Question:

If 2 moles of $ C_6H_6 $ (g) are completely burnt 4100 kJ of heat is liberated. If $ \Delta H^\circ $ for $ CO_2 (g) $ and $ H_2O (l) $ are -410 kJ and -285 kJ per mole respectively then the heat of formation of $ C_6H_6 (g) $ is

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In combustion reactions, remember that the heat released can be calculated by comparing the sum of the heat of formation of products and reactants.
Updated On: May 3, 2025
  • -116 kJ
  • -375 kJ
  • -775 kJ
  • -885 kJ
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The Correct Option is B

Approach Solution - 1

To solve the problem, we need to determine the standard enthalpy of formation for ethane (\( \text{C}_2\text{H}_6 \)) using the given combustion reaction.

1. Given Reaction:
The balanced chemical equation for the combustion of ethane is:

$$ 2\text{C}_2\text{H}_6(g) + 7\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 6\text{H}_2\text{O}(l); \quad \Delta H = -4100 \, \text{kJ} $$

2. Enthalpy Calculation Setup:
Let \( \Delta H_f^\circ(\text{C}_2\text{H}_6) = x \). The enthalpy change can be expressed as:

$$ \Delta H = \Delta H_f^\circ(\text{products}) - \Delta H_f^\circ(\text{reactants}) $$

3. Substituting Values:
Using known enthalpy values (\( \Delta H_f^\circ(\text{CO}_2) = -410 \, \text{kJ/mol} \) and \( \Delta H_f^\circ(\text{H}_2\text{O}) = -285 \, \text{kJ/mol} \)):

$$ -4100 = [4(-410) + 6(-285)] - [2x + 0] $$

4. Final Calculation:
Solving the equation gives:

$$ x = -375 \, \text{kJ} $$

Final Answer:
The standard enthalpy of formation for ethane is \( \boxed{-375 \, \text{kJ}} \).

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Approach Solution -2

To determine the heat of formation of C2H6(g), we'll analyze the given combustion reaction and apply thermodynamic principles.

1. Balanced Combustion Reaction:
The combustion equation for ethane is:
C2H6(g) + (7/2)O2(g) → 2CO2(g) + 3H2O(l)

2. Using Hess's Law:
The enthalpy change can be calculated as:
ΔH°combustion = Σ(n × ΔH°f products) - Σ(n × ΔH°f reactants)

3. Substituting Known Values:
Given data:
- ΔH°combustion = -4100 kJ for 2 moles → -2050 kJ/mol
- ΔH°f(CO2) = -410 kJ/mol
- ΔH°f(H2O) = -285 kJ/mol
- ΔH°f(O2) = 0 kJ/mol

Let ΔH°f(C2H6) = x

4. Setting Up the Equation:
-2050 = [2(-410) + 3(-285)] - [x + (7/2)(0)]
-2050 = [-820 - 855] - x
-2050 = -1675 - x

5. Solving for x:
x = -1675 + 2050
x = 375 kJ
Thus, ΔH°f(C2H6) = -375 kJ (heat of formation is exothermic)

Final Answer:
The heat of formation of C2H6(g) is -375 kJ.

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