Reversible reactions

In the “arrow and rate” picture for reversible reactions, arrow length is used as a visual shortcut for how fast each direction is happening.

• The forward arrow ($r_f$) shows how quickly reactants become products.
• The reverse arrow ($r_r$) shows how quickly products change back into reactants.

The prompt says the forward arrow is much longer than the reverse arrow.

That means: ${\text{Rate}}_{forward}>{\text{Rate}}_{reverse}$

So, at this moment, the reaction is moving toward the product side overall because the forward change is happening more quickly than the backward change.

Now we can reject choices that don’t match what unequal arrows mean:

• Option A: At dynamic equilibrium, the two rates are the same ($r_f=r_r$). In an arrow model, that would appear as matching arrow lengths. Since one arrow is clearly longer, this cannot be equilibrium.
• Option B: “Completion” would mean reactants are gone, so the forward process would not still be shown as strongly occurring. A model with two active arrows (even if unequal) indicates ongoing reversible change, not a finished one-way process.

Consider Option D. If the reverse reaction had truly stopped, its rate would be zero, and the reverse arrow would be missing (length zero). Because the diagram still includes a reverse arrow (just shorter), the reverse reaction is still happening.

With the forward rate larger than the reverse rate, products are being made faster than they are being converted back. The overall result is an increase in the amount (concentration) of products.

That matches the correct option.

In the “arrow and rate” dynamic model of chemical reactions, arrow length stands for the reaction rate in that direction.

• A longer arrow means a faster rate.
• A shorter arrow means a slower rate.
• Two opposing arrows indicate the reaction can run both ways (it is reversible).

Dynamic equilibrium happens when the forward and reverse rates are equal ($Rat{e}_{forward}=Rat{e}_{reverse}$). In the diagram, that would be shown by two arrows of the same length.

Because the forward arrow is stated to be much longer, the system is not at equilibrium.

If the reverse reaction had completely stopped, its rate would be zero. In the arrow diagram, that would mean no reverse arrow would be drawn.

Since there is still a reverse arrow present (even though it is short), the reverse reaction is still occurring.

If all reactants were gone, the forward reaction could not keep going because there would be nothing left to convert into products. Also, the two-arrow setup signals a reversible reaction, which typically does not simply “finish” the way an irreversible reaction might be described.

So Option B does not fit the model.

That leaves Option C. A longer forward arrow means $Rat{e}_{forward}>Rat{e}_{reverse}$. So reactants are turning into products more quickly than products are turning back.

Therefore, there is a net increase in products, which is exactly what Option C states.

We are dealing with a reversible reaction that has already reached equilibrium (the Haber process). The change described is that ammonia, $N{H}_3$, is being continuously removed from the mixture.

${\mathrm{N}}_{2(\mathrm{g})}+3{\mathrm{H}}_{2(\mathrm{g})}\rightleftharpoons 2{\mathrm{NH}}_{3(\mathrm{g})}$

Here, nitrogen and hydrogen are on the left as reactants, and ammonia is on the right as the product.

Le Chatelier’s Principle states that when an equilibrium system is disturbed (for example, by changing concentration), the system shifts in whatever direction reduces the effect of that disturbance.

Removing a product lowers its concentration. The equilibrium will respond by trying to replace what was taken away, meaning it will tend to make more of that product.

If the system needs to create more ammonia ($N{H}_3$), it must proceed more in the forward direction of the reaction (from reactants to products).

A greater extent of the forward reaction corresponds to the equilibrium position moving to the right.

Consider each statement:

• Option A cannot be correct because changing a concentration forces the equilibrium to adjust.
• Option B is wrong because a shift left would generate more $N_2$ and $H_2$ while using up $N{H}_3$, which would make the lack of ammonia even greater.
• Option D is incorrect because equilibrium is dynamic: both directions continue. Removing product does not halt the reaction; it changes the relative rates so more forward reaction occurs.

Shifting to the right helps replace some of the removed ammonia, which is exactly how an equilibrium system counteracts the removal of a product.

