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}*

**Reaction: 3 O _{2} + 2 CH_{3}OH → 2 CO_{2} + 4 H_{2}O**

H_{2}O is a product, the rate with respect to H_{2}O is *positive (+)* since we’re *gaining products*.

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

Given the following balanced chemical equation

3 O_{2} + 2 CH_{3}OH → 2 CO_{2} + 4 H_{2}O

How is the rate of appearance of H_{2}O related to the rate of disappearance of O_{2}?

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