Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is:
C2H2(g) + 2H2(g) === C2H6(g)
Given the following, what is the Kp for the reaction?
C2H2(g): 209.2 ΔG˚f (kJ/mol)
H2(g): 0 ΔG˚f (kJ/mol)
C2H6(g): -32.89ΔG˚f (kJ/mol)
ΔG˚ = -RTlnK
ΔG = ΔG˚RTlnQ
I don't what the answer is.
ΔG˚ for reaction = - RT lnK
242,090 J/mol = - (8.314 J/mol-K)(298.15 K) lnK
K = Kp = 3.85E-43
That's what I got but it wouldn't go through. Although that seems to be an extremely small number.
the delta G of the whole reaction you need to use is found by subtracting the delta G for your reactants from the delta G of your product
-32.89 (c2h6) - 209.2 (c2h2) + 0 (h2) = -242.09
then you have to plug it into the equation deltaG = -RTlnK, making sure to convert R to kJ by dividing it by 1000
8.314/1000 = 0.008314
and use the standard temp, 298.15K for T
-242.09 = -0.008314*298.15*lnK
this can be rewritten as
K=e^(G/-RT)
so
K=e^(-242.09/-0.008314*298.15)
K=e^97.713
the answer is 2.73E42
looks like your only problem was keeping your deltaG positive
To find the Kp for the reaction, we need to use the equation ΔG˚ = -RTlnK, 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 (298.15 K), and K is the equilibrium constant.
We can calculate the value of ΔG˚ for the reaction using the given ΔG˚f values for the reactants and products:
ΔG˚ = ΔG˚f(C2H6) + 2ΔG˚f(H2) - ΔG˚f(C2H2)
= (-32.89 kJ/mol) + 2(0 kJ/mol) - (209.2 kJ/mol)
= -32.89 kJ/mol - 209.2 kJ/mol
= -242.09 kJ/mol
Now, we can rearrange the equation to solve for K:
-242,090 J/mol = - (8.314 J/mol-K)(298.15 K) ln(K)
ln(K) = -242,090 J/mol / ((8.314 J/mol-K)(298.15 K))
ln(K) ≈ -31.248
K ≈ e^(-31.248)
K ≈ 1.3 x 10^(-14)
Therefore, the Kp for the reaction is approximately 1.3 x 10^(-14).
To calculate the Kp for the reaction, you need to use the equation ΔG˚ = -RTlnK, where ΔG˚ represents the standard Gibbs free energy change, R is the gas constant (8.314 J/mol-K), T is the temperature in Kelvin, and lnK is the natural logarithm of the equilibrium constant.
Given that ΔG˚f (Gibbs free energy of formation) for C2H2 is 209.2 kJ/mol, ΔG˚f for H2 is 0 kJ/mol, and ΔG˚f for C2H6 is -32.89 kJ/mol, you can calculate the ΔG˚ for the reaction.
ΔG˚ = Σ(ΔG˚f products) - Σ(ΔG˚f reactants)
ΔG˚ = (-32.89 kJ/mol) - (209.2 kJ/mol + 2(0 kJ/mol))
ΔG˚ = -32.89 kJ/mol - 209.2 kJ/mol
ΔG˚ = -242.09 kJ/mol
Now, convert ΔG˚ to J/mol:
ΔG˚ = -242.09 kJ/mol × 1000 J/kJ
ΔG˚ = -242,090 J/mol
Next, you can substitute the values into the equation ΔG˚ = -RTlnK and solve for K.
-242,090 J/mol = - (8.314 J/mol-K)(298.15 K) lnK
Rearranging the equation and solving for lnK:
lnK = (-242,090 J/mol) / ((8.314 J/mol-K)(298.15 K))
lnK ≈ -34.978
Using the natural logarithm ln, you can now calculate the value of K:
K = e^(lnK)
K ≈ e^(-34.978)
K ≈ 3.85 × 10^(-43)
So, the value of Kp for the reaction is approximately 3.85 × 10^(-43). It is indeed an extremely small number, which indicates that the reaction proceeds in the direction of the products to a very small extent.