Asked by Patrick Panasko
Calculate the standard entropy, ΔS°rxn, of the following reaction at 25.0 °C using the data in this table. The standard enthalpy of the reaction, ΔH°rxn, is –44.2 kJ·mol–1.
C2H4(G)+H20 ---> C5H5OH
ΔS°rxn= ______ J*K^-1*mol^-1
Then, calculate the standard Gibbs free energy of the reaction, ΔG°rxn.
ΔG°rxn.=_______ kJ*mol^-1
∆Hf° ∆Gf° S°
C2H4(g) 52.4 68.4 219.3
H2O(l) –285.8 –237.1 70.0
C2H5OH(l) –277.6 –174.8 160.7
---------------------------------
Hint
The standard entropy of the reaction, ΔS°rxn, is equal to the sum of the standard entropies, ΔS°, of the products multiplied by their coefficients from the balanced equation, n, minus the sum of the ΔS° of the reactants multiplied by their coefficients, m.
ΔS°rxn=Products - Reactants
Then,use the values of ΔH°rxn and ΔS°rxn to calculate the standard Gibbs free energy of the reaction, ΔG°rxn.
C2H4(G)+H20 ---> C5H5OH
ΔS°rxn= ______ J*K^-1*mol^-1
Then, calculate the standard Gibbs free energy of the reaction, ΔG°rxn.
ΔG°rxn.=_______ kJ*mol^-1
∆Hf° ∆Gf° S°
C2H4(g) 52.4 68.4 219.3
H2O(l) –285.8 –237.1 70.0
C2H5OH(l) –277.6 –174.8 160.7
---------------------------------
Hint
The standard entropy of the reaction, ΔS°rxn, is equal to the sum of the standard entropies, ΔS°, of the products multiplied by their coefficients from the balanced equation, n, minus the sum of the ΔS° of the reactants multiplied by their coefficients, m.
ΔS°rxn=Products - Reactants
Then,use the values of ΔH°rxn and ΔS°rxn to calculate the standard Gibbs free energy of the reaction, ΔG°rxn.
Answers
Answered by
DrBob222
I can't read the values you have in the table because of the problem with spacing on this forum but here is what you do.
dSo = (n*dS products) - (n*dS reactants)
dH you have but it can be done the same way.
After you have dSo rxn and dHo rxn, then'
dGorxn - dH - TdS.
dSo = (n*dS products) - (n*dS reactants)
dH you have but it can be done the same way.
After you have dSo rxn and dHo rxn, then'
dGorxn - dH - TdS.
Answered by
eve
-128.6
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