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 wrong.•
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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
If you have delta G, then
dG = -RTlnK
Substitute and solve for K
To find ΔH°rxn, ΔG°rxn, and ΔS°rxn at 298 K, we can use the following equations:
ΔH°rxn = ΣΔH°products - ΣΔH°reactants,
ΔG°rxn = ΣΔG°products - ΣΔG°reactants, and
ΔS°rxn = ΣΔS°products - ΣΔS°reactants.
Here's a step-by-step calculation:
(a) ΔH°rxn:
The product is styrene (C6H5--CH==CH2), and the reactant is ethylbenzene (C6H5--CH2CH3).
ΔH°rxn = ΔH°styrene - ΔH°ethylbenzene
= (103.8 kJ/mol) - (-12.5 kJ/mol)
= 116.3 kJ/mol
Therefore, ΔH°rxn at 298 K is 116.3 kJ/mol.
(b) ΔG°rxn:
Using a similar methodology:
ΔG°rxn = ΔG°styrene - ΔG°ethylbenzene
= (202.5 kJ/mol) - (119.7 kJ/mol)
= 82.8 kJ/mol
Therefore, ΔG°rxn at 298 K is 82.8 kJ/mol.
(c) ΔS°rxn:
Again, using the same method:
ΔS°rxn = ΔS°styrene - ΔS°ethylbenzene
= (238 J/mol*K) - (255 J/mol*K)
= -17 J/mol*K
Therefore, ΔS°rxn at 298 K is -17 J/mol*K.
To find the equilibrium constant (K) at 600 degrees, we can use the equation:
ΔG°rxn = -RT ln(K)
First, we need to convert the temperature to Kelvin:
600 degrees Celsius = 873 K.
Then, the equation becomes:
82.8 kJ/mol = -(8.314 J/mol*K)(873 K) ln(K)
Simplifying:
82.8 kJ/mol = -7243 J ln(K)
Dividing both sides by -7243 J:
ln(K) = -0.0114
Taking the exponential of both sides:
K = e^(-0.0114)
Calculating this:
K ≈ 0.988
Therefore, the equilibrium constant (K) at 600 degrees is approximately 0.988.