Heat of combustion of C(s), `H_2(g)` and `CH_4 (g)` respectively are -94, -68 and -213 kcal/mol, then`deltaH` for the reacton `[C(s) +2H_2(g) → CH_4(g)]` is

Options

A. -17 kcal
B. -34 kcal
C. -886 kcal
D. -51 kcal

Answer

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

\[ C(s) + 2H_2(g) \rightarrow CH_4(g) \]

we will use the heat of combustion values provided for each substance involved in the reaction. The formula to calculate ΔH for the reaction is:

\[ \Delta H = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \]

### Step 1: Identify the heat of combustion values
The heat of combustion values given are:
- For \( C(s) \): -94 kcal/mol
- For \( H_2(g) \): -68 kcal/mol
- For \( CH_4(g) \): -213 kcal/mol

### Step 2: Write the ΔH equation
Using the formula, we can write:

\[ \Delta H = \Delta H_{CH_4} - \left( \Delta H_C + 2 \times \Delta H_{H_2} \right) \]

### Step 3: Substitute the values into the equation
Substituting the values we have:

\[ \Delta H = (-213 \text{ kcal/mol}) - \left( (-94 \text{ kcal/mol}) + 2 \times (-68 \text{ kcal/mol}) \right) \]

### Step 4: Calculate the total heat of combustion for the reactants
Calculating the total heat of combustion for the reactants:

\[ \Delta H_{reactants} = -94 + 2 \times (-68) \]
\[ = -94 - 136 \]
\[ = -230 \text{ kcal/mol} \]

### Step 5: Calculate ΔH
Now substituting back into the ΔH equation:

\[ \Delta H = -213 - (-230) \]
\[ = -213 + 230 \]
\[ = 17 \text{ kcal/mol} \]

### Final Answer
Thus, the change in enthalpy (ΔH) for the reaction is:

\[ \Delta H = 17 \text{ kcal/mol} \]