2 N_{2}O(g) → 2 N_{2}(g) + O_{2}(g)

Recall that for a reaction **aA ****→**** bB**, the ** rate of a reaction** is given by:

$\overline{){\mathbf{Rate}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{a}}\frac{\mathbf{\Delta}\mathbf{\left[}\mathbf{A}\mathbf{\right]}}{\mathbf{\Delta t}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{b}}\frac{\mathbf{\Delta}\mathbf{\left[}\mathbf{B}\mathbf{\right]}}{\mathbf{\Delta t}}}$

where:

**Δ[A]** = change in concentration of reactants or products (in mol/L or M), *[A] _{final} – [A]_{initial}*

**Δt** = change in time, *t _{final} – t_{initial}*

Since N_{2}O is a reactant, the rate with respect to N_{2}O is *negative (–)* since we’re *losing reactants*.

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product.

2 N_{2}O(g) → 2 N_{2}(g) + O_{2}(g)