Stoichiometry & Reaction Rate (College Board AP® Chemistry): Study Guide

What is the rate of appearance of NO₂, if the rate of decomposition of N₂O₅ in the chemical reaction is 5.0 × 10^-8 M s^-1?

2N₂O₅ → 4NO₂ + O₂

Answer:

• Step 1: Identify the rate given

  • The rate of decomposition (disappearance) of N₂O₅ is 5.0 × 10^-8 M s^-1, a positive value

  • [N₂O₅] is falling, so its change in concentration per second is negative:

• Step 2: Set up an equation that links the rate of decomposition of N₂O₅ and the rate of appearance of NO₂

 $-\frac{1}{2}\frac{\mathrm{∆}\left[{\mathrm{N}}_2{\mathrm{O}}_5\right]}{\mathrm{∆}\mathrm{t}}=\frac{1}{4}\frac{\mathrm{∆}\left[{\mathrm{NO}}_2\right]}{\mathrm{∆}\mathrm{t}}$

• Step 3: Replace the values given and solve for the rate of appearance of NO₂

$-\frac{1}{2}(-5.0\times {10}^{-8}\mathrm{M}{\mathrm{s}}^{-1})=\frac{1}{4}\frac{\mathrm{∆}\left[{\mathrm{NO}}_2\right]}{\mathrm{∆}\mathrm{t}}$

$-\frac{1}{2}(-5.0\times {10}^{-8}\mathrm{M}{\mathrm{s}}^{-1})\times 4=\frac{\mathrm{∆}\left[{\mathrm{NO}}_2\right]}{\mathrm{∆}\mathrm{t}}$

$\frac{\mathrm{∆}\left[{\mathrm{NO}}_2\right]}{\mathrm{∆}\mathrm{t}}=1.0\times {10}^{-7}\mathrm{M}{\mathrm{s}}^{-1}$

To convert between rates, divide each rate of change by its coefficient in the balanced equation. In 2N₂O₅ → 4NO₂ + O₂, NO₂ appears twice as fast as N₂O₅ disappears, and O₂ appears at half the rate that N₂O₅ disappears

Remember: relative rates come from the balanced equation, not from the rate law