equilibrium constant answer check & help please I solved my homework problem but

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equilibrium constant answer check & help please

I solved my homework problem but I think I did them wrong... so please check them if its correct or not and for the last problem, could you solve it by step by step process?

question 1) Methanol, CH3OH, is manufactured industrially by the following reaction:

CO(g) + 2H2(g) <==> CH3OH(g) write the equilibrium constant expression, kc, for this reaction.

My answer: Kc = [CH3OH] / ([CO][H2]^2)

question 2) a mixture of 1.25 moles of CO and 0.848 moles of H2 is added to a 4.04 L vessel. Equilibrium is reached at 500K with an equilibrium constant, Kc, of 10.5. calculate the initial concentrations of the two reactants.

my answer: 1.25 moles CO / 4.04L = 0.309 M CO, and 0.848 moles H2/ 4.04L = 0.210 M H2

Question 3: Complete the ICE table below, letting x be the moles of CH3OH formed per liter ( do not solve for x) with equation: CO(g) + 2H2(g) <==> CH3OH(g)

my answer: initial 0.309 0.210 0

change -x -2x +x

Equilibrium 0.309-x 0.210-2x x

(btw for this question #3, i wasn't sure if I was suppose to use the datas from question 2... this is sapling question for pre lab chem if any one knows...)

question 4: at equilibrium, 0.15 moles of CH3OH are presented in the reaction vessel. determine the equilibrium concentrations of all components of the reaction mixture.

my answer: 0.309 - x for CO, 0.210 - 2x for H2, and 0.0371 + x for CH3OH.

question 5: substitute the equilibrium concentration calculated in Question 4 into the Kc expression and solve to compare with the value of Kc given in Question 1

My answer: (this is the one im having trouble with and need step by step solution.) when i tried to solve this, i would get 10.5 = [0.0371+x] / ([0.309-x] [0.210-2x]^2) which would result into x^3+x^2+x+y ... which means that solving for x would be difficult.... so im thinking i did this wrong...

I will only choose the best answer that checked my previous works (explaining if i solved them correctly or not) and gives a step by step solution for the question 5.

Explanation / Answer

1) Your answer is true.

2) & 3) are also correct

but you are wrong in 4 & 5 questions.

4)

at equillibrium [CH3OH] = 0.15/4.04 = 0.0371 M = x

Hence , [CO] = 0.309-x= 0.309-0.0371 = 0.272 M

[H2]=0.210-2x = 0.210-0.0742 = 0.136 M

5)

Kc = [CH3OH] / ([CO][H2]^2)

= 0.0371/0.272*0.136^2 = 9.55 (a bit less than 10 given Kc)

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