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1. The problem statement, all variables and given/known data

Consider a chemical reaction CO (g) + 2H2 (g) <-> CH3OH(g). When 2 moles of CO were reacting with 4 moles of H2 at temperature 500K and pressure p = 100bar, and chemical equilibrium was reached, n = 366.0 moles of CO was consumed.

Based on the given data determine the value of the chemical equilibrium constant K(T) for T = 500K and find the mass (m) of the product obtained in the reaction.

2. Relevant equations

K_{p}= K_{c}(RT)^{Δn}

PV = nRT

aA + bB <=> cC + dD

K_{p}= [P_{C}]^{c}[P_{D}]^{d}/ [P_{A}]^{a}[P_{B}]^{b}

K_{c}= [C]^{c}[D]^{d}/[A]^{a}^{b}

3. The attempt at a solution

Okay so here is what I did. First the ice chart

CO (g) + 2H2 (g) <-> CH3OH

Ini 2mol 4mol -

Ch -(.366) -2(.366) +(.366)

Eq. 1.364 3.268 0.366 ∴Total Moles = 4.998 mol

Eq. mol fraction = (Eq. mol / Total Mol)

∴ CO = 0.273 & H_{2}= 0.654 & CH_{3}OH = 0.073

Eq. partial pressure = Eq. mol frac X pressure

∴ CO = 26.9 atm & H_{2}= 64.5 atm & CH_{3}OH = 7.2 atm

K_{p}= [7.2]/([26.9][64.5]^{2}) => 6.43 E -5

∴ K_{c}= Kp/(RT)^{Δn}= 0.108

To calculate mass I treated it as ideal gas so PV = nRT

Where n = total mol = 4.998mol; T = 500K; P = 100 bar = 98.6 atm; R = 0.082 atm dm^{3}K^{-1}mol^{-1}

Total Volume = 2.08 dm^{3}= 2.08 L

Now using density we can find the mass where density is 791.80g/L and d = m/V

∴ m = d x V = 791.80 x 2.08 = 1646.944 g

Please let know if this problem was approached in the correct way or not.

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# Homework Help: Determine the Kc and mass of product obtained

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