Question

One mole of \( \text{C}_2\text{H}_5\text{OH(l)} \) was completely burnt in oxygen to form \( \text{CO}_2\text{(g)} \) and \( \text{H}_2\text{O(l)} \). The standard enthalpy of formation (\( \Delta_f H^\ominus \)) of \( \text{C}_2\text{H}_5\text{OH(l)} \), \( \text{CO}_2\text{(g)} \) and \( \text{H}_2\text{O(l)} \) is x, y, z kJ mol\(^{-1}\) respectively. What is \( \Delta_r H^\ominus \) (in kJ mol\(^{-1}\)) for this reaction?

• A. \( (2y + 3z + x) \)
• B. \( (2y - 3z + x) \)
• C. \( (x - 2y - 3z) \)
• D. \( (2y + 3z - x) \)

Hint:

1. Write and balance the chemical equation for the reaction. 2. Use Hess's Law in the form: \( \Delta_r H^\ominus = \sum \nu_P \Delta_f H^\ominus (\text{Products}) - \sum \nu_R \Delta_f H^\ominus (\text{Reactants}) \) where \( \nu \) are the stoichiometric coefficients. 3. The standard enthalpy of formation (\( \Delta_f H^\ominus \)) of an element in its most stable standard state is zero (e.g., O\(_2\)(g), C(graphite), H\(_2\)(g)).

Answer 1

The Correct Option is D

The combustion reaction for ethanol (\( \text{C}_2\text{H}_5\text{OH(l)} \)) is: \[ \text{C}_2\text{H}_5\text{OH(l)} + \text{O}_2\text{(g)} \longrightarrow \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)} \] Balance the equation: Carbons: 2 on left, so 2 CO\(_2\) on right.
Hydrogens: 5+1=6 on left, so 3 H\(_2\)O on right.
Oxygens: On right, \( 2 \times 2 (\text{from CO}_2) + 3 \times 1 (\text{from H}_2\text{O}) = 4+3 = 7 \) oxygen atoms.
On left, 1 oxygen in C\(_2\)H\(_5\)OH.
Need 6 more from O\(_2\).
So, 3 O\(_2\).
Balanced equation: \[ \text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \longrightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)} \] The standard enthalpy of reaction (\( \Delta_r H^\ominus \)) is calculated using standard enthalpies of formation (\( \Delta_f H^\ominus \)) of products and reactants: \[ \Delta_r H^\ominus = \sum (\text{stoichiometric coeff} \times \Delta_f H^\ominus)_{\text{products}} - \sum (\text{stoichiometric coeff} \times \Delta_f H^\ominus)_{\text{reactants}} \] Given standard enthalpies of formation: \( \Delta_f H^\ominus (\text{C}_2\text{H}_5\text{OH(l)}) = x \) kJ/mol \( \Delta_f H^\ominus (\text{CO}_2\text{(g)}) = y \) kJ/mol \( \Delta_f H^\ominus (\text{H}_2\text{O(l)}) = z \) kJ/mol The standard enthalpy of formation of an element in its standard state is zero.
So, \( \Delta_f H^\ominus (\text{O}_2\text{(g)}) = 0 \).
Now, calculate \( \Delta_r H^\ominus \): \[ \Delta_r H^\ominus = [2 \cdot \Delta_f H^\ominus(\text{CO}_2\text{(g)}) + 3 \cdot \Delta_f H^\ominus(\text{H}_2\text{O(l)})] - [1 \cdot \Delta_f H^\ominus(\text{C}_2\text{H}_5\text{OH(l)}) + 3 \cdot \Delta_f H^\ominus(\text{O}_2\text{(g)})] \] \[ \Delta_r H^\ominus = [2y + 3z] - [1x + 3(0)] \] \[ \Delta_r H^\ominus = 2y + 3z - x \] This can be written as \( (2y + 3z - x) \).
This matches option (4).