A chemical engineer is studying the following reaction: HCH 3 CO 2(aq)\" CH 3 NH
ID: 587445 • Letter: A
Question
A chemical engineer is studying the following reaction: HCH 3 CO 2(aq)" CH 3 NH 2(aq) CH 3 CO 2 (aq)+CH,NH;(aq) At the temperature the engineer picks, the equilibrium constant K for this reaction is 0.85 The engineer charges ("fills") four reaction vessels with acetic acid and methylamine, and lets the reaction begin. He then measures the composition of the mixture inside each vessel from time to time. His first set of measurements are shown in the table below Predict the changes in the compositions the engineer should expect next time he measures the compositions reaction vessel compound HCH3CO2 CH,NH expected change in concentration decrease decrease decrease decrease decrease decrease decrease decrease decrease decrease decrease decrease concentration 0.42 M 0.54 M 1.13 M 0.95 M (no change) (no change) (no change) (no change) (no change) (no change) (no change) (no change) (no change) (no change) (no change) (no change) increase increase CH,CO CHjNHj HCH3Co CH3NH2 CH,CO2 CH3NH3 HCH3CO2 CH3NH2 CH,CO CH3NH3 l increase increase 0.73 M l increase increase increase increase 0.85 M 0.82 M 0.64 M 1.02 M 1.19 M 1.15 M l increase l increase T increase 0.90 M l increaseExplanation / Answer
Reaction equilibrium
For the given reaction Kc is less than one, that means the concentration of products are lower than the concentration of reactants.
After certain time the observed changes would be,
A HCH3CO2 decrease
CH3NH2 decrease
CH3CO2- increase
CH3NH3+ increase
Kc = (0.95 x 1.13)/(0.42 x 0.54) = 4.73
So it would reduce concentration of substrates and increase concentration of products until Kc = 0.85
B HCH3CO2 no change
CH3NH2 no change
CH3CO2- no change
CH3NH3+ no change
Kc = (0.82 x 0.64)/(0.73 x 0.85) = 0.85
No change already at equilibrium
C HCH3CO2 no change
CH3NH2 no change
CH3CO2- no change
CH3NH3+ no change
Kc = (1.15 x 0.90)/(1.02 x 1.19) = 0.85
No change already at equilibrium
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.