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

We will use the heats of formation (\(\Delta H_f^0\)) of the reactants and products. The heat of reaction can be calculated using the formula:

\[ \Delta H_{reaction} = \Delta H_f^0 \text{(products)} - \Delta H_f^0 \text{(reactants)} \]

### Step 1: Identify the heat of formation values

From the provided data:
1. For \( \text{HCN}(g) \):
\[ \Delta H_f^0 \text{(2 moles HCN)} = 261.0 \text{ kJ} \]
Therefore, for 1 mole:
\[ \Delta H_f^0 \text{(HCN)} = \frac{261.0}{2} = 130.5 \text{ kJ/mol} \]

2. For \( \text{NH}_3(g) \):
The reaction given is:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g), \quad \Delta H_r^0 = -91.8 \text{ kJ} \]
Therefore, for 1 mole:
\[ \Delta H_f^0 \text{(NH}_3) = \frac{-91.8}{2} = -45.9 \text{ kJ/mol} \]

3. For \( \text{CH}_4(g) \):
The reaction given is:
\[ \text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g), \quad \Delta H_r^0 = +74.9 \text{ kJ} \]
Therefore:
\[ \Delta H_f^0 \text{(CH}_4) = 74.9 \text{ kJ/mol} \]

### Step 2: Substitute the values into the heat of reaction formula

Now we can substitute the values into the heat of reaction equation:

\[ \Delta H_{reaction} = [\Delta H_f^0 \text{(3H}_2) + \Delta H_f^0 \text{(HCN)}] - [\Delta H_f^0 \text{(CH}_4) + \Delta H_f^0 \text{(NH}_3)] \]

Since the heat of formation for \( \text{H}_2(g) \) (the most stable form) is 0:

\[ \Delta H_{reaction} = [0 + 130.5] - [74.9 + (-45.9)] \]

### Step 3: Calculate the heat of reaction

Now, calculate:

\[ \Delta H_{reaction} = 130.5 - (74.9 - 45.9) \]

\[ \Delta H_{reaction} = 130.5 - (74.9 + 45.9) \]

\[ \Delta H_{reaction} = 130.5 - 29.0 \]

\[ \Delta H_{reaction} = 101.5 \text{ kJ} \]

### Final Answer

The heat of reaction for the given reaction is:
\[ \Delta H_{reaction} = 101.5 \text{ kJ} \]

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