Chemical Equilibrium

Chemical equilibrium refers to the final mixture of a chemical reaction, where the reactants and products are done changing. In a chemical reaction, reactants are converted into products. A general belief is that all chemical reactions proceed to completion (where all reactants are converted into products). But this is not true in all cases. A lot of chemical reactions proceed only to a certain extent, i.e. the reactants are not fully converted into products and the resulting mixture contains both reactants and products. After some time, the concentration of reactants or products becomes constant and we get a state of equilibrium for the system.

For example, in Haber's process for ammonia $(\ce{NH3})$ manufacture, ammonia is only 15% of the equilibrium mixture, even at optimized temperature and pressure condition:

$$\ce{N2}+3\ce{H2} \rightleftharpoons 2\ce{NH3}.$$

Irreversible reaction

A chemical reaction which proceeds only in the forward direction so that the reactants are converted into products and products do not react with each other to reform reactants is called an irreversible reaction.

Neutralization reaction like $\ce{HCl}+\ce{NaOH} \to \ce{NaCl}+\ce{H_2O}$ is an example of an irreversible reaction.

Reversible reaction

A chemical reaction in which reactants react together to form products and products formed react with each other directly to reform the original reactant back is known as a reversible reaction.

Reaction of hydrogen with iodine $\ce{H2}+\ce{I2} \rightleftharpoons 2\ce{HI}$ is an example of a reversible reaction.

Consider an elementary reversible reaction taking place in a closed container:

$$a\ce{A}+b\ce{B} \rightleftharpoons c\ce{C}+d\ce{D}.$$

Where A and B react to produce C and D, while the products C and D can also react to produce A and B back.

Initially only the reactants A and B are present, and thus their concentration is maximum. As the reaction proceeds forward, C and D are produced. Thus in the beginning, the rate of forward reaction is very high. But as the reaction progresses, concentration of A and B decreases and on the other hand concentration of C and D increases. Hence the rate of forward reaction slows down and the rate of backward reaction goes on increasing.

Ultimately a state is reached when the rate of forward reaction becomes equal to the rate of backward reaction and the system attains equilibrium. After that, there will be no change in concentration of any of the species.

These are some of the characteristics of chemical equilibrium:

• The chemical reaction should be reversible.
• The equilibrium can be attained only if the system is closed, i.e. the reaction should be carried out in a closed vessel.
• The opposing processes (i.e. forward and backward reactions) occur at the same rate and there is a dynamic but stable condition.
• The observable properties of the system such as concentration, pressure, color, etc. becomes constant at equilibrium and remains unchanged thereafter.
• By bringing a change in conditions such as temperature, pressure or concentration, the equilibrium point can be shifted to the right or left hand side as required. Thus the reaction can be controlled to get more yield of products.
• The equilibrium can be approached from either direction.
• A catalyst does not alter the equilibrium point. It only increases the rate of reaction. The equilibrium is attained however.

Statement: For a homogeneous system at a constant temperature, the rate at which a substance reacts is proportional to its active mass and the rate of a chemical reaction is proportional to the product of active masses of the reacting substances at that moment.

The term active mass means molar concentration per unit volume, i.e. $\text{mol dm}^{-3}$.

The reaction quotient $(Q)$ measures the relative amounts of products and reactants. It is given by

$$Q_c=\frac{{C_c}^c{C_d}^d}{{C_a}^a{C_b}^b}.$$

The rate of forward reaction is $r_f=k_f{C_a}^a{C_b}^b$ and the rate of backward reaction is $r_b=k_{b}{C_c}^c{C_d}^d.$ Equating the forward and backward rates at equilibrium, we obtain

$$\begin{aligned} k_f{C_a}^a{C_b}^b&=k_{b}{C_c}^c{C_d}^d\\ \frac{k_f}{k_{b}}&=\frac{{C_c}^c{C_d}^d}{{C_a}^a{C_b}^b}. \end{aligned}$$

The ratio $\frac{k_f}{k_{b}}$ is called equilibrium constant of reaction and is denoted by $K_c:$

$$K_c=\frac{{C_c}^c{C_d}^d}{{C_a}^a{C_b}^b}.$$

The equilibrium constant is independent of initial concentrations and pressure and is a function of temperature alone. Higher values of $K_c$ or $K_p$ mean that more products are formed and equilibrium is tilted towards the right-hand side. Lower values of $K_c$ or $K_p$ mean reaction does not proceed much from left to right (equilibrium is tilted towards the left-hand side) even though equilibrium has been attained.

