If 2.50 moles of A and 0.500 moles of B are added to a 2.00 L container, an equi

Question

If 2.50 moles of A and 0.500 moles of B are added to a 2.00 L container, an equilibrium is established in which the [C] is found to be 0.250 M. Find [A] and [B] at equilibrium this is how far I got but I do not understand concentration letter C on the table can you please expaling me why the answer is -0.375, -0.125,+0.250

3A + B<----->2C this is the ICE method I used

I 1.25 0.25 0

C-0.375?, -0.125?,+0.250??

E0.875? 0.125 ? 0.250

Keq= [C]/[A]^3 [B] please let me know what I'm doing wrong in this equation Keq=[0.25]^2/[1.25-X]^3[0.25-x]

Explanation / Answer

molarity of A = 2.50 / 2 = 1.25 M

molarity of B = 0.500 / 2 = 0.25 M

3A    + B   <-------------------> 2C

1.25     0.25                             0   ---------------> I

-3x         -x                               +2x -------------> C

1.25-3x 0.25-x                         2x ---------------> E

at equilibrium:

2x =0.250 is given

x = 0.125

equilibrium concentrations :

[A] = 1.25 -3x

      = 0.875 M

[B] = 0.25-x  

      = 0.125 M

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