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chemistry

A + B -> C + D (reversible reaction)

a 2L container contains 1 mol of C, 1.6 mol of D, 0.8 mol of A snd 0.8 mol of B. Calculate the equillibrium composition of the reaction. Kc =0.065

I think C and D are forming A and B because there is higher concentration of C and D. So equillbrium concentrations will be C = 0.5 -x
D =0.800 -x , A = 0.400 + x, B = 0.400+x and then i can sovle for x by plugging these values into my equillibrum expession and setting it equal to Kc(.0065) I my approach correct?

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  1. Your approach is exactly correct. So Keq = [(0.5-x)(0.8-x)/(0.4+x)^2] =0.065
    [0.4-1.3x+x^2]/[0.16+.8x+x^2]= 0.065
    0.4-1.3x+x^2 = 0.0104+0.052+.065x^2
    Now just combine terms and solve the quadratic.

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