For a reaction involving gaseous reactants and products, a new equilibrium constant in terms of partial pressure can be expressed as

$$K_p=\frac{{p_c}^c{p_d}^d}{{p_a}^a{p_b}^b}.$$

If all gases are assumed to be ideal, then

$$p_aV=n_aRT.$$

Since $C_a=\frac{n_a}{V},$ it follows that

$$p_a=C_aRT.$$

This can be used for deriving the relationship between $K_p$ and $K_c$ as follows:

$$\begin{aligned} K_p &=\frac{{(C_cRT)}^c{(C_dRT)}^d}{{(C_aRT)}^a{(C_bRT)}^b}\\ &=\frac{{C_c}^c{C_d}^d}{{C_a}^a{C_b}^b}\times(RT)^{c+d-a-b}\\ &=K_c(RT)^{c+d-a-b} \\ &=K_c(RT)^{\Delta n_g}, \end{aligned}$$

where $\Delta n_g$ denotes the difference between the number of moles of products and the number of moles of the reactants.

Temperature dependence of $K_c:$

Van't Hoff's equation relates equilibrium constant of reaction with temperature as follows:

$$\frac{d(\ln K_c)}{dT}=\frac{\Delta H}{RT^2},$$

where $\Delta H$ is the heat of reaction at temperature $T$.

Homogeneous Equilibrium:

When in an equilibrium reaction, if all the reactants and products are present in the same phase (gaseous or liquid), then it is called homogeneous equilibrium.

• Example: $$\ce{PCl_5}(g) \rightleftharpoons \ce{PCl_3}(g)+\ce{Cl_2}(g)$$
• Equilibrium constant expression: $$\begin{array}{c}&K_C=\frac{[\ce{Cl_2}][\ce{PCl_3}]}{[\ce{PCl_5}]}, &K_P=\frac{P_{\ce{Cl_2}}P_{\ce{PCl_3}}}{P_{\ce{PCl_5}}}\end{array}$$

Note: $[X]$ denotes concentration of $X$.

Heterogeneous Equilibrium:

If reactants and products are found in two or more phases, the equilibria describing them are called heterogeneous equilibrium.

• Example 1: $$\begin{aligned} \ce{AgI}(s) &\rightleftharpoons \ce{Ag}^+(aq) +\ce{I}^- (aq)\\ K_C&=\big[\ce{Ag}^+(aq)\big]\big[\ce{I}^-(aq)\big], \ \ K_p \text{ not defined}. \end{aligned}$$
• Example 2: $$\begin{aligned} \ce{CaO}(s) +\ce{CO_2}(g) &\rightleftharpoons \ce{CaCO_3}(s)\\ K_C&=\frac{1}{[\ce{CO_2}(g) ]}, \ \ K_P=\frac{1}{P_{\ce{CO_2}}}. \end{aligned}$$

Note: $K_C$ and $K_P$ do not include concentration of solids. The concentration of solid substances is taken as unity.

As we know

$$K_c=\frac{{C_c}^c{C_d}^d}{{C_a}^a{C_b}^b}.$$

So, higher values of equilibrium constant indicate that the equilibrium concentration of the product(s) is high and its lower values indicate that the equilibrium concentration of products is low:

• Large values of $K_P$ or $K_C$, larger than about $10^3$, favor the products strongly.

• For intermediate values of $K_P$ or $K_C$, in range of $10^{-3}$to $10^{3}$ , concentration of reactions and products is comparable.

• Small values of $K\ \left(<10^{-3}\right)$ favor the reactants strongly.

Le Chatelier's principle tells us what will happen when equilibrium is disturbed qualitatively. This numerical problem is designed to explain the same principle quantitatively. This problem is large and its very purpose is to clear doubts regarding chemical equilibrium.

Consider an elementary reaction taking place in a closed vessel of volume $1\text{ m}^3$:

$$\ce{A}_{(g)} \rightleftharpoons \ce{B}_{(g)}.$$

We start with pure $\ce{A}$ with initial concentration of $100\text{ mol}$ in $1 \text{ m}^3$. Reaction temperature and pressure are $300\text{ K}$ and $1\text{ atm},$ respectively. Other details are shown below.

We start with a simple question.

