Question

Enthalpy of combustion of benzene, graphite, dihydrogen are \( -3260 \), \( -390 \) and \( -290\text{ kJ/mol} \). Find the enthalpy of formation of benzene

Hint:

A common trap is forgetting to multiply the combustion enthalpies of the elements by their stoichiometric coefficients from the balanced formation equation (6 for Carbon, 3 for Hydrogen). Always write out the balanced formation equation first!

Answer 1

Step 1: Understanding the Concept:
Hess's Law states that the total enthalpy change for a chemical reaction is independent of the pathway. The enthalpy of formation can be calculated using the enthalpies of combustion of the reactants and products.
Step 2: Key Formula or Approach:
The mathematical relationship is:
\[ \Delta H_f^\circ (\text{compound}) = \sum \Delta H_c^\circ (\text{reactants}) - \sum \Delta H_c^\circ (\text{products}) \] Alternatively, construct the formation reaction and use the constituent combustion equations.
Step 3: Detailed Explanation:
The target reaction is the formation of benzene (\( \text{C}_6\text{H}_6 \)) from its standard state elements:
\[ 6\text{C}_{(graphite)} + 3\text{H}_{2(g)} \rightarrow \text{C}_6\text{H}_{6(l)} \] We are given the standard enthalpies of combustion (\( \Delta H_c \)):
1) \( \Delta H_c (\text{C}_{graphite}) = -390 \text{ kJ/mol} \)
2) \( \Delta H_c (\text{H}_2) = -290 \text{ kJ/mol} \)
3) \( \Delta H_c (\text{C}_6\text{H}_6) = -3260 \text{ kJ/mol} \)
Using the formula derived from Hess's Law for formation from combustion data:
\[ \Delta H_f (\text{C}_6\text{H}_6) = [6 \times \Delta H_c (\text{C}) + 3 \times \Delta H_c (\text{H}_2)] - [\Delta H_c (\text{C}_6\text{H}_6)] \] Substitute the given values into the equation:
\[ \Delta H_f (\text{C}_6\text{H}_6) = [6(-390) + 3(-290)] - (-3260) \] Perform the multiplications:
\[ 6 \times (-390) = -2340 \text{ kJ} \] \[ 3 \times (-290) = -870 \text{ kJ} \] Add the reactant combustion enthalpies:
\[ -2340 + (-870) = -3210 \text{ kJ} \] Now, subtract the combustion enthalpy of the product (benzene):
\[ \Delta H_f (\text{C}_6\text{H}_6) = -3210 - (-3260) \] \[ \Delta H_f (\text{C}_6\text{H}_6) = -3210 + 3260 \] \[ \Delta H_f (\text{C}_6\text{H}_6) = +50 \text{ kJ/mol} \] Step 4: Final Answer:
The enthalpy of formation of benzene is \( +50 \text{ kJ/mol} \).