. The Haber process for the synthesis of ammonia is kinetically and thermodynamically controlled.

N2(g) + 3 H2(g) f 2 NH3(g)
for the reaction above: al-P = -92 kJ, AS° = -199 J K-I
a) Calculate aG° from the thermodynamic data and comment on the spontaneity of the reaction at room temperature.

b) Calculate the equilibrium constant at room temperature.

c) At what temperature does AG° become zero? What does this mean in regards to the temperature at which this reaction can be carried out at?

d) The rate of reaction at room temperature is very low. Give an explanation for this fact.

e) Propose a way to shift the rate of the reaction and also shift equilibrium towards products.
Refer to the system PCI3(g) + Cl)2(g f PCI5(g). To an empty 15.0-L cylinder, 0.500 mol of gaseous PC15 are added and allowed to reach equilibrium. The concentration of PCI3 is found to be 0.0220 M. Assume a temperature of 375 K.
a) How many mol of PCI5 remain at equilibrium?

b) Write the equilibrium constant expression for the above reaction.

c) Determine the value of K.

d) Determine the value of Kp for this same system at the same temperature

:/ I'm not really into chemistry :( Does anybody here have the answers?

keep reminding us that you can't help. We know that already.

@Jay.
a. dGo = dHo - TdSo
b. dGo = -RTlnK. Substitute and solve for K.
c. Set dGo = 0 and solve for T
d. The activation energy is high.
Post your work if you get stuck.

Okkk? :/ I said I'm better with math :\ Just giving a little hope :D

a) To calculate ΔG° (standard Gibbs free energy change), we can use the equation:

ΔG° = ΔH° - TΔS°
where ΔH° is the standard enthalpy change, ΔS° is the standard entropy change, and T is the temperature in Kelvin.

Given: ΔH° = -92 kJ, ΔS° = -199 J K^(-1) (Remember to convert to kJ K^(-1)), and room temperature is generally considered to be 298 K.

ΔG° = -92 kJ - (298 K * (-0.199 kJ K^(-1)))
= -92 kJ + 59.302 kJ
= -32.698 kJ

The reaction will be spontaneous at room temperature because ΔG° is negative.

b) The equilibrium constant (K) can be calculated using the formula:
K = e^(-ΔG°/RT)
where R is the gas constant (8.314 J K^(-1) mol^(-1)) and T is the temperature in Kelvin.

Given: ΔG° = -32.698 kJ (Remember to convert to J), and T = 298 K

K = e^((-32.698 kJ * 1000) / (8.314 J K^(-1) mol^(-1) * 298 K))
= e^(-109652.94 J / (2471.972 K J K^(-1) mol^(-1)))
= e^(-44.391 mol^(-1))
≈ 1.917

The equilibrium constant at room temperature is approximately 1.917.

c) To find the temperature at which ΔG° becomes zero (ΔG° = 0), we can rearrange the equation from part a:
ΔG° = ΔH° - TΔS°

0 = -92 kJ - T * (-0.199 kJ K^(-1))
T * (-0.199 kJ K^(-1)) = -92 kJ
T = -92 kJ / (-0.199 kJ K^(-1))
T ≈ 462.31 K

This means that at approximately 462.31 K, the reaction will have ΔG° = 0. Below this temperature, the reaction is spontaneous in the forward direction, while above this temperature, the reaction is spontaneous in the reverse direction.

d) The rate of reaction at room temperature being very low could be due to the high activation energy required for the reaction to proceed. The Haber process is known to be a slow reaction even at elevated temperatures and pressures. It requires a catalyst (typically iron) to increase the rate of reaction. Additionally, factors like reactant concentration and reaction conditions can also affect the reaction rate.

e) To shift the rate of the reaction and also shift the equilibrium towards products, there are a few approaches that can be considered:
1. Increase the reactant concentration: This can be done by adding more nitrogen and hydrogen gases to the reaction mixture.
2. Increase the temperature: According to Le Chatelier's principle, increasing the temperature favors the endothermic reaction, which is the forward reaction in this case.
3. Increase the pressure: In the Haber process, higher pressures are used to achieve higher yields of ammonia.
4. Use a catalyst: Adding a catalyst, such as iron, can increase the rate of the reaction without being consumed in the process. The presence of a catalyst provides an alternative reaction pathway that requires lower activation energy.
5. Remove ammonia as it forms: By continuously removing ammonia from the system, it drives the equilibrium towards the product side, resulting in a higher yield of ammonia.

Now, let's move on to the second part of your question.

a) The given equation is: PCI3(g) + Cl2(g) → PCI5(g)

The balanced chemical equation shows that the stoichiometry of the reaction is 1:1, meaning that one mole of PCI3 reacts with one mole of Cl2 to form one mole of PCI5. Therefore, if 0.500 mol of gaseous PCI5 are added, the same amount of PCI3 should remain at equilibrium.

b) The equilibrium constant expression for the given reaction is:
Kc = [PCI5] / [PCI3] * [Cl2]
= [PCI5] / ([PCI3] * [Cl2])

c) We have the concentration of PCI3 at equilibrium, which is 0.0220 M. However, we do not have the concentrations of PCI5 or Cl2. Without additional information, we cannot calculate the exact value of Kc.

d) The value of Kp can be determined from Kc using the ideal gas law. Kp is equal to Kc when all reactants and products are gases. Since we don't have the partial pressures of the gases, we cannot determine Kp without additional information.