Concentration-Time and Mole-Time Graphs

Concentration-time and mole-time graphs

A change in concentration of a substance would appear as a sharp increase or decrease in the concentration or number of moles of that substance and a gradual change in the other substances.
A change in temperature would affect both the forward and reverse reactions. However, one reaction would be affected more than the other. So the reactants and products would be affected gradually, in the opposite direction (one increased, the other decreased).
A change in pressure of the reaction would cause a sharp increase or decrease in all the reactants and products. For an increase in pressure, if the forward reaction is then favoured the reactant concentrations will decrease, and if the reverse reaction is then favoured the product concentrations will decrease.
The addition of a catalyst would increase both the forward and reverse reaction rates, meaning the equilibrium is reached faster. So if the reaction is already at equilibrium there will be no effect on a concentration-time or moles-time graph.

Example: Concentration-Time Graphs

Question

Consider the following chemical equilibrium and graph and answer the questions that follow.

$\text{CO}(\text{g}) + \text{Cl}_{2}(\text{g})$ $\rightleftharpoons$ $\text{COCl}_{2}(\text{g})$

How much time was necessary for the system to reach equilibrium for the first time?
How do the rates of the forward and reverse reactions compare at the following times:
- t = $\text{5}$ $\text{s}$
- t = $\text{17}$ $\text{s}$
- t = $\text{23}$ $\text{s}$
Determine the equilibrium constants for the system at t = $\text{17}$ $\text{s}$, $\text{25}$ $\text{s}$, and $\text{45}$ $\text{s}$.
What happens at t = $\text{20}$ $\text{s}$? Explain your answer by referring to Le Chatelier's principle.
What effect does the stress at t = $\text{20}$ $\text{s}$ have on $\text{K}_{\text{c}}$?

Step 1: Check the axes so that you know what the variables are on this graph

The axes are labelled concentration and time. Therefore this is a concentration-time graph.

Step 2: How much time was necessary for the system to reach equilibrium?

The concentration of all three compounds becomes constant at t = $\text{15}$ $\text{s}$. This means that the reaction has reached equilibrium.

Step 3: How can you determine which rate is faster from the concentration graphs?

If the concentrations of the reactants ($\text{CO}$ and $\text{Cl}_{2}$) are steadily decreasing or the concentration of the product ($\text{COCl}_{2}$) is increasing then the forward reaction is faster than the reverse reaction.

Similarly, if the concentrations of the reactants ($\text{CO}$ and $\text{Cl}_{2}$) are steadily increasing or the concentration of the product ($\text{COCl}_{2}$) is decreasing then the reverse reaction is faster than the forward reaction.

Step 4: Compare the forward and reverse reaction rates at t = $\text{5}$ $\text{s}$, t = $\text{17}$ $\text{s}$ and t = $\text{23}$ $\text{s}$

At t = $\text{5}$ $\text{s}$ and at t = $\text{23}$ $\text{s}$ the concentrations of the reactants are decreasing and the concentration of the product is increasing. Therefore the rate of the forward reaction is faster than the rate of the reverse reaction.

At t = $\text{17}$ $\text{s}$ the concentrations of the reactants and products are constant (unchanging). Therefore the reaction is in equilibrium and the rate of the forward reaction equals the rate of the reverse reaction.

Step 5: What are the concentrations of $\text{CO}$, $\text{Cl}_{2}$ and $\text{COCl}_{2}$ at t = $\text{17}$ $\text{s}$?

Reading off the graph you can see that:

$[\text{CO}]$ = $\text{1.75}$ $\text{mol.dm$^{-3}$}$

$[\text{Cl}_{2}]$ = $\text{0.6}$ $\text{mol.dm$^{-3}$}$

$[\text{COCl}_{2}]$ = $\text{0.75}$ $\text{mol.dm$^{-3}$}$

Step 6: Write an expression for the equilibrium constant of this reaction and calculate $\text{K}_{\text{c}}$.

$\text{K}_{\text{c}} = \dfrac{\text{[COCl}_{2}{\text{]}}}{\text{[CO][Cl}_{2}{\text{]}}}$

$\text{K}_{\text{c}} = \dfrac{\text{0.75}}{(\text{1.75})(\text{0.6})} =$ $\text{0.71}$

Step 7: What happens at t = $\text{20}$ $\text{s}$?

The concentration of $\text{CO}$ increases sharply. The concentrations of the other compounds do not change dramatically. Therefore $\text{CO}$ must have been added to the system.

