3. Methanol can be produced from methyl formate (HCOOCH3) by the following reaction: HCOOCH3 (g)+2 H2 (g) → 2CH3OH (g) Methyl formate can also undergo the following reaction: HCOOCH3 (g) → 2CH3OH (g) +CO (g) reaction 1 reaction 2 If the molar ratio of hydrogen to methyl formate is 4:1, estimate the gas composition at equilibrium at 40 bar if the temperature is: 500 K i. ii. 600 K iii. 800 K The equilibrium constants for the reactions are given by the following equations: 5772.5 In Ka1 - 4.81 lnT + 1.72 x 10-3 T – 6.79 x 10-7 T² + 19.1 T. 5130 In Ka2 + 3.16 InT – 4.023 x 10-3 T + 1.55 x 10-6 T2 – 14.15 T
3. Methanol can be produced from methyl formate (HCOOCH3) by the following reaction: HCOOCH3 (g)+2 H2 (g) → 2CH3OH (g) Methyl formate can also undergo the following reaction: HCOOCH3 (g) → 2CH3OH (g) +CO (g) reaction 1 reaction 2 If the molar ratio of hydrogen to methyl formate is 4:1, estimate the gas composition at equilibrium at 40 bar if the temperature is: 500 K i. ii. 600 K iii. 800 K The equilibrium constants for the reactions are given by the following equations: 5772.5 In Ka1 - 4.81 lnT + 1.72 x 10-3 T – 6.79 x 10-7 T² + 19.1 T. 5130 In Ka2 + 3.16 InT – 4.023 x 10-3 T + 1.55 x 10-6 T2 – 14.15 T
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Transcribed Image Text:3. Methanol can be produced from methyl formate (HCOOCH3) by the following reaction:
HCOOCH3 (g)+ 2 H2 (g) → 2CH3OH (g)
Methyl formate can also undergo the following reaction:
HCOOCH3 (g) → 2CH3OH (g) +CO (g)
reaction 1
reaction 2
If the molar ratio of hydrogen to methyl formate is 4:1, estimate the gas composition at equilibrium at 40 bar if the
temperature is:
i.
500 K
ii.
600 K
iii.
800 K
The equilibrium constants for the reactions are given by the following equations:
5772.5
In Ka1 =
T.
4.81 InT + 1.72 x 10-3 T – 6.79 x 10-7 T² + 19.1
5130
In Ka2
+ 3.16 lnT - 4.023 x 10-3 T + 1.55 x 10-6 T² – 14.15
T
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