Styrene is produced by catalytic dehydrogenation of ethyl- benzene at high temperature in the presence of superheated steam. (a) Find ÄH°rxn, ÄG °rxn, and ÄS °rxn, given these data at 298 K: (b) At what temperature is the reaction spontaneous?
What are Delta G and K at 600 degrees?
I only need help with K at 600 degrees. Thank you. I thought it was 9.9 x 10^2 but it was marked wrong.•Chemistry - Dalton, Saturday, November 7, 2015 at 3:38pm
data:
Given the following data, what is the delta Hrxn, delta Grxn and delta Srxn at 298 K. Ethylbenzene, C6H5--CH2CH3 : delta Ht = -12.5 kJ/mol, delta Gt = 119.7 kJ/mol and S = 255 J/mol*K Styrene, C6H5--CH==CH2 : delta Ht = 103.8 kJ/mol, delta Gt = 202.5 kJ/mol and S = 238 J/mol*K
To find the value of K at 600 degrees, we can use the Van't Hoff equation, which relates the equilibrium constant (K) with the change in enthalpy (ΔH) and temperature (T):
ln(K2/K1) = -(ΔH/R) * (1/T2 - 1/T1)
Where K1 is the equilibrium constant at temperature T1, K2 is the equilibrium constant at temperature T2, ΔH is the change in enthalpy, and R is the gas constant (8.314 J/mol*K).
In this case, we are given the values of ΔG and ΔS at 298 K, but we need ΔH to use the Van't Hoff equation. We can calculate ΔH using the equation:
ΔG = ΔH - TΔS
ΔH = ΔG + TΔS
Now let's calculate ΔH at 298 K:
ΔH = (103.8 kJ/mol) + (298 K) * (238 J/mol*K / 1000 J/kJ)
= 103.8 kJ/mol + 70.8 kJ/mol
= 174.6 kJ/mol
Now we can use the Van't Hoff equation to find the value of K at 600 degrees. We assume that the values of ΔH and ΔS don't change with temperature:
ln(K2/K1) = -(ΔH/R) * (1/T2 - 1/T1)
ln(K2/K1) = -(174.6 kJ/mol / (8.314 J/mol*K)) * (1/(600 K) - 1/(298 K))
ln(K2/K1) = -21.02
Now, we need to solve for K2/K1:
K2/K1 = e^(-21.02)
K2/K1 ≈ 5.84 x 10^(-10)
Therefore, K at 600 degrees is approximately 5.84 x 10^(-10).