Total:
\[ \Delta H = -74.9 + 45.9 + 130.5 = 101.5 \, \text{kJ} \]

### Final Answer:
The heat of reaction for the given reaction is:
\[ \Delta H = 101.5 \, \text{kJ/mol} \]

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