Stoichiometry & Reaction Rate (College Board AP® Chemistry): Revision Note

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?

2N₂O₅ → 4NO₂ + O₂

Answer:

• Step 1: Identify the rate given by the statement. Since the rate of decomposition is calculated using a reactant, the rate must have a negative sign

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

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

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

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

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

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

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