Given the thermodynamic data in the table below, calculate the equilibrium constant for the reaction:
2SO2(g)+O2--> 2SO3
Substance (DeltaH^o) (Delat S^o)
SO2 -297 249
O2 0 205
SO3 -395 256
Answer (it was given) 2.32x10^24
Even though the answer is given (2.32x10^24) I just can't get there. I calculated:
(Delta H^o)= ((2moles)(-395))-((2moles)(-297)+(1mole)(0))=-196
(Delta S^o)= ((2moles)(256))-((2moles)(249)+(1mole)(205)=-191
From here I thought: If I need to find the equilibrium constant K, I can use the formula (Delta G^0)= -RT ln K and thus K= e^(-(DeltaG^o)/RT)
To get there I figure I would need to find (Delta G^o)...which is possible because I do have (Delta H^o) and (Delta S^o) and I can use the formula
(Delta G^o)= (Delta H^o) - T(Delta S^o) ...I assumed that the temperature is 298K and that didn't work and then I calculated T (T= ((Delta H^o)/(Delta S^o))...I get 1026K and then I put that into (Delta G^o)= (Delta H^o) - T(Delta S^o)...and then solved all the rest but that doesn't work out either. Could you tell me were I am wrong or write down how you would solve the problem?
Thank You!!
I do not believe in the correctness ofyour assumption
T= ((Delta H^o)/(Delta S^o))
They should have told you what temperature to assume. delta S^o values are usually quoted for 298 K.
I agree with your
K= e^(-(DeltaG^o)/RT)
and your delta S and delta H calculations, but I am rusty on this subject after 45+ years
Our resident chemical expert is still away, and so it is possible that you won't be getting any other responses to this soon.
a little lat ebut i do believe you forgot to put delta s into kJ as the base unit is joules
To calculate the equilibrium constant (K) for the reaction, you can use the equation:
ΔG° = -RTln(K)
where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and ln represents the natural logarithm.
To determine ΔG°, you can use the equation:
ΔG° = ΔH° - TΔS°
where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change.
Let's break down the calculation step by step:
1. Calculate ΔH° for the reaction:
ΔH° = (2 mol)(ΔH°SO3) - [(2 mol)(ΔH°SO2) + (1 mol)(ΔH°O2)]
ΔH° = (2)(-395) - [(2)(-297) + (1)(0)]
ΔH° = -790 + 594
ΔH° = -196 kJ/mol
2. Calculate ΔS° for the reaction:
ΔS° = (2 mol)(ΔS°SO3) - [(2 mol)(ΔS°SO2) + (1 mol)(ΔS°O2)]
ΔS° = (2)(256) - [(2)(249) + (1)(205)]
ΔS° = 512 - 703
ΔS° = -191 J/(mol·K)
3. Calculate T:
T = (ΔH°) / (ΔS°)
T = (-196 kJ/mol) / (-191 J/(mol·K)) [Note: convert kJ to J]
T = 0.196/(0.191×10^-3 K)
T = 1026 K (rounded to the nearest whole number)
4. Substitute the values into the equation ΔG° = -RTln(K):
ΔG° = (-8.314 J/(mol·K))(1026 K) ln(K)
Now, let's solve for K:
ΔG° = -RTln(K)
-196,000 J/mol = (-8.314 J/(mol·K))(1026 K) ln(K)
Solve for ln(K):
ln(K) = -196,000 J/mol / (-8.314 J/(mol·K))(1026 K)
ln(K) = 196,000 J / (8.314 J·K)(1026 K)
Solve for K:
K = e^(ln(K)) [Note: e is the base of the natural logarithm]
K = e^(196,000 / (8.314×10^−3)(1026))
K ≈ 2.32 × 10^24 (rounded to two significant figures)
Therefore, the equilibrium constant (K) for the reaction: 2SO2(g) + O2 → 2SO3 is approximately 2.32 × 10^24.