Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is C2H2(g)+2H2(g)?C2H6(g)
Given the following data, what is the value of Kp for this reaction?
Substance ?G?f
(kJ/mol)
C2H2(g) 209.2
H2(g) 0
C2H6(g) ?32.89
The Correct Answer and Explanation is :
To calculate the equilibrium constant ( K_p ) for the hydrogenation of acetylene to ethane, we can use the Gibbs free energy change (( \Delta G )) of the reaction. The equation for the reaction is:
[
\text{C}_2\text{H}_2(g) + 2\text{H}_2(g) \rightleftharpoons \text{C}_2\text{H}_6(g)
]
Step 1: Calculate the Gibbs Free Energy Change (( \Delta G ))
The standard Gibbs free energy change for a reaction can be calculated using the following equation:
[
\Delta G^\circ = \sum \Delta G_f^\circ \text{(products)} – \sum \Delta G_f^\circ \text{(reactants)}
]
Where:
- ( \Delta G_f^\circ ) is the standard Gibbs free energy of formation.
Given:
- ( \Delta G_f^\circ ) for C₂H₂(g) = 209.2 kJ/mol
- ( \Delta G_f^\circ ) for H₂(g) = 0 kJ/mol (as it’s an element in its standard state)
- ( \Delta G_f^\circ ) for C₂H₆(g) = -32.89 kJ/mol
Plugging these values into the equation gives:
[
\Delta G^\circ = \Delta G_f^\circ \text{(C}_2\text{H}_6\text{)} – \left[ \Delta G_f^\circ \text{(C}_2\text{H}_2\text{)} + 2 \cdot \Delta G_f^\circ \text{(H}_2\text{)} \right]
]
[
\Delta G^\circ = (-32.89 \, \text{kJ/mol}) – \left[ 209.2 \, \text{kJ/mol} + 2 \cdot 0 \, \text{kJ/mol} \right]
]
[
\Delta G^\circ = -32.89 \, \text{kJ/mol} – 209.2 \, \text{kJ/mol} = -242.09 \, \text{kJ/mol}
]
Step 2: Calculate the Equilibrium Constant (( K_p ))
The relationship between the Gibbs free energy change and the equilibrium constant is given by the equation:
[
\Delta G^\circ = -RT \ln K_p
]
Where:
- ( R ) is the universal gas constant ( (8.314 \, \text{J/mol·K}) )
- ( T ) is the temperature in Kelvin (assuming standard conditions, ( T = 298 \, \text{K} ))
First, convert ( \Delta G^\circ ) to Joules:
[
\Delta G^\circ = -242.09 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -242090 \, \text{J/mol}
]
Now, rearranging the equation to solve for ( K_p ):
[
K_p = e^{-\Delta G^\circ / RT}
]
Substituting the values:
[
K_p = e^{242090 \, \text{J/mol} / (8.314 \, \text{J/mol·K} \cdot 298 \, \text{K})}
]
Calculating:
[
K_p = e^{242090 / 2478.572} = e^{97.685}
]
Calculating ( e^{97.685} ):
[
K_p \approx 1.305 \times 10^{42}
]
Conclusion
The equilibrium constant ( K_p ) for the hydrogenation of acetylene to ethane at standard conditions is approximately ( 1.305 \times 10^{42} ). This indicates that the reaction strongly favors the formation of ethane from acetylene and hydrogen, demonstrating the stability of ethane compared to the reactants under standard conditions.