Question

Consider the following reaction: 2{\rm{ N}}_2 {\rm{O(}}g)\; \rightarrow \;2{\rm{ N}}_2 (g)\; + \;{\rm{O}}_2 (g)

a. In the first 12.0 s of the reaction, 1.7×10−2 mol of {\rm{O}}_2 is produced in a reaction vessel with a volume of 0.240 L. What is the average rate of the reaction over this time interval?

b. Predict the rate of change in the concentration of {\rm{N}}_2 {\rm{O}} over this time interval. In other words, what is {\Delta [{\rm{N}}_2 {\rm{O}}]}/{\Delta t}?

Answer 1

a. 5.9 × 10⁻³ M/s

b. 0.012 M/s

Step-by-step explanation:

Let's consider the following reaction.

2 N₂O(g) → 2 N₂(g) + O₂(g)

a.

Time (t): 12.0 s

Δn(O₂): 1.7 × 10⁻² mol

Volume (V): 0.240 L

We can find the average rate of the reaction over this time interval using the following expression.

r = Δn(O₂) / V × t

r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s

r = 5.9 × 10⁻³ M/s

b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:

5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s