Question

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Answer 1

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Answer 2

Given reaction

SO2 + NO2 ⇔ SO3 + NO

Kc = 16 = [SO3][NO]/[SO2][NO2]

Given Volume = 1 liter

If taken 1 mole of four gases initally. Then

SO2 + NO2⇔ SO3 + NO

Initial concentration 1 1 1 1

At equalibrium 1-x 1-x 1+x 1+x

Kc at equlibrium:

Kc = [SO3][NO]/[SO2][NO2]

16 = (1+x)(1+x)/(1-x)(1-x)

16(1-x)(1-x) = (1-x)(1-x)

16x2 -32x+ 16 = x2 + 2x+1

15x2-34x+15 = 0

on simplification

x = ⅗ , 5/3

At equlibrium concentration of [NO2] = (1-x)

At equlibrium concentration of [NO] = (1+x)

on substituting values of "x" in equation you will get concentrations of [NO2] and [NO].

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