Consider the rxn during the Haber Process:
N2 + 3H2 ---> 2NH3
The equilibrium constant is 3.9 x 10^(5) at 300 K and 1.2 x 10^(-1) at 500 K.
Find Delta H reaction and Delta S for reaction
To find the ΔH (enthalpy change) and ΔS (entropy change) for the reaction, we can use the relationship between equilibrium constant (K), ΔH, and ΔS.
The equation relating these variables is:
ΔG = ΔH - TΔS
where ΔG is the change in Gibbs free energy, and T is the temperature in Kelvin.
At equilibrium, ΔG is equal to zero, so we can rewrite the equation as:
0 = ΔH - TΔS
To find ΔH and ΔS, we need to rearrange the equation as follows:
ΔH = TΔS
Now, let's calculate ΔH and ΔS for both temperatures given:
At 300 K:
K = 3.9 x 10^5
ΔH = 300K x ΔS
At 500 K:
K = 1.2 x 10^(-1)
ΔH = 500K x ΔS
Now, we need to calculate the ratio of equilibrium constants (K2/K1) using the data given:
K2 = 1.2 x 10^(-1)
K1 = 3.9 x 10^(5)
K2/K1 = 1.2 x 10^(-1) / 3.9 x 10^(5)
Now, we can substitute this ratio into the equation:
ΔH = TΔS
(500K/300K) = (1.2 x 10^(-1) / 3.9 x 10^(5))
Simplifying, we get:
2.00 = (1.2 x 10^(-1) / 3.9 x 10^(5))
Now, we can solve for ΔS:
ΔS = 2.00
Since ΔH = TΔS, we can calculate ΔH using one of the temperatures given. Let's use 300 K:
ΔH = 300K x ΔS = 300K x 2.00
ΔH = 600 J
Therefore, the ΔH (enthalpy change) for the reaction is 600 J, and the ΔS (entropy change) for the reaction is 2.00.