Question

The reaction of nitrogen dioxide with fluorine 2 NO2 F22 NO2F is first order in NO2 and first order in F2. Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n... , where '1' is understood for m, n ... (don't enter 1) and concentrations taken to the zero power do not appear.
In an experiment to determine the rate law, the rate constant was determined to be 1.58E-4 M-1s-1. Using this value for the rate constant, the rate of the reaction when [NO2] = 2.84 M and [F2] = 3.60 M would be Ms-1.

Answer 1

Answer:

1.62 × 10⁻³ M s⁻¹

Explanation:

Let's consider the following reaction.

2 NO₂ +  F₂ = 2 NO₂F

The generic rate law is:

rate = k × [NO₂]ᵃ × [F₂]ᵇ

where,

• rate: reaction rate

• k: rate constant

• a and b: reaction orders with respect to the reactants

The reaction is first order in NO₂ and first order in F₂ and the rate constant is k = 1.58 × 10⁻⁴ M⁻¹ s⁻¹.

The rate law is:

rate = 1.58 × 10⁻⁴ M⁻¹ s⁻¹ × [NO₂] × [F₂]

When [NO₂] = 2.84 M and [F₂] = 3.60 M, the reaction rate is:

rate = 1.58 × 10⁻⁴ M⁻¹ s⁻¹ × 2.84 M × 3.60 M

rate = 1.62 × 10⁻³ M s⁻¹

Answer 2

Answer:

a) The rate of reaction is $Rate=k[NO_{2}] [F_{2} ]$

b) The rate of reaction is 0.00161 Ms⁻¹

Explanation:

a) The reaction is:

2NO₂ + F₂ = 2NO₂F

The rate of the reaction is equal:

$Rate=k[NO_{2}]^{m} [F_{2} ]^{n}$

The rate of reaction (m) respect to NO₂ is 1 and the rate of reaction (n) respect to F₂ (n) is 1, thus the rate is:

$Rate=k[NO_{2}] [F_{2} ]$

b) Using the rate of part a) and

k = 1.58x10⁻⁴M⁻¹s⁻¹

[NO₂] = 2.84 M

[F₂] = 3.6 M

$Rate=1.58x10^{-4} *(2.84)*(3.6)=0.00161M s^{-1}$