a. Calculate the delta H, delta G, and equilibrium constant for the reaction of methane with Cl2 to give CH3Cl and HCl. Use the following information for your calculations:

Bond Dissociation Energies:
CH3-H (104.8kcal/mol), Cl-Cl(59.0 kcal/mol), CH3-Cl(85.0 kcal/mol), and H-Cl (103.2 kcal/mol).
Thermodynamic data:
DeltaSrxn= + 0.29 x 10^.3 kcal/mol K, T=0 degrees celcius, and R= 1.987 cal/mol


find what K=

I got that delta H=-24.4 and delta G=-24.3... But keep getting the equilibrium constant (K) wrong... Can someone please help...

To calculate the equilibrium constant (K) for the reaction, we can use the equation:

ΔG° = -RT ln K

where ΔG° is the standard Gibbs free energy change, R is the gas constant (1.987 cal/mol K), T is the temperature in Kelvin, and K is the equilibrium constant.

First, let's calculate ΔH° for the reaction by summing the bond dissociation energies of the bonds broken and subtracting the energies of the bonds formed:

ΔH° = ∑(bond energies of bonds broken) - ∑(bond energies of bonds formed)

For the reaction:

2CH4 + Cl2 → 2CH3Cl + H2

ΔH° = (2 × 104.8 kcal/mol) + (1 × 59.0 kcal/mol) - (2 × 85.0 kcal/mol) - (1 × 103.2 kcal/mol)
= 209.6 kcal/mol + 59.0 kcal/mol - 170.0 kcal/mol - 103.2 kcal/mol
= -4.6 kcal/mol

Next, we can calculate ΔG° using the equation:

ΔG° = ΔH° - TΔS°

where ΔS° is the standard entropy change.

Given ΔS° = +0.29 × 10^3 kcal/mol K, T = 0 degrees Celsius = 273.15 K, and R = 1.987 cal/mol K, we can convert the units and calculate ΔG°:

ΔG° = -0.0046 kcal/mol - (273.15 K) × (0.29 × 10^(-3) kcal/mol K)
= -0.0046 kcal/mol - 0.079 kcal/mol
= -0.0836 kcal/mol

Finally, we can use the equation ΔG° = -RT ln K to solve for K:

K = e^(-ΔG° / RT)

K = e^(-(-0.0836 kcal/mol) / (1.987 cal/mol K × 273.15 K))
≈ 2.90 × 10^(-6)

Therefore, the equilibrium constant (K) for the reaction is approximately 2.90 × 10^(-6).