8oP. Consider the following reaction: 2O3(g)---------------> 3O2(g) The rate law

Question

8oP. Consider the following reaction: 2O3(g)---------------> 3O2(g) The rate law for this reaction is as follows: Rate= k ([O3]^2/[O2]) Suppose that a 1.0 l reaction vessel initially contains1.0 mol of O3 and 1.0 mol of O2. What fraction of the O3 will havereacted when the rate falls to on-half of its initial value? 8oP. Consider the following reaction: 2O3(g)---------------> 3O2(g) The rate law for this reaction is as follows: Rate= k ([O3]^2/[O2]) Suppose that a 1.0 l reaction vessel initially contains1.0 mol of O3 and 1.0 mol of O2. What fraction of the O3 will havereacted when the rate falls to on-half of its initial value?

Explanation / Answer

[O3] = 1 M - x [O2] = 1 M +1.5x initial rate = k*[O3]2/[O2] = k*(1^2)/1 =k final rate = k*[1 - x]2/(1 + 1.5x) = 0.5*k 0.5 + 0.75x = 1 - 2x + x2 x2 - 2.75x + 0.5 = 0 x = {2.75 ± [2.752 - 4*0.5]1/2 }/(2) x = 0.195752358 (the amount of O3 reacted) [O3]_reacted/[O3_init] = 0.1957 M / 1 M = 0.196 ~ 19.6%

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