Please answer Part A and B with some work! Thank you so much!! brium expression

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Please answer Part A and B with some work! Thank you so much!!

brium expression ample, in tric equilibrium t equilibrium t it is called he equilibrium ed at some previous l 7 of 40 l next Part A A mixture initially contains A, B, and C in the following concentrations: Al 0.700 M. Bl 1.10 M. and [C] 0.700 M.The following reaction occurs and equilibrium is established A 2B C At equilibrium, LA] 0.500 Mand Cl -0.900 M. Calculate the value of the equilibrium constant, K Express your answer numerically. Submit Hints My Answers Give Up Review Part Part B This question will be shown after you complete previous question(s) eedback Continue vide

Explanation / Answer

PART A

A + 2B < --------> C

Kc = [C]/[A][B]^2

[A] at equilibrium = (0.7-X)M

[B] at equilibrium = (1.10-2X)M

[C] at quillibrium = X M

Kc =X/(0.7 - X) ( 1.10 -2X)^2

At equilibrium [A] = 0.5M

0.70 - X = 0.5

X = 0.20

B= 1.10 - 2X = 1.10 - 2(0.2) = 0.7M

At equilibrium [ A ] = 0.5M , [B] = 0.7M, [C] = 0.9M

Therefore, Kc = 0.9/(0.5 ×(0.7)^2) = 3.67

Part B

Volume of reaction vessel = 3.25litre

mole of H2O(g) =13.2 mole

[H2O] = 13.2/3.25 =4.06M

mole of CO(g) = 3.40

[CO] = 3.40/ 3.25 = 1.05M

mole of H2(g) = 6.10

[ H2] = 6.10 /3.25 = 1.88M

Reaction Quotient ,Qreaction = [CO][H2]/[H2O]

(Since C is solid , it will not include in reaction Quotient and in equilibrium constant)

Therefore, Qreaction =1.05M×1.88M/4.06M

= 0.486

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