To attack this problem, we write the expression for equilibrium concentration $K_c:$

$$\begin{aligned} K_c&=\frac{C_{be}}{C_{ae}}\\ K_c&=\frac{k_f}{k_b}\\&=\frac{2}{1}=2\\ \Rightarrow \frac{C_{be}}{C_{ae}}&=2. \end{aligned}$$

To solve for $C_{ae}$, we need to express $C_{be}$ in terms of $C_{ae}$. This can be done as follows:

$$\begin{aligned} C_{be}&=C_{a0}-C_{ae}\\ \frac{C_{a0}-C_{ae}}{C_{ae}}&=2. \end{aligned}$$

Solving for $C_{ae},$ we have

$$\begin{aligned} C_{ae} &=33.33 \text{ mol/m}^3\\ C_{be} &=100-33.33=66.67 \big(\text{mol/m}^3\big). \end{aligned}$$

If this reaction would be irreversible, we would have $0\text{ mol}$ of $\ce{A}$ and $100\text{ mol}$ of $\ce{B}.$

Remember Van't Hoff equation! That will give equilibrium constant at a particular temperature. For this, we need to solve the following differential equation:

$$\frac{d(\ln K_c)}{dT}=\frac{\Delta H}{RT^2}.$$

Separating variables, we have

$$d(\ln K_c)=\frac{\Delta H}{R}\frac{dT}{T^2}.$$

Integrating gives

$$\ln \frac{K_2}{K_1}=\frac{\Delta H}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right).$$

We will use this relation to find $K_c$ at $T=273\text{ K}:$

$$\begin{aligned} \Delta H&=-10^4\\ K_1&=2\quad &&\text{at} \quad T_1=300\text{ K}\\ K_2&=?\quad &&\text{at} \quad T_2=273\text{ K}. \end{aligned}$$

This will give $K_2=2.97$.

From equilibrium constant formula

$$\begin{aligned} \frac{C_{a0}-C_{ae}}{C_{ae}}&=2.97\\ C_{ae}&=25.17\text{ mol/m}^3 \\ C_{be}&=100-25.17=74.83 \big(\text{mol/m}^3\big). \end{aligned}$$

We see that $C_{be}$ is increased when temperature is lowered.

Similarly, one can calculate $K_c$ at $T=323\text{ K}:$

$$\begin{aligned} K_{323}&=1.50\\ C_{ae}&=39.95 \text{ mol/m}^3 \\ C_{be}&=100-39.95=60.05 \big(\text{mol/m}^3\big). \end{aligned}$$

We notice that $C_{be}$ is increased when temperature is lowered.

These results prove Le Chatelier's principle: An exothermic reaction is favored at low temperature as equilibrium concentration of $\ce{B}$ increases from $66.67 \text{ mol/m}^3$ to $74.83 \text{ mol/m}^3$ when temperature is lowered from $300 \text{ K}$ to $273 \text{ K}.$

We know at $T=300\text{ K},$ $C_{ae}=33.33\text{ mol/m}^3.$ If we can relate concentration of $\ce{A}$ with time, the problem can be easily solved by finding $t$ for this $C_{ae}$. For this we proceed as follows:

Material balance for $\ce{A}:$

$$(\text{Rate of accumulation of }A)=\text{(input-output)}+\text{(generation-consumption)}$$

Since reaction is taking place in a closed vessel, both input=0 and output=0.

• $\text{Rate of accumulation of }A=\frac{dC_a}{dt}$
• $\text{Generation}=k_bC_b$
• $\text{Consumption}=k_fC_a$

Putting these terms in material balance equation gives

$$\frac{dC_a}{dt}=k_bC_b-k_fC_a.$$

Writing $C_b$ in terms of $C_a,$ we have

$$\begin{aligned} C_{be}&=C_{a0}-C_{ae}\\ \frac{dC_a}{dt}&=k_b(C_{a0}-C_{ae})-k_fC_a\\ \frac{dC_a}{dt}&=k_bC_{a0}-(k_f+k_b)C_a\\ \frac{dC_a}{dt}+(k_f+k_b)C_a&=k_bC_{a0}. \end{aligned}$$

