Stoichiometry & Reaction Rate (College Board AP® Chemistry)

Study Guide

Written by: Martín

Reviewed by: Stewart Hird

$aA + bB \to cC + dD$

The rate of reaction must be independent of the chemical substance used to calculate the rate
There is only one positive value for the rate of reaction
The stoichiometry of the reaction can be used to establish general equations for the rate of reaction
- The rate depends on the coefficient for each substance in the chemical equation
- The negative sign is added to the reactants since the rate is a positive value
- The brackets represent molarity

$Rate = - \frac{1}{a} \frac{∆ [A]}{∆ t} = - \frac{1}{b} \frac{∆ [B]}{∆ t} = \frac{1}{c} \frac{∆ [C]}{∆ t} = \frac{1}{d} \frac{∆ [D]}{∆ t}$

What is the rate of appearance of NO₂, if the rate of decomposition of N₂O₅ in the chemical reaction is 5.0 ✕ 10^-8M/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

$\frac{∆ [N_{2} O_{5}]}{∆ t} = - 5.0 \times 10^{- 8} M s^{- 1^{}}$

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

$- \frac{1}{2} \frac{∆ [N_{2} O_{5}]}{∆ t} = \frac{1}{4} \frac{∆ [{NO}_{2}]}{∆ t}$

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

$- \frac{1}{2} \frac{∆ [N_{2} O_{5}]}{∆ t} = \frac{1}{4} \frac{∆ [{NO}_{2}]}{∆ t}$

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

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

$\frac{∆ [{NO}_{2}]}{∆ t} = 1.0 \times 10^{- 7} M s^{- 1}$

Last updated: 22 August 2024

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