Question

Consider the following reaction.

Useful Information: 1. (5 pts) Consider the following reaction formation reaction below at 450 *C in a 375. mL container. The following species were found at equilibrium. 2.00 mol C (gr, s), H2 = 3.00 moles. O2 = 5.00 moles, CH3OH =2.00 moles. Calculate Kc at 450. "C. C(s) + 2 H2 (g) + 12 O2 (g) = CH3OH (1)

Answer 1

The equilibrium reaction is an example of heterogenous equilibrium which means that the components involved in the equilibrium are of different phases. The product CH₃OH is in liquid phase while on the reactant side we have hydrogen and oxygen in their gaseous phase and carbon in its solid phase.

We know that the equilibrium constant for the chemical reaction, Kc is given by the equation

Kc =

[concentration of products coefficiennt in the balanced equation [concentration of reactants coefficeint in the balanced equation

But in the case of heterogenous equilibria while calculating Kc concentration of pure solid and liquid species is considered as unity or is omitted from the equation for calculation of Kc.

This can be explained as follows,

we know that chemical potential is given by μ = μ° + RT ln a

When the compound is pure, μ = μ° so that the activity is 1. Thus, we can replace activity of each pure substance with 1.

Therefore, the activity of C_(s) and CH3OH_(l) in the equilibrium is 1.

Now the concentration of H₂ in Molarity (M) = (3 mol) × 0.375 L = 1.125 M

The concentration of O₂ (M) = (5 mol) × 0.375 l = 1.875 M

Now we know that PV =nRT

P = (n/V)RT but (n/V) is molarity or M

Therefore, P = MRT

For H₂

P_H2 = M_H2RT,

here R = 0.0821 LatmK^-1mol^-1and T = 450°C = 723 k and M_H2 = 1.125 M = 1.125 molL^-1

Substituting

P_H2 = 66.78 atm

Similarly for O₂

P_O2 = M_O2RT, where M_O2 = 1.875 M = 1.875 molL^-1

P_O2 = 111.13 atm

Now from the equilibrium equation given in the question

Kp = [1/ (P_H2)² × (P_O2)^0.5]

Kp = [1/(66.78)² × (111.13)^0.5] = [1/4459.57× 10.55] =

70年

Kp = 2.125 × 10^-5

Now Kc = Kp/(RT)^Δn

Here Δn = number of moles of product -number of moles of reactants = -2.5

Kc = (2.125 × 10^-5)/( 0.0821 × 723)^-2.5

Kc = (2.125 × 10^-5)/(59.36)^-2.5 = (2.125 × 10^-5)/(3.683 × 10^-5)

Kc = 0.577