Question

Consider the reaction: 2NO(g) O2(g) --> 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of 0.0022 M/s.
a. At what rate is NO2 being formed?
b. At what rate is molecular oxygen reacting?

Answer 1

Answer:

a. 0.0022 M/s

b. 0.0011 M/s

Explanation:

Let's consider the following reaction.

2 NO(g) + O₂(g) → 2 NO₂(g)

The rate of disappearance of NO is 0.0022 mol NO. L⁻¹.s⁻¹

a. At what rate is NO₂ being formed?

The molar ratio of NO to NO₂ is 2:2. The rate of formation of NO₂ is:

$\frac{0.0022molNO}{L.s} .\frac{2molNO_{2}}{2molNO} =\frac{0.0022molNO_{2}}{L.s}$

b. At what rate is molecular oxygen reacting?

The molar ratio of NO to O₂ is 2:1. The rate of disappearance of O₂ is:

$\frac{0.0022molNO}{L.s} .\frac{1molO_{2}}{2molNO} =\frac{0.0011molO_{2}}{L.s}$

Answer 2

Answer:

NO2 is beign formed at a rate of 0.0022 M/s

O2 is disappearing at a rate of 0.0011 M/s

Explanation:

Step 1: Data given

The rate of 0.0022 M/s.

-1/2*[NO]/ΔT = -[O2]/ΔT = 1/2*[NO2]/ΔT

Step 2: The balanced equation

2NO(g) + O2(g) ⇆ 2NO2(g)

Step 3: Calculate the rate of (dis)appearance

[NO2]/ΔT = 0.0022/2 M/s = 0.0011 M/s (Positive because NO2 is being formed)

-1/2[O2]/ΔT = -1/2*0.0022/ΔT = -0.0011 M/s (negative because O2 is disappearing)

NO2 is beign formed at a rate of 0.0022 M/s

O2 is disappearing at a rate of 0.0011 M/s