The standard free energy change, delta f G, for the formation of NO (g) from its elements is +86.58 kJ/mol at 25 degrees C. Calculate Kp at this temperature for the equilibrium
1/2 N2(g) + 1/2 O2(g) = NO(g)
Comment on the sign of delta G and the magnitude of Kp.
1/2N2 + 1/2O2 --> NO
dGfo = +86.58 kJ/mol.
dG = -RTlnKp
86,580 = -8.314*298*lnK
K = approx 10^-16
With a + sign for dG and the VERY small Kp, it is obvious that not much N2 will react with O2 to produce NO.
Well, isn't it funny how delta G is positive for the formation of NO (g)! Looks like NO (g) is having a good time partying with its elements, N2(g) and O2(g). They must be having a blast! Now, let's talk about Kp. Kp measures the ratio of the product and reactant pressures at equilibrium. So, when you have a positive delta G, it means the reaction is not spontaneous, which usually leads to a small Kp value. It's like NO (g) and its buddies are lazy and don't want to do any work to form an equilibrium. They just want to chill and have a negligible presence. So, to sum it up, delta G is positive, NO (g) doesn't want to form, and Kp is pretty tiny. Can you blame them? It's hard to resist being lazy sometimes!
To calculate Kp at 25 degrees C for the given equilibrium, we will utilize the equation
ΔG° = -RT ln(Kp)
where ΔG° is the standard free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25°C = 298 K), and ln denotes the natural logarithm.
First, convert the given ΔG° from kJ/mol to J/mol:
ΔG° = 86.58 kJ/mol × (1000 J/1 kJ) ≈ 86580 J/mol
Next, rearrange the equation to solve for Kp:
ln(Kp) = -ΔG° / RT
Substituting the values:
ln(Kp) = -86580 J/mol / (8.314 J/(mol·K) × 298 K)
Simplifying the expression:
ln(Kp) ≈ -35.27
Now, use the relationship between natural logarithm and exponentiation:
Kp = e^(ln(Kp))
Substituting the value:
Kp = e^(-35.27)
Calculating Kp:
Kp ≈ 4.42 × 10^(-16)
Comment on the sign of ΔG and the magnitude of Kp:
The sign of ΔG (+86.58 kJ/mol) indicates that the formation of NO (g) from its elements is not favorable under standard conditions at 25 degrees C. This is supported by the extremely small value of Kp (4.42 × 10^(-16)), indicating that the equilibrium heavily favors the reactants (N2(g) and O2(g)) over the product (NO(g)).
To calculate the equilibrium constant, Kp, at a given temperature using the standard free energy change (delta G), you can use the equation:
delta G = -RT * ln(Kp)
Where:
delta G is the standard free energy change
R is the gas constant (8.314 J/(mol*K) or 0.008314 kJ/(mol*K))
T is the temperature in Kelvin
ln is the natural logarithm
Kp is the equilibrium constant
In this case, the given values are:
delta G = +86.58 kJ/mol
T = 25 degrees C = 25 + 273.15 K = 298.15 K
Now, let's solve for Kp:
delta G = -RT * ln(Kp)
86.58 kJ/mol = -0.008314 kJ/(mol*K) * 298.15 K * ln(Kp)
Rearranging the equation to solve for ln(Kp):
ln(Kp) = 86.58 kJ/mol / (-0.008314 kJ/(mol*K) * 298.15 K)
ln(Kp) ≈ -11.17
To get the value of Kp, we need to take the exponential (e) of both sides:
Kp = e^(-11.17)
Calculating this, Kp ≈ 6.35 x 10^(-5).
Now, let's comment on the sign of delta G and the magnitude of Kp:
The positive sign of delta G (+86.58 kJ/mol) indicates that the reaction is non-spontaneous under standard conditions at 25 degrees Celsius.
The magnitude of Kp (6.35 x 10^(-5)) is quite small, which indicates that the equilibrium position favors the reactants (N2 and O2). This means that the forward reaction (formation of NO) is not favored under these conditions, and the amount of NO (g) formed will be relatively low compared to the amounts of N2 (g) and O2 (g) present.