# Method for getting correct answer? Thank you!

What is the value of Kc at 226.85 °C for the reaction below if Kp is 6.24e-3 at this temperature?
CO(g) + 2H2(g) ↔ CH3OH(g)

a. 3.70e-6

b. 2.1614

c. 1.08e5

d. 10.5000

## To find the value of Kc at 226.85 °C for the given reaction, we can use the formula relating Kp and Kc. The formula is:

Kp = Kc(RT)^(∆n)

Where:
- Kp is the equilibrium constant in terms of partial pressures.
- Kc is the equilibrium constant in terms of concentrations.
- R is the ideal gas constant (0.0821 L·atm/(mol·K)).
- T is the temperature in Kelvin.
- ∆n is the difference in the number of moles of gaseous products and gaseous reactants.

In this case, the reaction is:
CO(g) + 2H2(g) ↔ CH3OH(g)

We can see that there is 1 mole of CO and 2 moles of H2 as reactants, and 1 mole of CH3OH as a product. Therefore,
∆n = (1) - (1+2) = -2

Given:
Kp = 6.24e-3
T = 226.85 °C = (226.85 + 273.15) K = 500 K

Now we can substitute the values into the formula:
6.24e-3 = Kc(0.0821 L·atm/(mol·K))(500 K)^(-2)

To solve for Kc, we rearrange the formula:
Kc = Kp / (R * T)^∆n

Substituting the values:
Kc = 6.24e-3 / (0.0821 L·atm/(mol·K))(500 K)^(-2)
Kc = 6.24e-3 / (0.0821 * 500^(-2))

Calculating the value of Kc gives:
Kc ≈ 10.5000

Therefore, the correct answer is (d) 10.5000.

## To find the value of Kc at 226.85 °C for the given reaction, we need to use the relationship between Kp and Kc. The equation for this relationship is:

Kp = Kc(RT)^(Δn)

Where Kp is the equilibrium constant in terms of partial pressures, Kc is the equilibrium constant in terms of molar concentrations, R is the gas constant, T is the temperature in Kelvin, and Δn is the change in the number of moles of gas particles.

In this case, we are given the value of Kp as 6.24e-3 at 226.85 °C, but we need to convert the temperature to Kelvin first.

To convert from degrees Celsius to Kelvin, we use the formula:

T(K) = T(°C) + 273.15

So, T(K) = 226.85 + 273.15 = 500 K.

Now, we need to determine the value of Δn, which is the difference in the number of moles of gas particles on the product side minus the number of moles of gas particles on the reactant side.

In the given reaction: CO(g) + 2H2(g) ↔ CH3OH(g)

The reactant side has 1 mole of CO(g) and 2 moles of H2(g), while the product side has 1 mole of CH3OH(g). So, Δn = (1-0) = 1.

Next, we substitute the given values into the equation and solve for Kc:

Kp = Kc(RT)^(Δn)
6.24e-3 = Kc(0.0821 L·atm/(mol·K))(500 K)^(1)
6.24e-3 = Kc(0.0821)(500)
6.24e-3 = 41.05 Kc

Now, we can solve for Kc by dividing both sides of the equation by 41.05:

Kc = 6.24e-3 / 41.05 = 1.519e-4

Therefore, the value of Kc at 226.85 °C for the given reaction is approximately 1.519e-4.

None of the answer choices provided match this value.