The reaction described by the equation O 3 ( g ) + NO ( g ) ⟶ O 2 ( g ) + NO 2 ( g ) has, at 310 K, the rate law rate of reaction = k [ O 3 ] [ NO ] k = 3.0 × 10 6 M − 1 ⋅ s − 1 Given that [ O 3 ] = 7.0 × 10 − 4 M and [ NO ] = 5.0 × 10 − 5 M at t = 0 , calculate the rate of the reaction at t = 0 .

Explanation:

Equation of the reaction:

O3(g) + NO(g) ⟶ O2(g) + NO2(g)

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, [O3] = 7.0×10^-4M

Concentration of NO, [NO] = 5.0×10^-5M

Rate constant, k = 3.0 × 10^6M^-1s^-1

R = 3.0×10^6 × 7.0×10^-4 × 5.0×10^-5

= 0.105 M/s

16

explanation:

the marks go by 4s