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For the reaction a+2b=c+d the equilibrium constant kc =10^9 if the initial moles of a, b, c and d are .5, 1, 3.5, .5 moles respectively in a one litre vessel, what is the equilibrium concentration of b at equilibrium?

For the reaction a+2b=c+d the equilibrium constant kc =10^9 if the initial moles of a, b, c and d are .5, 1, 3.5, .5 moles respectively in a one litre vessel, what is the equilibrium concentration of b at equilibrium?

Grade:12

1 Answers

Manas Satish Bedmutha
22 Points
7 years ago
Kc = [c][d]/[a][b]^2 Now, as volume is 1 ltr, Molarity = No. of moles So, [b]^2= [c][d]/[a]Kc = (3.5 x 0.5 / 0.5) x 10^-9 Hence, [b] = (35 x 10^-10) ^ 0.5 = 5.91 x 10^-5 M

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