The standard enthalpy of combustion for naphthalene, C10H8(s), is
-5156.8 kJ/mol. Using this data and the standard enthalpies of formation, ∆H: H2O(l) = -285.9 kJ mol; CO2(g)= -393.5 kJ mol, calculate the standard enthalpy of formation of C10H8(s), in kJ mol.
a) +78.2kj
b) +935.9 kJ
C) -1065.4 kJ
d) +3619.7 kJ
e) -10235.4 kj
The correct answer and explanation is :
To calculate the standard enthalpy of formation ((\Delta H_f^\circ)) for naphthalene ((C_{10}H_8(s))), we will use the following formula derived from Hess’s law:
[
\Delta H_{\text{combustion}}^\circ = \sum (\Delta H_f^\circ \, \text{of products}) – \sum (\Delta H_f^\circ \, \text{of reactants})
]
Step 1: Write the combustion reaction for naphthalene
The combustion of naphthalene can be represented by the following balanced chemical equation:
[
C_{10}H_8(s) + 12 O_2(g) \rightarrow 10 CO_2(g) + 4 H_2O(l)
]
Step 2: Apply the given data
We are given:
- The standard enthalpy of combustion of naphthalene is (\Delta H_{\text{combustion}}^\circ = -5156.8 \, \text{kJ/mol}).
- The standard enthalpy of formation for water: (\Delta H_f^\circ \, \text{H}_2\text{O(l)} = -285.9 \, \text{kJ/mol}).
- The standard enthalpy of formation for carbon dioxide: (\Delta H_f^\circ \, \text{CO}_2(g) = -393.5 \, \text{kJ/mol}).
Step 3: Calculate the sum of the standard enthalpies of formation of the products
Using the balanced combustion equation, we can calculate the enthalpy change for the products:
[
\Delta H_f^\circ \, (\text{products}) = 10 \times \Delta H_f^\circ \, \text{CO}_2(g) + 4 \times \Delta H_f^\circ \, \text{H}_2\text{O(l)}
]
[
\Delta H_f^\circ \, (\text{products}) = 10 \times (-393.5) + 4 \times (-285.9)
]
[
\Delta H_f^\circ \, (\text{products}) = -3935 + (-1143.6) = -5078.6 \, \text{kJ/mol}
]
Step 4: Apply Hess’s law to find the standard enthalpy of formation of naphthalene
Now, using Hess’s law, the standard enthalpy of formation of naphthalene is:
[
\Delta H_f^\circ \, (C_{10}H_8(s)) = \Delta H_{\text{combustion}}^\circ – \Delta H_f^\circ \, (\text{products})
]
[
\Delta H_f^\circ \, (C_{10}H_8(s)) = -5156.8 – (-5078.6)
]
[
\Delta H_f^\circ \, (C_{10}H_8(s)) = -5156.8 + 5078.6 = -78.2 \, \text{kJ/mol}
]
Thus, the standard enthalpy of formation of naphthalene is (-78.2 \, \text{kJ/mol}), which corresponds to the answer a).
Conclusion
The standard enthalpy of formation of naphthalene is (-78.2 \, \text{kJ/mol}), making the correct answer a).