The standard Gibbs free energy of formation of ammonia at 25 ° c is - 16.5kj per mole.calculate the value of equilibrium constant Kp at this temperature for this reaction. ( 5marks )

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dG = -RTlnKp
Substitute and solve for Kp. By the way, remember dG goes in the equation as J/mol.

To calculate the value of the equilibrium constant Kp at 25 °C for the reaction of ammonia formation, we need to use the equation:

ΔG° = −RTln(Kp)

Where:
ΔG° is the standard Gibbs free energy of reaction
R is the gas constant (8.314 J/(mol⋅K))
T is the absolute temperature in Kelvin (25 °C = 298.15 K)
Kp is the equilibrium constant

First, convert the given value of the standard Gibbs free energy of formation from kilojoules (kJ) to joules (J) per mole:
ΔG° = (-16.5 kJ/mol) * (1000 J/1 kJ) = -16,500 J/mol

Substitute the values into the equation and solve for Kp:
-16,500 J/mol = (-8.314 J/(mol⋅K)) * (298.15 K) ln(Kp)

Divide both sides of the equation by (-8,314 J/(mol⋅K) * (298.15 K):
⇒ ln(Kp) = -16,500 J/mol / (-8,314 J/(mol⋅K) * (298.15 K)

Calculate the right side:
⇒ ln(Kp) = 6.2572

Exponentiate both sides to eliminate the natural logarithm:
⇒ Kp = e^(6.2572)

Using a calculator, evaluate the exponent:
⇒ Kp ≈ 518.34

Therefore, at 25 °C, the value of the equilibrium constant Kp for the formation of ammonia is approximately 518.34.

To calculate the equilibrium constant Kp for the reaction, we need to use the equation:

ΔG = -RTln(Kp)

Where:
- ΔG is the change in Gibbs free energy
- R is the ideal gas constant (8.314 J/(mol⋅K))
- T is the temperature in Kelvin
- ln is the natural logarithm

First, we need to convert the given temperature of 25 °C to Kelvin. The conversion equation is:

T(K) = 25 + 273.15

By substituting the values into the equation, we get:

ΔG = -16.5 kJ/mol
R = 8.314 J/(mol⋅K)
T = (25 + 273.15) K = 298.15 K

Now we can calculate the value of Kp. Rearranging the equation, we have:

ln(Kp) = -ΔG / (RT)

Plugging in the values, we get:

ln(Kp) = -(-16.5 × 10^3) J / (8.314 J/(mol⋅K) × 298.15 K)

Simplifying the equation, we get:

ln(Kp) = 66,170 J / (24.71 J/(mol⋅K))

ln(Kp) ≈ 2676.1

Finally, we can calculate Kp by taking the exponential of both sides of the equation:

Kp = e^(ln(Kp))

Kp = e^(2676.1)

Kp ≈ 1.67 × 10^116

Therefore, the value of the equilibrium constant Kp at this temperature for the reaction is approximately 1.67 × 10^116.