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What is a reversible reaction?
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All Flashcards in Topic 5.3
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5.3.112 cards
What is a reversible reaction?
A reaction that can go in **both directions** — reactants can form products and products can re-form reactants. Shown with the **⇌** symbol.
Define dynamic equilibrium.
The state, in a **closed system**, where the **forward and reverse reactions occur at equal rates**, so the **concentrations of reactants and products stay constant**.
Why is equilibrium called 'dynamic'?
Because **both** the forward and reverse reactions are **still happening** — the reaction has **not** stopped; the opposite changes simply cancel out.
At equilibrium, are the concentrations equal?
**No** — they are **constant** (unchanging), but generally **not equal** to one another.
What is true about the rates at equilibrium?
The **rate of the forward reaction = the rate of the reverse reaction**.
Why must the system be closed for equilibrium?
So that **nothing is added or escapes** (no reactant/product/heat leaves); an open system could never settle to constant concentrations.
Name three macroscopic properties that stay constant at equilibrium.
**Colour (absorbance)**, **pressure** (for gases) and **pH** — all remain constant because the concentrations are constant.
What does 'equilibrium position' mean?
**How far** a reaction has gone — the **relative amounts** of reactants and products. To the **right** = mostly products; to the **left** = mostly reactants.
How could you tell experimentally that equilibrium has been reached?
Measure a **macroscopic property** over time (e.g. colour intensity); when it **levels off to a constant value**, equilibrium has been reached.
Most common misconception about equilibrium?
That the reaction has **stopped** — in fact both reactions continue (it is **dynamic**); the concentrations are merely constant.
On a rate–time graph, what happens to the forward and reverse rates?
The **forward rate falls** and the **reverse rate rises** until they **meet (become equal)** — that point is equilibrium.
On a concentration–time graph, how do you spot equilibrium?
Both the reactant and product curves **level off** (become flat) and stay constant — at **different** values.
5.3.212 cards
State Le Châtelier's principle.
If a system at equilibrium is disturbed, the position shifts in the direction that **opposes** (partly cancels) the change.
Adding more of a reactant shifts the position…
…towards the **products** (right) — the system uses up the added reactant.
Removing a product shifts the position…
…towards the **products** (right) — the system replaces the lost product.
Effect of increasing pressure on a gas equilibrium?
The position shifts to the side with **fewer moles of gas**, to reduce the pressure.
What if both sides have equal moles of gas?
Changing the pressure causes **no shift** in the position.
Effect of raising the temperature?
The position shifts in the **endothermic** direction (it absorbs the added heat).
Which change is the only one that alters K_{c}?
A change in **temperature** — concentration, pressure and a catalyst leave K_{c} unchanged.
Effect of a catalyst on equilibrium?
**No shift** in position and **no change** in K_{c}; it just reaches equilibrium **sooner** (speeds up both directions equally).
Exothermic forward reaction: what does raising T do to K_{c}?
K_{c} **decreases** (the position shifts towards the reactants).
How do you predict the pressure effect quickly?
**Count the moles of gas** on each side; the position shifts towards the side with **fewer** gas moles when pressure rises.
Trick for the temperature direction?
Write **heat** as a species (exo: products + heat; endo: reactants + heat), then treat adding heat like adding that species.
If heating shifts the position towards the products, is the forward reaction exo- or endothermic?
**Endothermic** — adding heat favours the heat-absorbing direction.
5.3.312 cards
What is the equilibrium constant K_{c}?
The **fixed ratio** of product to reactant concentrations at equilibrium, at a given temperature — it shows **how far** a reaction goes.
How do you write the K_{c} expression?
**Products over reactants**, each concentration **raised to the power of its balancing coefficient**; use [ ] for equilibrium concentration in mol dm⁻³.
Write K_{c} for N_{2}(g) + 3H_{2}(g) ⇌ 2NH_{3}(g).
$K_{c} = \dfrac{[\text{NH}_{3}]^{2}}{[\text{N}_{2}][\text{H}_{2}]^{3}}$ — the 2 and 3 become powers.
What does a large K_{c} (>> 1) mean?
**Products are favoured** — the equilibrium lies to the **right** (mostly products).
What does a small K_{c} (<< 1) mean?
**Reactants are favoured** — the equilibrium lies to the **left** (mostly reactants).
What is K_{c} for the reverse reaction?
The **reciprocal**: K_{reverse} = **1 / K_{forward}**.
Which is the only change that alters K_{c}?
A change in **temperature** — concentration, pressure and a catalyst all leave K_{c} unchanged.
Endothermic forward reaction: what happens to K_{c} as T rises?
K_{c} **increases** (the position shifts towards products).
Exothermic forward reaction: what happens to K_{c} as T rises?
K_{c} **decreases** (the position shifts towards reactants).
First step in calculating K_{c} from amounts in a flask?
Convert each **amount (mol)** to a **concentration (mol dm⁻³)** using **c = n/V**, then substitute.
Which species are left out of a K_{c} expression?
Pure **solids** and pure **liquids** — only gases and dissolved (aqueous) species appear.
Does a large K_{c} mean the reaction is fast?
**No** — K_{c} describes the **extent** (how far), not the **rate** (how fast).
Topic 5.3 study notes
Full notes & explanations for How far? The extent of chemical change
Chemistry exam skills
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