Question

For a reversible reaction where A + B (both arrows) C + D, the equilibrium constant is represented by kc = 5.3 = (c)(d)/(A)[B]. Initially A and B are present at 2.00 M each. What is the final concentration of A once equilibrium is reached.

I made an ICE table and came up with 5.3 = x^2/ (2-x)^2
i foiled the denominator and got
5.3 = x^2/x^2-4x-4

i am having trouble rearranging this into quadratic form because i am doing some arithmetic wrong. could you help me with how to rearrange this expression? i have 5.3 = x^2/x^2-4x-9.3 but i don't think that is correct.

Answer 1

[x^2/(2-x)^2] =5.3
[x^2/(4-4x+x^2)]=5.3 <b>Note the sign of +4---not -4. </b>
x^2 = 5.3*(x^2-4x+4)
x^2 = 5.3x^2 -21.2x + 21.2
0 = 4.3x^2 - 21.2x + 21.2
And solve that quadratic. I get something close to 3.5 and 1.4 and you know 3.5 doesn't make sense because the initial concn was 2 M. Hope this helps.