Chemistry / Unit 10: Solutions, Acids and Bases
10.7 Reaction Rates and Equilibrium
Notes
Collision theory: particles must collide with enough energy (the activation energy) and the right orientation to react.
- Rates increase with higher temperature, higher concentration, greater surface area and a catalyst (which lowers the activation energy without being used up).
- A reversible reaction reaches equilibrium when the forward and reverse rates are equal. Amounts stay constant, but both reactions keep going.
- For $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$: $\displaystyle K_{eq} = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$. Leave out pure solids and liquids. $K_{eq} > 1$ favors products.
- Le Chatelier's principle: a system at equilibrium shifts to undo a stress. Add reactant: shift right. Remove product: shift right. Raise pressure: shift toward fewer gas moles. Heat an exothermic reaction: shift left.
Practice
Which change will NOT increase the rate of a reaction?
Fewer particles per volume means fewer collisions and a slower rate.
For $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$, what is $K_{eq}$?
Products over reactants, each raised to its coefficient.
For the exothermic reaction $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$, which change shifts the equilibrium to make more $\text{NH}_3$?
Higher pressure favors the side with fewer gas moles (2 on the right versus 4 on the left). A catalyst speeds both directions equally.
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