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X + 2Y ---> XY2In order to determine the order of the reaction represented above, the initial rate of formation of XY2 is measured using different initial values of [X] and [Y]. The results of the experiment are shown in the table below.Trial [X] [Y] Rate of formation [XY2]1 .5 M .5 M 8.0 x 10^-3 M/s2 1.0 M .5 M 3.2 x 10^-2 M/s3 1.0 M 1.0 M 6.4 x 10^-2 M/sIn trial 2 which of the reactants would be consumed more rapidly and why?X, because the reaction is second order with respect to X.Y, because the reaction is second order with respect to Y.X, because it has a higher molar concentration.Y, because the rate of disappearance will be double that of X.

Answer :

Explanation:

The reaction given is ,

  • [tex]X+2Y[/tex]⇒[tex]XY_2[/tex]

[tex]\left[\begin{array}{ccc}[X]&[Y]&rate\\1.5M&0.5M&8*10^{-3}Ms^{-1}\\21M&0.5M&3.2*10^{-2}Ms^{-1}\\31M&1M&6.4*10^{-2}Ms{-1}\end{array}\right][/tex]

let's assume that the rate [tex]r[/tex] is ,

[tex]r \alpha [X]^a[Y]^b[/tex]

to find [tex]a[/tex] and [tex]b[/tex]

we use the given information from the table ;

[tex]8*10^{-3}=(1.5)^a(1.5)^b\\32*10^{-3}=(21)^a(0.5)^b\\[/tex]

dividing them ,

  • [tex]\frac{1}{4}=(\frac{1.5}{21} )^a(\frac{0.5}{0.5})^b\\\\4=14^a\\a=\frac{log4}{log14}[/tex]
  • [tex]8*10^{-3}=1.5^a0.5^b\\64*10^{-3}=31^a\\\\\frac{31}{1.5} ^\frac{log4}{log14}*2^b = 8\\[/tex]
  • a=0.5253
  • b=0.7048

Thus , the order with respect to Y is more.

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