The standard molar enthalpy of formation of ethane, carbon dioxide and liquid water are -21.1, -91.1 and -68.3 kcal respectively. Calculate the standard enthalpy change of the following reaction :
`2C_(2)H_(5)(g) + 7O_(2)(g) to 4CO_(2)(g) + 6H_(2)O(l)`.

Explanation

To calculate the standard enthalpy change (ΔH) for the reaction:

\[ 2C_{2}H_{6}(g) + 7O_{2}(g) \rightarrow 4CO_{2}(g) + 6H_{2}O(l) \]

we will use the standard molar enthalpy of formation values provided for each substance involved in the reaction.

### Step 1: Write down the standard molar enthalpy of formation values
- Ethane (C₂H₆): ΔH_f = -21.1 kcal/mol
- Carbon dioxide (CO₂): ΔH_f = -91.1 kcal/mol
- Liquid water (H₂O): ΔH_f = -68.3 kcal/mol
- Oxygen (O₂) is in its elemental form, so ΔH_f = 0 kcal/mol

### Step 2: Write the formula for standard enthalpy change
The standard enthalpy change of the reaction can be calculated using the formula:

\[ \Delta H = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \]

### Step 3: Calculate the enthalpy of the products
For the products:
- For 4 moles of CO₂:
\[ 4 \times (-91.1 \text{ kcal/mol}) = -364.4 \text{ kcal} \]
- For 6 moles of H₂O:
\[ 6 \times (-68.3 \text{ kcal/mol}) = -409.8 \text{ kcal} \]

Adding these together gives:
\[ \text{Total for products} = -364.4 \text{ kcal} + (-409.8 \text{ kcal}) = -774.2 \text{ kcal} \]

### Step 4: Calculate the enthalpy of the reactants
For the reactants:
- For 2 moles of C₂H₆:
\[ 2 \times (-21.1 \text{ kcal/mol}) = -42.2 \text{ kcal} \]
- For 7 moles of O₂:
\[ 7 \times (0 \text{ kcal/mol}) = 0 \text{ kcal} \]

Adding these together gives:
\[ \text{Total for reactants} = -42.2 \text{ kcal} + 0 \text{ kcal} = -42.2 \text{ kcal} \]

### Step 5: Calculate the standard enthalpy change (ΔH)
Now we can substitute the totals into the ΔH formula:
\[ \Delta H = (-774.2 \text{ kcal}) - (-42.2 \text{ kcal}) \]
\[ \Delta H = -774.2 \text{ kcal} + 42.2 \text{ kcal} = -732 \text{ kcal} \]

### Final Answer
The standard enthalpy change for the reaction is:
\[ \Delta H = -732 \text{ kcal} \]

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