What is the standard change in Gibbs free energy of this reaction at 25 °C?

Steps to Solve

1. Write the expression for Kp

The equilibrium constant in terms of partial pressures, $K_p$, for the reaction ( 3A(g) + 3B(g) \rightleftharpoons C(g) + 3D(g) ) is given by the formula:

$$ K_p = \frac{P_C \cdot P_D^3}{P_A^3 \cdot P_B^3} $$

2. Substitute the equilibrium partial pressures

Now, substitute the given partial pressures into the ( K_p ) expression:

• ( P_A = 4.94 , \text{atm} )
• ( P_B = 5.93 , \text{atm} )
• ( P_C = 5.65 , \text{atm} )
• ( P_D = 4.73 , \text{atm} )

The equation becomes:

$$ K_p = \frac{5.65 \cdot (4.73)^3}{(4.94)^3 \cdot (5.93)^3} $$

3. Calculate the value of Kp

Calculate the value of ( K_p ):

First, calculate ( (4.73)^3 ), ( (4.94)^3 ), and ( (5.93)^3 ):

• ( (4.73)^3 \approx 105.4 )
• ( (4.94)^3 \approx 120.8 )
• ( (5.93)^3 \approx 207.4 )

Now, substitute these back into the equation:

$$ K_p = \frac{5.65 \cdot 105.4}{120.8 \cdot 207.4} $$

Calculate ( K_p ):

$$ K_p \approx \frac{596.01}{25041.92} \approx 0.0238 $$

4. Calculate the standard Gibbs free energy change

Use the relationship between the standard Gibbs free energy change and the equilibrium constant:

$$ \Delta G^\circ = -RT \ln K_p $$

Where:

• ( R = 0.0821 , \text{L atm/(K mol)} )
• ( T = 25 , \text{°C} = 298 , \text{K} )

Now, calculate ( \Delta G^\circ ):

First, calculate ( \ln K_p ):

$$ \ln(0.0238) \approx -3.758 $$

Then plug into the equation:

$$ \Delta G^\circ = - (0.0821)(298)(-3.758) $$

5. Final calculation

Now, compute ( \Delta G^\circ ):

$$ \Delta G^\circ \approx 82.1 \cdot 298 \cdot 3.758 \approx 98.307 , \text{kJ/mol} $$