Determine the rate law and the value of k for the balanced chemical reaction of
ID: 893455 • Letter: D
Question
Determine the rate law and the value of k for the balanced chemical reaction of nitrogen monoxide with oxygen to produce nitrogen dioxide, using the data provided.
[NO]i (M) [O2]i (M) Initial Rate (M-1s-1)
0.030 0.0055 8.55 x 10-3
0.030 0.0110 1.71 x 10-2
0.060 0.0055 3.42 x 10-2
Rate = 57 × 103 M-1s-1[NO][O2]
Rate = 3.8 M-1/2s-1[NO][O2]1/2
Rate = 9.4 × 103 M-2s-1[NO][O2]2
Rate = 3.1 × 105 M-3s-1[NO]2[O2]2
Rate = 1.7 × 103 M-2s-1[NO]2[O2]
Rate = 57 × 103 M-1s-1[NO][O2]
Explanation / Answer
2NO + O2 .......>2 NO2
rate =K[NO]^x [O2]^y
From the data provided.
By observing first set values
8.55*10^-3 = K[0.03]^x [0.0055]^y .................1
By observing second set values
1.71*10^-2 = K[0.03]^x [0.0110]^y .................2
By observing third set values
3.42*10^-2 = K[0.06]^x [0.0055]^y .................3
by doing eq.1 / eq.2
(8.55*10^-3) /(1.71*10^-2) = (0.0055/0.011)^y
(0.5 )^1 = (0.5)^y
Bases are equal powers are equated
so y = 1
order w.r.t. O2 is 1
Now by dowing eq.1 / eq.3
(8.55*10^-3) /(3.42*10^-2) = (0.03/0.06)^x
0.25 = (0.5)^x
(0.5)^2 = (0.5)^x
Bases are equal powers are equated
so x = 2
order w.r.t. NO is 2
overal order of the reaction is = (1+2) = 3 (third order)
there fore the order is
rate = K [NO]2 [O2]1
consider first set of values
8.55*10^-3 = K[0.03]^2 [0.0055]^1
K = 1.7*10^3 M^-2 S^-1
so the last option is correct
Rate = 1.7 × 103 M-2s-1[NO]2[O2]
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.