### Learning check

Once you have watched the video, check your learning with this quiz.

## Reactions may go both ways

Most of the reactions we have looked at so far go completely to the right: All the reactants are converted into products. This is the case in many redox reactions.

• Combustion of sugar:
C12H22O11 + 12O2 → 12CO2 + 11H2O
• Dissolving magnesium in hydrocloric acid:
Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)

When sugar is burnt, it is converted completely into carbon dioxide and water.

When magnesium reacts with hydrochloric acid, it is converted completely into magnesium chloride and hydrogen gas.

In a saturated table salt solution, there are two reactions balancing each other.

• Saturated = no more salt can be dissolved, and a solid precipitate of NaCl(s) is formed.

One reaction:

• NaCl(s) → Na+(aq) + Cl(aq)

At exactly the same rate, the opposite reaction takes place:

• Na+(aq) + Cl(aq) → NaCl(s)

The system is at equilibrium, because:

• Na+(aq) + Cl(aq) is formed at exactly the same rate as NaCl(s).
• The reaction to the right is exactly as fast as the reaction to the left: $$\overrightarrow{v} = \overleftarrow{v}$$

The reactions that take place in a saturated table salt solution.

## What happens when a reaction reaches equilibrium?

An imaginary reaction: A + B ⇌ C + D

We start with some concentrations A and B, and no C and D at all.

• Because [A] and [B] are quite large, the reaction rate to the right, $$\overrightarrow{v}$$ is quite high.
• Because [C] and [D] equal zero, $$\overleftarrow{v} = 0$$.

The reaction A + B ⇌ C + D reaches equilibrium.

As some C + D is formed, $$\overleftarrow{v}$$ increases. At the same time, [A] and [B] decrease, which causes $$\overrightarrow{v}$$ to decrease, too.

At some point, $$\overleftarrow{v}$$ has increased so much, and $$\overrightarrow{v}$$ has decreased so much, that $$\overrightarrow{v} = \overleftarrow{v}$$.

• The two reactions, to the left and to the right, balance each other perfectly.
• From this point forward, the system is at equilibrium.

## At equilibrium, the concentrations remain constant

[A], [B], [C], and [D] don’t change any more.

• Expressed mathematically:

$K= \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}$

## What affects the value of the equilibrium constant $$K$$?

For a given reaction:

• Only the temperature
• NOT the initial concentrations or concentration change
• NOT any catalysts
• Only the temperature
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