This approach compares the reaction quotient ($Q_c$) with the equilibrium constant ($K_c$). For the Haber process:

${\mathrm{N}}_{2(\mathrm{g})}+3{\mathrm{H}}_{2(\mathrm{g})}\rightleftharpoons 2{\mathrm{NH}}_{3(\mathrm{g})}$

The reaction quotient expression is:

$Q_c=\frac{[{\mathrm{NH}}_3{]}^2}{[{\mathrm{N}}_2][{\mathrm{H}}_2{]}^3}$

At equilibrium, the key condition is $Q_c=K_c$. This fits with what you learn in Reversible reactions.

The statement “continuously removed” means the product ammonia, ${\mathrm{NH}}_3$, is being taken out of the system repeatedly, so its concentration $[{\mathrm{NH}}_3]$ drops.

In the expression

$Q_c=\frac{[{\mathrm{NH}}_3{]}^2}{[{\mathrm{N}}_2][{\mathrm{H}}_2{]}^3}$

a decrease in $[{\mathrm{NH}}_3]$ makes the numerator smaller. That makes the whole value of $Q_c$ fall below its equilibrium value.

So after removal begins:

$Q_c<K_c$

When $Q_c<K_c$, the mixture must change until $Q_c$ increases to match $K_c$ again.

To increase $Q_c$, the system must:

1. Raise the amount of products (increase the numerator).
2. Reduce the amount of reactants (decrease the denominator).

That overall change happens through a net forward reaction, so the equilibrium position moves to the right.

• Option A is not possible because removing product changes $Q_c$, so the system cannot remain at the same equilibrium position.
• Option B is incorrect because moving left would lower $[{\mathrm{NH}}_3]$ further and reduce $Q_c$ even more.
• Option D is incorrect because taking away product encourages more forward reaction; it doesn’t cause the equilibrium reaction to stop.

Because removing ${\mathrm{NH}}_3$ makes $Q_c<K_c$, the system responds by shifting right to form more ammonia until equilibrium is re-established.

An equilibrium mixture is being disturbed because ammonia is continuously removed.

Le Chatelier’s Principle says the equilibrium position will shift in the direction that opposes the change being made.

Option A claims nothing happens. But removing a substance changes its concentration, so the system must adjust to reduce the effect of that removal. With less ammonia present, the equilibrium cannot simply remain as it was.

Option D suggests the reaction ends. At equilibrium, reactions are dynamic—both forward and reverse processes continue. Taking away product does not stop the chemistry; it shifts the balance so the forward reaction becomes more favored. Industries often remove product for exactly this reason.

Option B says the equilibrium goes left. From ${\mathrm{N}}_{2(\mathrm{g})}+3{\mathrm{H}}_{2(\mathrm{g})}\rightleftharpoons 2{\mathrm{NH}}_{3(\mathrm{g})}$

a left shift would use up ${\mathrm{NH}}_3$ and make more ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$. Since ammonia is already being removed, shifting left would push ammonia even lower—this would support the disturbance instead of opposing it.

After eliminating A, D, and B, Option C remains. A right shift means more forward reaction, using ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$ to produce additional ammonia, which counteracts the removal of ${\mathrm{NH}}_3$.

Here, the container is made larger: we are increasing the volume while keeping temperature the same.

From Boyle’s Law/ideal gas ideas, when volume goes up (at constant temperature), the total pressure drops. Also, because the same amount of gas is now spread through a bigger space, the concentration of each gas decreases right away.

Le Chatelier’s Principle says an equilibrium system adjusts to reduce the effect of a change.

Since the pressure has fallen because the volume increased, the system will tend to raise the pressure again. It can do this by shifting toward the side with the larger number of gas moles.

Check the balanced equation:

${\text{N}}_2{\text{O}}_4(\text{g})\rightleftharpoons 2{\text{NO}}_2(\text{g})$

• Left: 1 mole of gas (${\text{N}}_2{\text{O}}_4$)
• Right: 2 moles of gas ($2{\text{NO}}_2$)

Because the right side has more gas particles, the equilibrium shifts to the right.

