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. I thought it was 9.9 x 10^2 but it was marked wrong.
Data:
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
thank you. I only need help with K at 600 degrees.
To find the value of K at 600 degrees, we can use the equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
ΔS is the change in entropy
T is the temperature in Kelvin
First, let's calculate the values of ΔH and ΔS for the reaction:
ΔHrxn = ΔHproducts - ΔHreactants
= (103.8 kJ/mol) - (-12.5 kJ/mol)
= 116.3 kJ/mol
ΔSrxn = ΔSproducts - ΔSreactants
= (238 J/mol*K) - (255 J/mol*K)
= -17 J/mol*K
Now we can calculate ΔG at 298 K using the equation:
ΔG = ΔH - TΔS
= (116.3 kJ/mol) - (298 K)(-17 J/mol*K)
= 116.3 kJ/mol + 5074 J/mol
= 121.4 kJ/mol
To find K at 600 degrees, we can use the equation:
ΔG = -RTln(K)
Where:
R is the gas constant (8.314 J/(mol*K))
T is the temperature in Kelvin
We need to convert 600 degrees to Kelvin:
T = 600 + 273.15
= 873.15 K
Now we can rearrange the equation to solve for K:
K = e^(-ΔG/RT)
Plugging in the values:
K = e^(-(121.4 kJ/mol)/(8.314 J/(mol*K))(873.15 K))
= e^(-145.94)
Using a calculator, the value of e^(-145.94) is approximately 1.34 x 10^(-63).
Therefore, the correct value for K at 600 degrees is approximately 1.34 x 10^(-63).