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mart
1 month ago
9

The reaction described by the equation O 3 ( g ) + NO ( g ) ⟶ O 2 ( g ) + NO 2 ( g ) O3(g)+NO(g)⟶O2(g)+NO2(g) has, at 310 K, the

rate law rate of reaction = k [ O 3 ] [ NO ] k = 3.0 × 10 6 M − 1 ⋅ s − 1 rate of reaction=k[O3][NO]k=3.0×106 M−1⋅s−1 Given that [ O 3 ] = 3.0 × 10 − 4 M [O3]=3.0×10−4 M and [ NO ] = 8.0 × 10 − 5 M [NO]=8.0×10−5 M at t = 0 , t=0, calculate the rate of the reaction?
Chemistry
2 answers:
lorasvet [2.7K]1 month ago
8 0

Answer:

The reaction rate, R= 0.072Ms^-1

Explanation:

Based on the Rate Law:

R=k[O3][NO]

Here, the order for NO is 1 and for O3 is also 1,[ [TAG_15]]

Thus, the total order of reaction =1+1=2

Given:

Concentration of O3 = [03]= 3.0×10^-4M

Concentration of NO= [NO]= 8.0×10^-5M

Rate constant K=3.0×10^6M^-1s^-1

Substituting the given values into the rate law equation

R= 3.0×10^6 × 3.0×10^-4 ×

8.0×10^-5

R= 0.072Ms^-1

Thus, the reaction rate is 0.072Ms^-1

KiRa [2.9K]1 month ago
6 0

Explanation:

Reaction equation:

O3(g) + NO(g) ⟶ O2(g) + NO2(g)

Employing the Rate Law:

R = k[O3][NO]

where,

k = rate constant

R = rate of reaction

Given:

[O3] = 3.0×10^-4 M

[[NO] = 8.0×10^-5M

k = 3.0×10^6 /M.s

R = 3 × 10^6 × 3 × 10^-4 × 8 × 10^-5

= 0.072.

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