Calculate ∆G (in kJ) at 541 K for the following reaction under the given conditions. Use data obtained from the table below and assume that ∆Hfo and So do not vary with temperature. Report your answer to three significant figures in scientific notation (i.e. 1.23E4).
2 H2S (g) + 3 O2 (g) ⇌ 2 SO2 (g) + 2 H2O (l)
Compound ÄHfo So P (atm)
H2S (g) -20.6 205.81 2.3
O2 (g) 0 205.7 9.9
SO2 (g) -296.81 248.2 6.3
H2O (l) -285.83 69.91 ---
I know how to do the question, I was just confused about the Q value, it turns out to be a negative and so I can't take the ln of it. What do I do?
Can you turn the reaction around (which will change the sign of K?) and work it that way. Then after finding delta G for the reverse direction, that can be reversed again for delta G in the original direction.
I didn't post three of the same answers on purpose.
To calculate ∆G (the change in Gibbs free energy) for the given reaction at 541 K, you can use the equation:
∆G = ∆G° + RT * ln(Q)
Where:
∆G° is the standard Gibbs free energy change
R is the gas constant (8.314 J/(mol*K) or 0.008314 kJ/(mol*K))
T is the temperature in Kelvin (541 K in this case)
Q is the reaction quotient
First, let's calculate the reaction quotient (Q) for the reaction at the given conditions. Q is calculated by substituting the partial pressures of the reactants and products into the reaction equation and taking their ratios, each raised to the power of their stoichiometric coefficients:
Q = (P(SO2)^2 * P(H2O)^2) / (P(H2S)^2 * P(O2)^3)
From the table, we can find the partial pressures for each compound at the given conditions:
P(H2S) = 2.3 atm
P(O2) = 9.9 atm
P(SO2) = 6.3 atm
Since P(H2O) is not provided, we can assume its partial pressure to be 1 atm as it is in the liquid state. Thus,
Q = (6.3^2 * 1^2) / (2.3^2 * 9.9^3)
Next, calculate the ∆G° (standard Gibbs free energy change) for the reaction. ∆G° is calculated using the standard enthalpy of formation (∆H°f) values of the compounds involved in the reaction:
∆G° = Σ(n * ∆H°f(products)) - Σ(n * ∆H°f(reactants))
Where n is the stoichiometric coefficient of each compound.
For the given reaction:
∆G° = (2 * ∆H°f(SO2) + 2 * ∆H°f(H2O)) - (2 * ∆H°f(H2S) + 3 * ∆H°f(O2))
∆H°f(SO2) = -296.81 kJ
∆H°f(H2O) = -285.83 kJ
∆H°f(H2S) = -20.6 kJ
∆H°f(O2) = 0 kJ
Substituting the values, ∆G° can be calculated.
With ∆G° and Q, you can now substitute these values into the equation:
∆G = ∆G° + RT * ln(Q)
To address your specific concern, if Q turns out to be negative and taking the natural logarithm (ln) of a negative number is undefined. However, in this case, if Q is negative, it implies that the reaction is not at equilibrium. In this scenario, you need to go back and check your calculations. Double-check the partial pressure values and ensure they are correct.
Once you have corrected any errors and obtained the correct Q value, you can proceed with plugging it into the equation to calculate ∆G.