For the reaction below, the enthalpy of reaction is ΔrH° = 29.36 kJ mol−1 and the thermodynamic equilibrium constant is K = 2.17 at 298.15 K.
Cl2(g) + Br2(l) ⟶ 2 BrCl(g)
What is the value of K at 476 K ?
Use the van't Hoff equation.
The Van't Hoff equation, ln(K2/K1) = ΔHo/R (1/T1 - 1/T2)
Post your work if you get stuck.
To find the value of K at 476 K, we can use the Van 't Hoff equation, which relates the equilibrium constant at different temperatures:
ln(K2/K1) = ΔH°/R * (1/T1 - 1/T2)
Where:
K1 is the equilibrium constant at temperature T1,
K2 is the equilibrium constant at temperature T2,
ΔH° is the enthalpy of reaction,
R is the gas constant (8.314 J/mol·K),
T1 and T2 are the temperatures in Kelvin.
Given:
ΔH° = 29.36 kJ/mol
K1 = 2.17 (at 298.15 K)
T1 = 298.15 K
T2 = 476 K
Converting ΔH° to J/mol:
ΔH° = 29.36 kJ/mol * 1000 J/kJ = 29,360 J/mol
Substituting the values in the equation:
ln(K2/2.17) = (29,360 J/mol)/(8.314 J/mol·K) * (1/298.15 K - 1/476 K)
Solving for ln(K2/2.17):
ln(K2/2.17) = 35.333
Now, we can exponentiate both sides to solve for K2:
K2/2.17 = e^(35.333)
K2 = 2.17 * e^(35.333)
Using a calculator:
K2 ≈ 3.707 x 10^15
Therefore, the value of K at 476 K is approximately 3.707 x 10^15.
To determine the value of K at a different temperature (476 K), we can use the Van 't Hoff equation. The Van 't Hoff equation relates the equilibrium constant (K) at one temperature to the equilibrium constant at a different temperature. The equation is given as:
ln(K2/K1) = -ΔH°/R * (1/T2 - 1/T1)
Where:
K1 = equilibrium constant at temperature T1 (298.15 K in this case)
K2 = equilibrium constant at temperature T2 (476 K in this case)
ΔH° = enthalpy of reaction
R = gas constant (8.314 J/(mol*K))
T1 = temperature 1 (298.15 K in this case)
T2 = temperature 2 (476 K in this case)
First, let's convert the given ΔrH° from kJ/mol to J/mol:
ΔH° = 29.36 kJ/mol = 29.36 * 1000 J/mol = 29360 J/mol
Now, let's plug in the values into the Van 't Hoff equation:
ln(K2/K1) = -(29360 J/mol)/(8.314 J/(mol*K)) * (1/476 K - 1/298.15 K)
Simplifying the equation:
ln(K2/K1) = -35.3 * (0.00216 - 0.00336)
ln(K2/K1) = -35.3 * (-0.0012)
ln(K2/K1) = 0.04236
Now, let's solve for K2/K1:
K2/K1 = e^(0.04236)
K2 = K1 * e^(0.04236)
Finally, plug in the given K1 value:
K2 = 2.17 * e^(0.04236)
Calculating this expression will give you the value of K2 at 476 K.