Question

A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O 3 ( g ) + NO ( g ) ⟶ O 2 ( g ) + NO 2 ( g ) The rate law for this reaction is rate of reaction = k [ O 3 ] [ NO ] Given that k = 3.91 × 10 6 M − 1 ⋅ s − 1 at a certain temperature, calculate the initial reaction rate when [ O 3 ] and [ NO ] remain essentially constant at the values [ O 3 ] 0 = 2.35 × 10 − 6 M and [ NO ] 0 = 7.74 × 10 − 5 M, owing to continuous production from separate sources. initial reaction rate: M ⋅ s − 1 Calculate the number of moles of NO 2 ( g ) produced per hour per liter of air. NO 2 produced: mol ⋅ h − 1 ⋅ L − 1

Answer 1

(a) 7.11x10⁻⁴ M/s

(b) 2.56 mol.L⁻¹.h⁻¹

Step-by-step explanation:

(a) The reaction is:

O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)

The reaction rate of equation (1) is given by:

`rate = k*[O_(3)][NO]` (2)

We have:

k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹

[O₃]₀ = 2.35x10⁻⁶ M

[NO]₀ = 7.74x10⁻⁵ M

Hence, to find the inital reacion rate we will use equation (2):

`rate = k*[O_(3)]_(0)[NO]_(0) = 3.91 \cdot 10^(6) M^(-1)s^(-1)*2.35\cdot 10^(-6) M*7.74 \cdot 10^(-5) M = 7.11 \cdot 10^(-4) M/s`

Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s

(b) The number of moles of NO₂(g) produced per hour per liter of air is:

t = 1 h

V = 1 L

`(\Delta[NO_(2)])/(\Delta t) = rate`

`(\Delta[NO_(2)])/(\Delta t) = 7.11 \cdot 10^(-4) M/s*(3600 s)/(1 h) = 2.56 mol.L^(-1).h{-1}`

Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹

I hope it helps you!