Question

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?

Answer 1

Answer:

Rate of reaction R= $0.072Ms^-1$

Explanation:

From the Rate Law:

R=k[O3][NO]

The order of NO is 1 and that of O3 is also 1,

Therefore 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

Insert the given values into the rate law equation

R= 3.0×10^6 × 3.0×10^-4 ×

8.0×10^-5

R= 0.072$Ms^-1$

Therefore the rate of reaction is 0.072$Ms^-1$

Answer 2

Explanation:

Equation of reaction:

O3(g) + NO(g) ⟶ O2(g) + NO2(g)

Using 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.