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Dhamnekar Winod
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Equilibrium constant questions

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Consider the reaction:
CO(g) + 2 H2(g) → CH3OH(g)

a]When 0.20 mole CO are mixed with 0.40 mole H2 in an empty 1.0 L flask, 0.10 mole of CH3OH are obtained when equilibrium is established. Calculate the equilibrium constant.

b]If 0.10 mole of CH3OH are added to the equilibrium mixture of part a, how much CH3OH would be present when the equilibrium is reestablished?

c]How much CH3OH would be obtained at equilibrium if the reaction were started with 0.40 mole CO and 0.40 H2?

d]How much H2 must be mixed with 0.20 mole of CO in an empty 1.0 L flask in order to obtain 0.20 mole of CH3OH at equilibrium?

e]At a higher temperature, 0.05 mole of CH3OH is obtained when the reaction is started with 0.20 mole CO and 0.40 mole of H2 in a 1.0 L flask. Is the reaction exothermic or endothermic? Explain.

f]If the pressure is reduced by increasing the volume of the flask to 2.0 L, what will happen to the amount of CH3OH present at equilibrium?

I am working on these questions. What would be answers to these questions?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Equilibrium constant questions

Post by ChenBeier »

Show your ideas first.
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Re: Equilibrium constant questions

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Answers:-
CO(g) + 2 H2(g) <=> CH3OH(g)
a. When 0.10 mole of CH3OH(Methanol) are obtained, 0.1 mole of CO(g) and 0.2 mole of H2(g) must be consumed.

Hence \(K_c= \frac{CH_3OH}{(CO) (H_2)^2} \Rightarrow \frac{0.10}{(0.1)(0.2)^2}=25\)

b. If 0.10 mole of CH3OH is added to equilibrium mixture of part a, let X mole of CH3OH would be present when equilibrium is reestablished.
\(K_c=25=\frac{(0.2-x)}{(0.1+x)(0.2+2x)^2}\Rightarrow x=0.0213412 M\) So, CH3OH =0.2-0.0213412=0.1787 M would be present when the equlbrium is reestablished



c. 0.13044 M of CH3OH would be obtained at equilibrium if the reaction were started with 0.40 mole of CO and 0.40 mole of H2. It is computed as follows:-\(25=\frac{x}{(0.4-x)(0.4-2x)^2}\Rightarrow x=0.13044 M\)

d. Infinite amount of H2. Formation of 0.2 M CH3OH, requires the consumption of 0.2 M CO. That won't happen even with large amount of H2.

e. \(K_c= \frac{0.05}{0.15* 0.3^2}=3.7037\) as \(K_c\) is smaller, reaction is exothermic.

f. If the pressure is reduced by increasing the volume of the flask to 2.0 L, the equilibrium will shift to the side with more moles of gas, so the amount of CH3OH will decrease.
These are my workings on all the six questions. Please express your remarks, reviews on these workings.
Last edited by Dhamnekar Winod on Mon Mar 08, 2021 7:42 am, edited 1 time in total.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Equilibrium constant questions

Post by ChenBeier »

d I dont think so. Develop the equation.
The others look ok for me.
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Re: Equilibrium constant questions

Post by Dhamnekar Winod »

Let the initial concentration of H2 is x. Then equation would be \(K_c= \frac{[CH_3OH]}{[CO][H_2]^2}\Rightarrow 25= \frac{[0.2]}{[0]*[x]^2} \Rightarrow 25=\infty\) is practically impossible.
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Equilibrium constant questions

Post by ChenBeier »

Yes correct
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