Shifting right makes extra ${\text{NO}}_2$, which is the brown gas. However, increasing volume also spreads the gases out more.

For this kind of question, the expansion (dilution) effect outweighs the extra ${\text{NO}}_2$ made by the shift. So compared with the original equilibrium state, the ${\text{NO}}_2$ concentration ends up lower, and the colour looks lighter brown.

Check each statement:

• Option A is wrong: even if temperature stays the same, changing pressure/volume can still move the equilibrium position.
• Option C is wrong: when volume increases, concentration drops at first because $c=n/V$.
• Option D is wrong: the shift is toward the side with more gas moles (right), not left.

Increasing volume lowers pressure, so the equilibrium moves right (toward more moles of gas). But because the gases are more spread out overall, the ${\text{NO}}_2$ concentration is lower than before, so the mixture appears lighter brown.

The key change is increasing the volume. With temperature constant, a larger volume means a lower total pressure.

Now compare gas moles in:

${\text{N}}_2{\text{O}}_4(\text{g})\rightleftharpoons 2{\text{NO}}_2(\text{g})$

Left has 1 mole of gas; right has 2 moles. When pressure decreases, Le Chatelier predicts a shift toward the side with more gas moles to push pressure back up.

Option A claims no shift happens because temperature is unchanged. While $K_c$ depends on temperature, the equilibrium position can still move when pressure/volume changes. Since pressure changed here, Option A must be rejected.

Concentration is

$c=\frac{n}{V}$

When $V$ increases, $c$ for every gas immediately decreases because you are dividing by a bigger number. So both gases cannot become more concentrated at the same time.

Option D says the shift is left. But lower pressure favours the side with more gas moles. The right side has 2 moles vs 1 on the left, so the shift must be to the right, not to the left.

Only Option B fits everything:

1. The equilibrium shifts right toward $2{\text{NO}}_2$.
2. Even though shifting right forms more brown ${\text{NO}}_2$, the larger volume dilutes the gases so the final mixture looks lighter brown.

Start from the balanced reversible reaction (from Topic: Types of chemical reaction - Reversible reactions):

${\text{N}}_2{\text{O}}_4(\text{g})\rightleftharpoons 2{\text{NO}}_2(\text{g})$

So the equilibrium constant is:

$K_c=\frac{[{\text{NO}}_2{]}^2}{[{\text{N}}_2{\text{O}}_4]}$

If the container volume increases, each gas concentration immediately drops because concentration is moles per volume ($c=n/V$).

If the volume becomes $k$ times larger, then every concentration changes to:

$[\text{X}{]}_{new}=\frac{[\text{X}{]}_{old}}{k}$

Immediately after expansion (before the system re-equilibrates), substitute the diluted concentrations into the expression:

$Q_c=\frac{{\left(\frac{[{\text{NO}}_2]}{k}\right)}^2}{\frac{[{\text{N}}_2{\text{O}}_4]}{k}}=\frac{\frac{[{\text{NO}}_2{]}^2}{k^2}}{\frac{[{\text{N}}_2{\text{O}}_4]}{k}}=\frac{1}{k}\cdot \frac{[{\text{NO}}_2{]}^2}{[{\text{N}}_2{\text{O}}_4]}$

Because the volume increased, $k>1$, so:

$Q_c=\frac{1}{k}{K}_c<K_c$

Since $Q_c<K_c$, the mixture has too little product relative to reactant compared with equilibrium.

To get back to $Q_c=K_c$, the system must increase the numerator and decrease the denominator. That means shifting to the right, forming more ${\text{NO}}_2$.

Even though shifting right makes more ${\text{NO}}_2$, the initial expansion spreads the gases out a lot.

To keep $K_c$ constant after $[{\text{N}}_2{\text{O}}_4]$ has dropped, the final $[{\text{NO}}_2]$ must also end up below its original equilibrium value. Because ${\text{NO}}_2$ is the brown gas, a lower $[{\text{NO}}_2]$ means the mixture looks lighter brown.

