Question

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

Answer 1

Answer: x = 2, y = 1

Step-by-step explanation:

solving for x using trial 2 & 3:

`(trial 3)/(trial 2) = (45.0)/(5.0) =(k[0.030]^(x)[0.020]^(y))/(k[0.010]^(x)[0.020]^(y)) \\9 = (k[0.030]^(x))/(k[0.010]^(x)) \\\\9 = 3^(x)\\x = 2`

• 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:

`(trial 2)/(trial 1) = (5.0)/(2.5) = (k[0.10]^(x) [0.020]^(y))/(k[0.10]^(x)[0.010]^(y)) \\2 = (k[0.020]^(y))/(k[0.010]^(y)) \\\\2 = 2^(y)\\y = 1`