nH2 nCO nCO2

nN2nH2O nCH3OH

Methanol Synthesis Reactor

75-mol-% H215-mol-% CO5-mol-% CO25-mol-% N2

T=550KP=100 bar

Reaction 1: 2H2(g) + CO(g) CH3OH(g)K1 = 6.749 x 10-4, v = -2

Reaction 2: H2(g) + CO2(g) CO(g) + H2O(g)K2 = 0.01726, v = 0

75 moles feed gas

The feed gas to a methanol synthesis reactor is composed of 75-mol-% H2, 15-mol-% CO, 5-mol-% CO2, and 5-mol-% N2. The system comes to equilibrium at 550K and 100 bar with respect to the reactions.

2H2(g) + CO(g) CH3OH(g)H2(g) + CO2(g) CO(g) + H2O(g)

Assuming ideal gases, determine the composition of the equilibrium mixture.

Given:K1 = 6.749 x 10-4

K2 = 0.01726

The plant has a capacity of 75 moles per operation of feed gas.

Block Diagram

Basis1 mole feed gas

Mass Balance Equations

Let: no = 1 molnoH2 = 0.75 molnoCO = 0.15 molnoCO2 = 0.05 molnoN2 = 0.05 molε1 = equilibrium conversion of reaction 1ε2 = equilibrium conversion of reaction 2

H2 Balance: nH2 = 0.75 - 2 ε1 - ε2 (1)

CO Balance: nCO = 0.15 - ε1 + ε2 (2)

CO2 Balance: nCO2 = 0.05 - ε2 (3)

N2 Balance: nN2 = 0.05 (4)

H2O Balance: nH2O = ε2 (5)

CH3OH Balance: nCH3OH = ε1 (6)

Total Balance: n = 1 - 2 ε1 (7)

Solution

Assumption: Input and exit streams are ideal gases.

For multiple reactions: Π yvi,j = (P/Po)-vj Kj

For reaction 1:

yCH3OH / [yCO (yH2)2] = (100/1)2 (6.749 x 10-4)

nCH3OH (n2) / [nCO (nH2)2] = 6.749

Equation 1: ε1 (1-2 ε1)2 / [(0.15 - ε1 + ε2) (0.75 - 2 ε1 – ε2)2] = 6.749

For reaction 2:

yH2O (yCO) / [yH2 (yCO2)] = (100/1)0 (0.01726)

nH2O (nCO) / [nH2 (nCO2)] = 0.01726

Equation 2: ε2 (0.15 - ε1 + ε2) / [(0.75 - 2 ε1 – ε2) (0.05 - ε2)] = 0.01726

Solve 2 equations and 2 unknowns using Microsoft Office Excel Solver Tool.

Snapshot of Excel Solver Tool with initial guesses of ε1 = 0.01 and ε2 =0.01.

With the constraints that the right hand sides of equations 1 and 2 are equal to their respective numerical left hand sides. We use an objective function where F(x) = f1

2 + f22. The

target value of this function is 0. The functions f1 and f2

are defined as:

f1(x) = ε1 (1-2 ε1)2 / [(0.15 - ε1 + ε2) (0.75 - 2 ε1 – ε2)2] - 6.749

f2(x) = ε2 (0.15 - ε1 + ε2) / [(0.75 - 2 ε1 – ε2) (0.05 - ε2)] - 0.01726

By using Microsoft Office Excel Solver Tool:

The computed ε1 = 0.1186 and the ε2 = 0.0089.

Thus, the equilibrium composition of exit stream is:

yH2 = nH2 / n = (0.75 - 2 ε1 - ε2) / (1-2 ε1) = 0.6606

yCO = nCO / n = (0.15 - ε1 + ε2) / (1-2 ε1) = 0.0528

yCO2 = nCO2 / n = (0.05 - ε2) / (1-2 ε1) = 0.0539

yN2 = nN2 / n = 0.05 / (1-2 ε1) = 0.0655

yH2O = nH2O / n = ε2 / (1-2 ε1) = 0.0116

yCH3OH = nCH3OH / n = ε1 / (1-2 ε1) = 0.1555

The molar flow rates of the exit streams using a scale-up ratio of 75 are:

nH2 = 0.6606 x 75 = 49.55 mol

nCO = 0.0528 x 75 = 3.96 mol

nCO2 = 0.0539 x 75 = 4.04 mol

nN2 = 0.0655 x 75 = 4.91 mol

nH2O = 0.0116 x 75 = 0.87 mol

nCH3OH = 0.1555 x 75 = 11.66 mol

Summary of Answers

Exit Stream Component Equilibrium Composition Molar Flow Rate (moles)

H2 0.6606 49.55

CO 0.0528 3.96

CO2 0.0539 4.04

N2 0.0655 4.91

H2O 0.0116 0.87

CH3OH 0.1555 11.66

Total 0.9999 ≈ 1 74.99 ≈ 75