• Option A is wrong: volume/pressure changes can shift equilibria when gas moles differ.
• Option C is wrong: increasing volume causes an immediate concentration decrease for all gases.
• Option D is wrong: the shift is toward the side with more moles (right), not left.

The system shifts right after the volume increase, but dilution means the brown ${\text{NO}}_2$ is less concentrated than before, so the colour becomes lighter brown.

Before talking about how any rate changes, we need to identify what particles are available at the start of the reaction. The prompt tells us the container begins with reactants only and no products yet.

Collision theory says reactions happen when particles collide with enough energy and the right orientation.

For a reversible reaction, the reverse direction can only occur when product particles collide in a way that reforms reactants. At the beginning there are no product particles, so there cannot be any product–product collisions. That means the reverse reaction rate at the start must be exactly zero.

As the forward reaction runs, it produces product particles, so the product concentration increases.

$\text{Concentration of Products}\uparrow ⟹\text{Frequency of Collisions}\uparrow$

With more product particles in the same space, collisions between product particles happen more often, so the number of successful collisions for the reverse reaction rises. Therefore the reverse rate increases from its initial value of zero.

Using the ideas above, we can remove the wrong statements:

• Option B is wrong because the reverse rate depends on how much product there is, not on how much reactant remains. While products are forming, the reverse rate increases rather than falling.
• Option C is wrong because at the start the forward rate is greatest (reactants are most concentrated), but the reverse rate is zero.
• Option D is wrong because rates cannot stay fixed if the concentrations of the reacting particles are changing over time.

Because there are no products at first, the reverse reaction cannot occur initially. As products appear and their concentration rises, collisions between product molecules become more frequent, so the reverse rate rises from zero. Therefore, the correct choice is that it increases from zero as products are formed.

At the moment the reaction begins, the system contains reactants only. That starting detail determines what the forward and reverse curves must do at the beginning.

In reversible reactions, rate depends on concentration, which is often shown on a rate–time graph:

1. Forward Rate: At the start it is largest because reactant concentration is highest, and it then drops as reactants are used.
2. Reverse Rate: At $t=0$ there are no products, so the reverse rate must start at exactly zero.

As time passes, products form and their concentration increases. With more product particles available, successful collisions between product particles become more likely, so the reverse rate rises.

From the expected rate–time behavior:

• Option B: The rate that falls because reactants are being used up is the forward rate, not the reverse rate.
• Option C: Initially the forward rate is high and the reverse rate is zero, so the reverse cannot be larger.
• Option D: Before equilibrium, both rates change as concentrations change, so they cannot stay constant in the early stage.

Since product concentration starts at zero and then increases, the reverse rate must also start at zero and then increase. This is exactly what standard rate–time graphs show: the reverse-rate curve rises toward the equilibrium rate.

The key information is the state of the mixture right at the beginning. We are told the reaction begins with reactants present and no products in the container.

For the reverse direction to happen, product particles must collide and react to make reactants.

But at the start, there are no products ($[Products]=0$). With no product particles available, the reverse direction cannot proceed at all, so its initial rate must be zero.

Once the reaction starts, the forward reaction occurs quickly at first because reactants are most concentrated initially. This forward process begins creating product particles.

Collision Theory (covered under the Types of chemical reaction content) tells us reaction rate depends on how many reacting particles are available.

As $[Products]$ rises from zero, collisions between product particles become more frequent, so the reverse reaction rate must rise from zero as well.

• Option B: The reverse rate cannot go down at first because it begins at zero; it can only increase once products exist.
• Option C: The reverse rate is zero initially, so it cannot exceed the forward rate, which starts at its maximum.
• Option D: The rate cannot stay unchanged while concentrations are changing during the approach to equilibrium.

Because the reverse direction starts at zero (no products) and then increases as products are produced, the option describing an increase from zero is the correct one.