After this addition of $\text{CO}$ there is a shift to reduce the amount of $\text{CO}$, that is in the forward direction. Therefore the concentration of the product increases while the concentrations of the reactants decrease.

Step 8: What effect will this have on $\text{K}_{\text{c}}$?

Only a change in temperature affects $\text{K}_{\text{c}}$, therefore this will have no effect on $\text{K}_{\text{c}}$.

Example: Concentration-Time Graphs

Question

Consider the following chemical equilibrium and graph and answer the questions that follow.

$\text{H}_{2}(\text{g}) + \text{I}_{2}(\text{g})$ $\rightleftharpoons$ $2\text{HI}(\text{g})$

After how many seconds does the system reach equilibrium?
Calculate the value of the equilibrium constant.
Explain what happened at t = $\text{20}$ $\text{s}$.
If the change at t = $\text{35}$ $\text{s}$ is due to an increase in temperature, is the reaction exothermic or endothermic? Explain your answer.

Step 1: Check the axes so that you know what the variables are on this graph

The axes are labelled concentration and time. Therefore this is a concentration-time graph.

Step 2: After how many seconds does the system reach equilibrium?

The concentration of all three compounds becomes constant at t = $\text{10}$ $\text{s}$. This means that the system takes $\text{10}$ $\text{s}$ to reach equilibrium.

Step 3: What are the concentrations of $\text{H}_{2}$, $\text{I}_{2}$, and $\text{HI}$ at t = $\text{10}$ $\text{s}$, $\text{25}$ $\text{s}$, and $\text{45}$ $\text{s}$?

Reading off the graph you can see that at $\text{10}$ $\text{s}$:

$[\text{H}_{2}]$ and $[\text{I}_{2}]$ = $\text{0.75}$ $\text{mol.dm$^{-3}$}$, $[\text{HI}]$ = $\text{6.0}$ $\text{mol.dm$^{-3}$}$

At $\text{25}$ $\text{s}$:

$[\text{H}_{2}]$ and $[\text{I}_{2}]$ = $\text{0.6}$ $\text{mol.dm$^{-3}$}$, $[\text{HI}]$ = $\text{4.8}$ $\text{mol.dm$^{-3}$}$

At $\text{45}$ $\text{s}$:

$[\text{H}_{2}]$ and $[\text{I}_{2}]$ = $\text{1.5}$ $\text{mol.dm$^{-3}$}$, $[\text{HI}]$ = $\text{2.75}$ $\text{mol.dm$^{-3}$}$

Step 4: Write an expression for the equilibrium constant of this reaction and calculate $\text{K}_{\text{c}}$.

$\text{K}_{\text{c}} = \dfrac{\text{[HI]}^{2}}{\text{[H}_{2}{\text{][I}_{2}{\text{]}}}}$

At t = $\text{10}$ $\text{s}$:

$\text{K}_{\text{c}} = \dfrac{\text{6.0}^{2}}{(\text{0.75})(\text{0.75})} =$ $\text{64.0}$

At t = $\text{25}$ $\text{s}$:

$\text{K}_{\text{c}} = \dfrac{\text{4.8}^{2}}{(\text{0.6})(\text{0.6})} =$ $\text{64.0}$

At t = $\text{45}$ $\text{s}$:

$\text{K}_{\text{c}} = \dfrac{\text{2.75}^{2}}{(\text{1.5})(\text{1.5})} =$ $\text{3.4}$

The different $\text{K}_{\text{c}}$ at $\text{45}$ $\text{s}$ means that the event at t = $\text{35}$ $\text{s}$ must be a change in temperature.

Step 5: Explain what happened at t = $\text{20}$ $\text{s}$?

The concentration of $\text{HI}$ decreases sharply, as a result there is a slight decrease in the concentration of $\text{H}_{2}$ and $\text{I}_{2}$. Therefore $\text{HI}$ must have been removed from the system.

After this there is a shift to increase the amount of $\text{HI}$, that is in the forward direction.

Step 6: If the change at t = $\text{35}$ $\text{s}$ is due to an increase in temperature, is the reaction exothermic or endothermic? Explain your answer.

An increase in temperature caused the concentration of the product to decrease and the concentrations of the reactants to increase. This means that the reverse reaction has been favoured.

An increase in temperature will favour the reaction that takes heat in and cools the reaction vessel (endothermic). Therefore the reverse reaction must be endothermic and the forward reaction must be exothermic. The reaction is exothermic.

Concentration-Time and Mole-Time Graphs