Standard Change in Free Energy

In summary, the first conversation discusses how to calculate the free energy change for the formation of methanol from carbon monoxide and hydrogen gas at 298K. The standard free energy of formation for methanol is given and the equilibrium constant is calculated using the partial pressures of the reactants and products. In the second conversation, the equilibrium constant for the formation of AB from A and B is determined using the free energy of formation of AB and the temperature.
  • #1
Soaring Crane
469
0
Are my final values correct?

1. Determine the free energy change associated with the formation of methanol from carbon monoxide and hydrogen gas if the partial pressure of CO is 1.0700 atm, H2 is 2.2800 atm, and CH3OH is 0.0600 atm at 298K. The standard free energy of formation for methanol is - 24.70 kJ. Give your answer in kilojoules.

CO (g) + 2H2 (g) --> CH3OH (g)

deltaG* = -24.70 kJ = -24,700 J

deltaG = G* + RTlnQ

Q = [(CH3O)]/[(CO)*(H2)^2] = [0.06 atm]/[(2.28 atm)^2*(1.07 atm)]
= 0.010786928

delta G = -24,700 J + (298 K)(8.314)ln(0.010786928)
= -35921.965 J
= -35.92 kJ




2. The compound AB can be formed from A and B according to the following reaction:A2 (g) + B2 (g) --> 2 AB (g) If the free energy of formation for AB is 8.66 kJ/mol, what is the equilibrium constant for the reaction at 181C? Give your answer to four decimal places.


deltaG* = 2 mol(8.66 kJ/mol) = 17.32 kJ = 17,320 J

deltaG* = RTlnK

K = e^-(G/RT)
= e^-(17,320 J/(8.314 J *454.15 K))
= e^-(4.5871)
= 0.10182 = 0.1018 ?




Thanks.
 
Physics news on Phys.org
  • #2
Are my answers valid?

Thanks.
 
  • #3
Your final values appear to be correct. When calculating deltaG, it is important to make sure that all values are in the correct units (in this case, joules and kelvin) and to use the natural logarithm (ln) function. Your calculations show that the equilibrium constant for the formation of AB is 0.1018 at 181C, which means that the reaction favors the formation of AB. Good job!
 

1. What is the definition of standard change in free energy?

The standard change in free energy, denoted as ΔG°, is a measure of the energy released or absorbed during a chemical reaction under standard conditions. It takes into account the change in enthalpy (ΔH°) and the change in entropy (ΔS°) of the system.

2. How is the standard change in free energy calculated?

The standard change in free energy is calculated using the equation ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin. This equation takes into account the energy released or absorbed (ΔH°) and the change in disorder (ΔS°) of the system at a constant temperature.

3. What is the significance of the standard change in free energy?

The standard change in free energy provides important information about the spontaneity and direction of a chemical reaction. A negative ΔG° indicates that the reaction is spontaneous and will proceed in the forward direction, while a positive ΔG° indicates a non-spontaneous reaction that will proceed in the reverse direction.

4. How does the standard change in free energy relate to equilibrium?

At equilibrium, the standard change in free energy is equal to zero (ΔG° = 0). This means that the reaction is at a state of maximum stability, and no further changes will occur. If ΔG° is negative, the reaction will proceed towards equilibrium in the forward direction, while a positive ΔG° indicates that the reaction will proceed towards equilibrium in the reverse direction.

5. Can the standard change in free energy be used to predict the rate of a reaction?

No, the standard change in free energy does not provide information about the rate of a reaction. It only indicates the direction and spontaneity of the reaction. The rate of a reaction is determined by factors such as the concentration of reactants, temperature, and the presence of a catalyst.

Similar threads

  • Biology and Chemistry Homework Help
Replies
7
Views
2K
  • Biology and Chemistry Homework Help
Replies
3
Views
2K
  • Biology and Chemistry Homework Help
Replies
7
Views
2K
  • Thermodynamics
Replies
4
Views
994
  • Biology and Chemistry Homework Help
Replies
9
Views
2K
  • Biology and Chemistry Homework Help
Replies
6
Views
2K
  • Biology and Chemistry Homework Help
Replies
13
Views
3K
  • Biology and Chemistry Homework Help
Replies
4
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
Back
Top