The data below were collected for the following reaction at acertain temperature

Question

The data below were collected for the following reaction at acertain temperature:
X2Y-->2 X+Y. It is a second order reaction
k=1.682 M-1 hr-1

What is theconcentration of X after 10.0 hours? (NOT[X2Y]) Time (Hours) [X2Y] (M) 0.0 0.100 1.0 0.0856 2.0 0.0748 3.0 0.0664 4.0 0.0598 5.0 0.0543 The data below were collected for the following reaction at acertain temperature:
X2Y-->2 X+Y. It is a second order reaction
k=1.682 M-1 hr-1

What is theconcentration of X after 10.0 hours? (NOT[X2Y]) What is theconcentration of X after 10.0 hours? (NOT[X2Y]) Time (Hours) [X2Y] (M) 0.0 0.100 1.0 0.0856 2.0 0.0748 3.0 0.0664 4.0 0.0598 5.0 0.0543

Explanation / Answer

d[X2Y]/dt =-k*[X2Y]^2            Second order eqauation -d[X2Y]/[X2Y]^2 =k*dt                 1/[X2Y] - 1/[X2Y]0 =k*t            Integration 1/[X2Y] = 1/[X2Y]0 + kt = (1 + kt*[X2Y]0)/[X2Y]0 [X2Y] = [X2Y]0 / ( 1 + kt*[X2Y]0) plug in t = 10 hours, k = 1.682 M-1 hr-1   and[X2Y]0 = 0.100 [X2Y] ( t= 10) = 0.100/ (1 + 1.682*10*0.10) = 0.0372856078 M From stoichiometry, 2 moles of X forms for every X2Y consumed. amount of X2Y consumed = [X2Y]0 - [X2Y] = 0.10 - 0.0372856078= 0.0627143922 M reacted Since 2 moles of X form for every X2Y consumed, [X] = 2*0.0627143922 M = 0.125428784 M ~ 0.125 M of X

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