Determine ∆G° (in kcal) for the reaction
CCl4(ℓ) + H2(g) HCl(g) + CHCl3(ℓ)
at 25 °C under standard conditions. The following information is available:
At 25 °C
C(graphite) + 2Cl2(g) CCl4(ℓ) ∆S° = -56.23 cal K-1
C(graphite) + Cl2(g) + H2(g) CHCl3(ℓ)
∆S° = -48.71 cal K-1
H2(g) + Cl2(g) HCl(g) ∆S° = 2.398 cal K-1
CCl4(ℓ) + H2(g) HCl(g) + CHCl3(ℓ) ∆H° = -21.83 kcal
In class, we learned that deltaG standard=deltaH-TdeltaS. I have T and deltaH, but how do I calculate deltaS from all these equations? How do I start this problem?
To calculate the standard Gibbs free energy of the reaction, you need to use the equation ΔG° = ΔH° - TΔS°. You already have the value of ΔH°, which is -21.83 kcal. To calculate ΔS°, you need to use the equations given to you and the equation ΔS° = ΔS°products - ΔS°reactants.
For the reaction CCl4(ℓ) + H2(g) → HCl(g) + CHCl3(ℓ), the ΔS°products is the sum of the ΔS° of the products, which is 2.398 cal K-1 + -48.71 cal K-1 = -46.312 cal K-1. The ΔS°reactants is the sum of the ΔS° of the reactants, which is -56.23 cal K-1. Therefore, ΔS° = -46.312 cal K-1 - (-56.23 cal K-1) = 9.918 cal K-1.
Now that you have all the values, you can calculate the standard Gibbs free energy of the reaction. ΔG° = -21.83 kcal - (25°C)(9.918 cal K-1) = -21.83 kcal - 247.95 cal = -269.78 kcal.
To calculate ∆G° for the reaction, you need to determine the entropy change (∆S°) for each step in the reaction and then apply the equation ∆G° = ∆H° - T∆S°, where T is the temperature in Kelvin.
To start the problem, you can follow these steps:
1. Write out the balanced chemical equation for the reaction:
CCl4(ℓ) + H2(g) → HCl(g) + CHCl3(ℓ)
2. Examine the given information to identify the relevant equations and their associated entropy changes (∆S°). In this case, you are given two equations with corresponding entropy changes:
a) C(graphite) + 2Cl2(g) → CCl4(ℓ), ∆S° = -56.23 cal K-1
b) C(graphite) + Cl2(g) + H2(g) → CHCl3(ℓ), ∆S° = -48.71 cal K-1
c) H2(g) + Cl2(g) → HCl(g), ∆S° = 2.398 cal K-1
3. Notice that equation (a) involves the formation of CCl4(ℓ), which is present on the reactant side of the given reaction equation. Therefore, you can reverse the equation (a) to obtain the correct sign for the entropy change: CCl4(ℓ) → C(graphite) + 2Cl2(g), ∆S° = +56.23 cal K-1.
4. Combine equations (a) and (b) to cancel out the C(graphite) term and obtain the net equation for the formation of CHCl3(ℓ):
2CCl4(ℓ) + H2(g) → 2CHCl3(ℓ), ∆S° = -48.71 cal K-1
5. Add equations (c) and (a) to get the net equation for the formation of HCl(g):
CCl4(ℓ) + H2(g) → HCl(g) + CHCl3(ℓ), ∆S° = 2.398 cal K-1 + 56.23 cal K-1 = 58.628 cal K-1
6. Now that you have determined the ∆S° values for each step of the reaction, you can calculate the ∆G° using the equation ∆G° = ∆H° - T∆S°, where T is the temperature in Kelvin (25 °C = 298 K).
Substitute the values into the equation:
∆G° = -21.83 kcal - (298 K) * (58.628 cal K-1)
Convert the units of cal to kcal:
1 kcal = 1000 cal
∆G° = -21.83 kcal - (298 K) * (58.628 cal/1000 cal kcal-1)
Calculate the value to get the final answer for ∆G° in kcal.
To calculate ∆S° for the reaction, you can use Hess's Law which states that if a reaction can be expressed as the sum of two or more chemical equations, then the ∆S° for the overall reaction is equal to the sum of the ∆S° values for the individual reactions.
In this case, you have two given reactions:
1) C(graphite) + 2Cl2(g) CCl4(ℓ) ∆S° = -56.23 cal K-1
2) C(graphite) + Cl2(g) + H2(g) CHCl3(ℓ) ∆S° = -48.71 cal K-1
By looking at these two equations, it is clear that the overall reaction is obtained by combining them in a way that cancels out the common species:
CCl4(ℓ) + H2(g) HCl(g) + CHCl3(ℓ)
To do this, we can reverse the second equation and multiply both equations by the appropriate coefficients:
1) C(graphite) + 2Cl2(g) CCl4(ℓ) ∆S° = -56.23 cal K-1
2) -[C(graphite) + Cl2(g) + H2(g)] -[CHCl3(ℓ)] ∆S° = 48.71 cal K-1
Now we can add these two equations together to get the overall reaction:
CCl4(ℓ) + H2(g) HCl(g) + CHCl3(ℓ)
∆S° overall = -56.23 cal K-1 + 48.71 cal K-1
∆S° overall = -7.52 cal K-1
Now that you have ∆H° = -21.83 kcal and ∆S° = -7.52 cal K-1, you can use the equation ∆G° = ∆H° - T∆S° to calculate ∆G°. As you mentioned, you have the temperature (25 °C) to calculate ∆G° under standard conditions. Just make sure to convert the units to match (kcal and K).
Hope this helps!