At a certain temperature, this reaction establishes an equilibrium with the give

Question

At a certain temperature, this reaction establishes an equilibrium with the given equilibrium constant Kc. 3A + 2B -><- 4C kc=2.33x10^23 If at this temperature, 1.40 mol of A and 3.60 mol of B are placed in a 1.00L container, what are the concentrations of A, B, C at equilibrium? I know the answer to B and C but system is keep telling me A is wrong The answer I have is [A]=0 <--- wrong [B]=2.67 <--- correct [C}=1.87 <---- correct The explanations says " Note above that the final concentration of A is approximately 0M. We initially assumed that the reaction went to competion, however, though this assumption was made, C will still dissociate, or back-react, to produce A and B in order to reach equilibriym. Therefore, set up an ICE table to calculate the equilibrium values for each specis where the final concentration above are now the initial values in ICE table. So I did try the ICE table and I got 1.40 for A but it is still wrong Please help me find the correct A

Explanation / Answer

The equilibrium constant Kc value of 2.33 x 1023 is extremely large. The reaction goes essentially to completion.Then we have to treat the question as a limiting reactant problem.
Assuming all A is consumed, the quantity of B required is:
moles of B = 1.40 moles A (2moles B)/3molesC)= 0.933 moles B

So... A is the limiting reagent and B is in ecxess:

Moles of B that react:
1.4 moles A x 2 moles B/3 moles A = 0.933 mole B
Moles of B that remain = 3.60 – 0.933 = 2.67 moles B

Moles of C produced = 1.40moles (4 molesC)/(3 moles A) = 1.87 moles

After reaction is complete: [A] = 0, [B] = 2.67 moles/liter ; [C] = 1.87 moles/ liter

After this, the equilibrium is stablished :

3A + 2B <====> 4C

(0 + 3x) (2.67 + 2x ) (1.87 - 4x)

and Kc = (1.87 - 4x)4 / (3x)3(2.67 + 2x)2 = 2.33 x 1023 .

Kc is very large so , assuming x<<1.87 and x<<2.67 we can write:

(1.87)4 / (3x)3(2.67)2  = 2.33 x 1023 . And calculate X:

(3x)3 = (1.87)4 / (2.33 x 1023 )((2.67)2

27 X3 = 7.36 x 10-24

X3 = 2.73 x 10-25 .

X = 6.49 x 10-9 .

In equilibrium:

A = (0 + 3x) = 3(6.49 x 10-9) = 1.95 x 10-8 mol/L (This is almost "0")

B =(2.67 + 2x ) = 2.67 mol/L   

C = (1.87 - 4x) = 1.87 mol/L

I hope this could help you !!!

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