Substituting in the values $k_f=2 s^{-1}, k_b=2 s^{-1}$ and $C_{a0}=100 \text{ mol/m}^3$ gives

$$\frac{dC_a}{dt}+3C_a=100.$$

The above equation can be solved using the method of integration factor:

$$C_a=\frac{100}{3}+Ne^{-3t},$$

where $N$ is the constant of integration. Its value can be found out by the following initial condition:

$$\text{at } t=0, C_a=C_{a0}=100 \text{ mol/m}^3,$$

which gives $N=\frac{200}{3}.$ Therefore,

$$C_a=\frac{100}{3}+\frac{200e^{-3t}}{3}.$$

Now, we have the relation between $C_a$ and $t.$

To find time required for equilibrium, put $C_{ae}=\frac{100}{3},$ then

$$\frac{100}{3}=\frac{100}{3}(1+2e^{-3t}) \Rightarrow e^{-3t}=0,$$

which implies $t=\infty.$

This is not good. We did such a lengthy calculation and got infinity as a result. This may predict that equilibrium cannot be achieved in finite time. But if we find time required for $C_a$ to become 33.34, it will be only $t=3.07 s.$ This strange behavior can be explained by the nature of the function $C_a=\frac{100}{3}+\frac{200e^{-3t}}{3}$ which becomes asymptotic to the line $C=\frac{100}{3}$. For practical purpose, time required for $C_{ae}=33.34$ is enough.

We know a catalyst increases rate of reaction without itself actually getting consumed. But here the question is whether it will change the equilibrium concentrations of $\ce{A}$ and $\ce{B}.$ The answer is simply no. It increases both rate constants to the same extent so that the ratio of rate constants that is $K_c$ is the same as in non-catalytic reaction. For a proof see How does catalyst work.

For finding time required for equilibrium, we proceed as follows:

We have increased rate constants for both forward and backward reactions in catalytic reaction as

$$k_{fc}=4 s^{-1} \space \text{ and } \space k_{bc}=2 s^{-1} .$$

Now writing the material balance for $\ce{A},$ we have

$$\frac{dC_a}{dt}+(k_{fc}+k_{bc})C_a=k_bC_{a0}.$$

Substituting in the values $k_{fc}=2 s^{-1}, k_{bc}=2 s^{-1}$ and $C_{a0}=100 \text{ mol/m}^3$ gives

$$\frac{dC_a}{dt}+6C_a=200.$$

Try to solve this equation by integration factor method as we did above and get the final equation

$$C_a=\frac{100}{3}+\frac{200e^{-6t}}{3}.$$

Putting $C_{ae}=\frac{100}{3}$ and solving for t, we will get same result as we did above in case of non-catalytic reaction. i.e. $t=\infty$. This time we know how to avoid this problem. We proceed to find time for $C_{a}=33.34$ and get

$$t=1.56 s.$$

We see that time required for concentration to reach $33.34$ is half of what we get in case of non-catalytic reaction.

For non-catalytic equation, we had

$$C_a=\frac{100}{3}+\frac{200e^{-3t}}{3}.$$

In general, one can easily show that time required will be less in catalytic reaction than non-catalytic reaction for the same concentration.

Here is the graph of concentration of $\ce{A}$ in two cases.

Since one mole of $\ce{A}$ gives one mole of $\ce{B},$ there is no increase or decrease of pressure during reaction.Hence changing pressure at equilibrium will not change the equilibrium concentration of $\ce{A}$ and $\ce{B}.$

At equilibrium, we have $33.33\text{ mol}$ of $\ce{A}$ and $66.67\text{ mol}$ of $\ce{B}.$ Now we are adding another $50\text{ mol}$ of $\ce{A}.$ How will this disturb the equilibrium? Adding this extra $\ce{A}$ at previous equilibrium will result in same equilibrium as if it will be if we start with $100+50=150\text{ mol}$ of $\ce{A}.$ Another way to think is that $50\text{ mol}$ of $\ce{A}$ will give $33.33\text{ mol}$ $\ce{B}$ and $16.667\text{ mol}$ $\ce{A}$ will remain unreacted. So in total we will have $66.67+33.33=100\text{ mol}$ $\ce{B}$ and $33.33+16.6667=50\text{ mol}$ $\ce{A}.$ Hence we will have more $\ce{B}$ at equilibrium.

You can point out that the percentage of $\ce{B}$ in equilibrium mixture is the same in both cases ($66.67$%). But this is not the case in general.

You can look into some related topics. Though these are related to physics, you can understand the meaning of equilibrium in different branches of science.

• Equilibrium deals with the general aspect of equilibrium.

• Acid/Base Equilibrium deals with neutralization of acids and bases.

• Torque - Equilibrium

• Zeroth law of thermodynamics explains how two systems get into equilibrium with a third system.

• Second Law of Thermodynamics