Hydrocarbons containing halogens are very useful molecules in synthetic chemistry. Calculate S° for the reaction product in the equation shown below if ΔS° for the reaction is -213.0 J K-1.
C2H4(g) + HI(g) --> C2H5I(l) (for HI(g) S°=206 J/mol/K)
I looked up in my text So for C2H4(g) (200.8 J/mol/K) since you didn't include that in your post. You should use the number in your text/notes and not the one I've used.
C2H4(g) + HI(g) --> C2H5I(l)
dSrxn = (n*So products) - (n*So reactants)
dSrxn = (n*So for C2H5I) - [(n*So for C2H4) + (n*So for HI)]
-213.0 = (1*So for C2H5I) - [(1*200.8) + (1*206)]
Solve for So for C2H5I. Post your work if you get stuck.
To calculate ΔS° for the reaction of C2H4(g) + HI(g) -> C2H5I(l), we need to use the "sum of products minus sum of reactants" formula for entropy change. The equation is as follows:
ΔS° = ΣS°(products) - ΣS°(reactants)
First, let's identify the number of each species in the reaction:
Reactants:
C2H4(g) - 1 mole
HI(g) - 1 mole
Products:
C2H5I(l) - 1 mole
Now, we need to determine the standard entropy (S°) value for each species.
Given:
S° for HI(g) = 206 J/mol/K
ΔS° for the reaction = -213.0 J/K
Using the information provided, we can calculate the ΔS° of the reaction:
ΔS° = [S°(C2H5I(l))] - [S°(C2H4(g)) + S°(HI(g))]
We know that S°(C2H4(g)) = 0 J/mol/K because it is an elemental form, and S° for an elemental form is 0.
Now, let's substitute in the given values and calculate:
ΔS° = [S°(C2H5I(l))] - [0 J/mol/K + 206 J/mol/K]
ΔS° = S°(C2H5I(l)) - 206 J/mol/K
Since the value of ΔS° for the reaction is -213.0 J K-1:
-213.0 J/K = S°(C2H5I(l)) - 206 J/mol/K
Rearranging the equation to solve for S°(C2H5I(l)):
S°(C2H5I(l)) = -213.0 J/K + 206 J/mol/K
S°(C2H5I(l)) = -7.0 J/mol/K
Therefore, the standard entropy (S°) for the product C2H5I(l) in the given reaction is -7.0 J/mol/K.
To calculate the standard entropy change (ΔS°) for the reaction product, you need to use the standard entropy values of the reactants and products.
The equation you provided is:
C2H4(g) + HI(g) → C2H5I(l)
First, let's calculate the standard entropy change for the reaction using the given ΔS° value:
ΔS° = -213.0 J K^-1
Now, you need to find the standard entropies of the reactants and the products:
S°(C2H4(g)) = ? (unknown)
S°(HI(g)) = 206 J/mol/K
S°(C2H5I(l)) = ? (unknown)
Since the reactants are in the gas phase and the product is in the liquid phase, we need to account for the difference in physical states. We can use the standard molar entropy of vaporization (ΔS°vap) to calculate the entropy change due to the change in physical state.
ΔS°vap = S°(C2H5I(l)) - S°(HI(g))
To find the value of ΔS°vap, we need to look up the standard molar entropy of vaporization for C2H5I(l). However, if that value is not provided, we will not be able to accurately calculate the standard entropy change for the reaction product.
Once you have the value for ΔS°vap, you can substitute it into the equation below to calculate S° for the reaction product:
ΔS° = S°(C2H5I(l)) - S°(HI(g)) - ΔS°vap
If ΔS°vap is not provided, you would need to make additional assumptions or approximations to determine the standard entropy change for the reaction product.