Calculate ΔG° for the following reaction at 298 K.?
A+B --> 2D ΔH° = 775.0 kJ ΔS°=296.0 J/K
C --> D ΔH° = 446.0 kJ ΔS°=-218.0 J/K
A+B --> 2C
ΔG° = ? kJ
I know I am suppose to reverse the (C --> D) and add a coefficient of 2 to both the enthalpy and entropy.
I got this but it doesn't seem right.
ΔH° = 2(-446.0) - (-722.0) = -1674 kJ
ΔS° = 2(218) - (296) = .14kJ/k
ΔG° = -1674 - (.14)(298) =-1688.7 a really large number that does not seem right,
Please help and explain this to me.
You didn't change to the delta S into kJ.
To calculate the standard free energy change (ΔG°) for the reaction A+B → 2C, you need to consider the free energy changes for the individual reactions and apply the appropriate calculations.
Given:
Reaction 1: A+B → 2D
ΔH° = 775.0 kJ
ΔS° = 296.0 J/K
Reaction 2: C → D
ΔH° = 446.0 kJ
ΔS° = -218.0 J/K
To combine these reactions to get A+B → 2C, you can use the following steps:
Step 1: Reverse Reaction 2
Flipping the direction of reaction 2 gives: D → C
Step 2: Multiply the ΔH° and ΔS° values for Reaction 2 by 2
Since we want to combine two molecules of C, we need to double the enthalpy (ΔH°) and entropy (ΔS°) changes. Therefore:
ΔH° for 2D → 2C = 2 * (-446.0) kJ = -892.0 kJ
ΔS° for 2D → 2C = 2 * (-218.0) J/K = -436.0 J/K
Step 3: Combine the reactions to get the overall reaction A+B → 2C
Adding up the enthalpy (ΔH°) and entropy (ΔS°) changes from the individual reactions:
ΔH° overall = ΔH° for A+B → 2D + ΔH° for 2D → 2C
= 775.0 kJ + (-892.0 kJ) = -117.0 kJ
ΔS° overall = ΔS° for A+B → 2D + ΔS° for 2D → 2C
= 296.0 J/K + (-436.0 J/K) = -140.0 J/K
Step 4: Convert ΔS° to kJ/K
Since ΔH° is in kJ, you should convert ΔS° to kJ/K by dividing it by 1000:
ΔS° overall = -140.0 J/K ÷ 1000 = -0.14 kJ/K (note the change in sign)
Step 5: Calculate ΔG° using the equation:
ΔG° = ΔH° - TΔS°
Plugging in the values:
ΔG° = -117.0 kJ - (298 K * -0.14 kJ/K) = -117.0 kJ + 41.72 kJ = -75.28 kJ
Therefore, the standard free energy change (ΔG°) for the reaction A+B → 2C at 298 K is -75.28 kJ.
To calculate the standard Gibbs free energy change (ΔG°) for a reaction, you can use the equation:
ΔG° = ΔH° - TΔS°
where ΔH° is the change in enthalpy and ΔS° is the change in entropy, both expressed in the same units (kJ). T represents the temperature in Kelvin (K), which in this case is given as 298 K.
Let's break down the calculations step by step for the given reaction A+B --> 2C:
1. Reverse the reaction C --> D to obtain D --> C, and adjust the enthalpy and entropy:
ΔH°(D --> C) = - ΔH°(C --> D) = - (-446.0 kJ) = 446.0 kJ
ΔS°(D --> C) = - ΔS°(C --> D) = - (-218.0 J/K) = 218.0 J/K
2. Multiply the enthalpy and entropy values of the reversed reaction by 2 to account for the stoichiometric coefficients:
ΔH°(2D --> 2C) = 2 * 446.0 kJ = 892.0 kJ
ΔS°(2D --> 2C) = 2 * 218.0 J/K = 436.0 J/K
3. Now calculate ΔG° using the equation:
ΔG° = ΔH° - TΔS°
Substituting the values:
ΔG° = 892.0 kJ - (298 K * 0.436 kJ/K) = 892.0 kJ - 128.728 kJ = 763.272 kJ
So, with the correct calculations, the ΔG° for the reaction A+B --> 2C at 298 K is 763.272 kJ.
I agree with your approach with the coefficients of 2 and reverse the C==>D. But aren't you supposed to add? You're adding the equations to get the final equation; therefore, you add dH and dS. And where did the 722 come from?
I believe you are confusing this with dH rxn = (n*dHproducts) - (n*dHreactants)
dH = 775 kJ + (2*-446) = ?
dS = 296 J + (2*218) = ?
Then dGo = dH - TdS
Don't forget to change dH to J; it's in kJ now.