Ag(s) + Cl2(g) --> 2AgCl(s)
∆H and ∆S are 254.14kJ and –115.0J/K respectively. Calculate equilibrium constant, K for the reaction at 500K?
K = e^(-∆S/R*T) = e^(-(-115.0J/K)/(8.314J/K*mol)*500K) = 1.9 x 10^-3
To calculate the equilibrium constant (K) for the given reaction at 500K, we will use the formula:
ΔG = ΔH - TΔS
where ΔG is the standard Gibbs free energy change, ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin.
First, convert the units of ΔS from J/K to kJ/K:
ΔS = -115.0 J/K = -0.115 kJ/K
Now, substitute the given values into the formula:
ΔG = 254.14 kJ - (500 K)(-0.115 kJ/K)
= 254.14 kJ + 57.5 kJ
= 311.64 kJ
Next, use the equation ΔG = -RT ln(K) to find the equilibrium constant, K. Rearranging the equation gives:
ln(K) = -ΔG / RT
where R is the ideal gas constant (8.314 J/(mol K)). Since the reaction involves 2 moles of AgCl, we need to multiply the ΔG value by 2:
ln(K) = -(2)(311.64 kJ) / ((8.314 J/(mol K))(500 K))
ln(K) = -623.28 kJ / (8.314 J/(mol K))(500 K)
ln(K) = -149.18 mol
Finally, we can exponentiate both sides to solve for K:
K = e^(-149.18 mol)
Using a scientific calculator or computer, the equilibrium constant, K, is approximately 1.20 x 10^(-65).
To calculate the equilibrium constant (K) for the given reaction, you can use the equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in Gibbs free energy,
ΔH is the change in enthalpy,
T is the temperature in Kelvin,
ΔS is the change in entropy.
The relationship between ΔG and K is given by:
ΔG = -RTln(K)
Where:
R is the gas constant (8.314 J/mol·K),
T is the temperature in Kelvin,
K is the equilibrium constant.
By equating the two equations, we can solve for K:
-RTln(K) = ΔH - TΔS
Now let's substitute the given values into the equation:
R = 8.314 J/mol·K
ΔH = 254.14 kJ = 254140 J
T = 500 K
ΔS = -115.0 J/K
-8.314 J/mol·K * 500 K * ln(K) = 254140 J - 500 K * (-115.0 J/K)
Simplifying further:
-4157 J·K/mol * ln(K) = 254140 J + 57500 J
-4157 J·K/mol * ln(K) = 311640 J
Now, divide both sides of the equation by -4157 J·K/mol:
ln(K) = -311640 J / -4157 J·K/mol
ln(K) = 74.95 mol/K
Finally, we can solve for K by taking the exponential of both sides:
K = e^(74.95 mol/K)
Using a scientific calculator or an online calculator, the value of K at 500 K can be calculated.