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:
[tex]\frac{0.0022molNO}{L.s} .\frac{2molNO_{2}}{2molNO} =\frac{0.0022molNO_{2}}{L.s}[/tex]
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:
[tex]\frac{0.0022molNO}{L.s} .\frac{1molO_{2}}{2molNO} =\frac{0.0011molO_{2}}{L.s}[/tex]
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
