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Consider the initial rates measured for reaction 2 NO + O 2 → 2 NO 2

Trial 1 2 3
[NO] (in M) 0.010 0.010 0 .030
[O2] (in M) 0.010 0.020 0 .020
Initial rate (M/s) 2.5 5.0 45 .0
Determine the value of x and y in the rate law for this reaction: r = k[NO] x[O 2] y
x = 0, y = 2
x = 2, y = 1
x = 1, y = 1
x = 2, y = 2
x = 1, y = 2

Respuesta :

misspeaches

Answer: x = 2, y = 1

Explanation:

solving for x using trial 2 & 3:

[tex]\frac{trial 3}{trial 2} = \frac{45.0}{5.0} =\frac{k[0.030]^{x}[0.020]^{y}}{k[0.010]^{x}[0.020]^{y}} \\9 = \frac{k[0.030]^{x}}{k[0.010]^{x}} \\\\9 = 3^{x}\\x = 2[/tex]

  • using the basic rate law equation r = k [A]^x[B]^y
  • set the initial rate = k[NO]^x[O2]^y and plug values in
  • you can cancel out the O2 because they have the same value -> you should be selecting the trials where only one independent value (in this case, NO or O2) changes
  • this is basically the same thing for y

solving for y using trial 1 & 2:

[tex]\frac{trial 2}{trial 1} = \frac{5.0}{2.5} = \frac{k[0.10]^{x} [0.020]^{y}}{k[0.10]^{x}[0.010]^{y}} \\2 = \frac{k[0.020]^{y}}{k[0.010]^{y}} \\\\2 = 2^{y}\\y = 1[/tex]

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