Chemical Forums

Please login or register.

Login with username, password and session length

Sponsored links

Pages: [1]   Go Down

Author Topic: Determine the Kc and mass of product obtained ??  (Read 445 times)

0 Members and 1 Guest are viewing this topic.

rash219

  • Very New Member
  • *
  • Mole Snacks: +0/-0
  • Offline Offline
  • Posts: 1
Determine the Kc and mass of product obtained ??
« on: December 05, 2012, 09:30:48 PM »

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

Kp = Kc (RT)Δn
PV = nRT

aA + bB <=> cC + dD

Kp and Kc standard formulae

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 & H2 = 0.654 & CH3OH = 0.073
Eq. partial pressure = Eq. mol frac X pressure
∴ CO = 26.9 atm & H2 = 64.5 atm & CH3OH = 7.2 atm

Kp = [7.2]/([26.9][64.5]2) => 6.43 E -5
∴ Kc = 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 dm3 K-1 mol-1
Total Volume = 2.08 dm3 = 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.
Logged

Pages: [1]   Go Up
 

Mitch Andre Garcia's Chemical Forums 2003-Present.

Page created in 0.05 seconds with 23 queries.