Equilibrium Lecture Notes

In general we can write the equilibrium expression for a reaction. Consider the following general chemical equation;

aA(g) + bB(g) cC(g) + dD(g)

The equilibrium expression for the reaction is,

K measures the extent of a reaction.

There are two different equilibrium constants in gas phase equilibrium systems. K_c is the equilibrium constant when the amounts of each substance at equilibrium is expressed in mol per liter (M). K_p is the equilibrium constant when the amounts of each substance at equilibrium is expressed in atmospheres (atm). In either case the equilibrium constant is report as a unitless number.

Consider the following reactions;

H_2(g) + Cl_2(g) 2HCl_(g)

The equilibrium constant expression for this reaction is

For the reaction

3H_2(g) + N_2(g) 2NH_3(g)

The equilibrium constant expression for this reaction is

Notice the first equilibrium constant is very large and the second is very small. Since K is a measure of the extent of the reaction it is important to recognize that a large equilibrium constant, K, means the reaction 'favors' the products. That is, at equilibrium the concentration of the products is greater than the concentration of the reactants. We also say the position of the equilibrium lies on the right side of the reaction. Similarly, a small equilibrium constant means the reaction does not go very far towards the products, and the concentration of the products is small compared to the concentration of the reactants. The equilibrium lies on the left side of the reaction.

To summarize;

large K:

position of equilibrium lies on the products side;

concentration of the products is greater than the concentration of the reactants;

small K:

position of equilibrium lies on the reactants side;

concentration of the reactants is greater than the concentration of the products;

Sample Problems;

Write the equilibrium constant expressions for the following reactions;

a) PCl_5(g) PCl_3(g) + Cl_2(g)

b) 4NH_3(g) + 3O_2(g) 2N_2(g) + 6H₂O_(g)

For each of the following reactions which side (reactants or products) will have the higher concentrations at equilibrium?

a) C_(s) + CO_2(g) + 2Cl_2(g) 2COCl₂_(g) K = 4.68 x 10⁹

b) H₂O_(l) H⁺_(aq) + OH^-_(aq) K = 1.0 x 10^-14

There are some additional 'properties' of the equilibrium constant and the equilibrium constant expression that are important.

1) reversing the reaction;

2) multiplying the chemical equation by a number.

In the first case consider the reaction;

H_2(g) + Cl_2(g) 2HCl_(g)

The equilibrium constant expression for this reaction is

If the reaction is reversed

2HCl_(g) H_2(g) + Cl_2(g)

The equilibrium constant expression for this reaction is

The equilibrium constant

K'_c = 1/K_c

K'_c = 1/(2 x 10⁷) = 5 x 10^-8

When the chemical equation is reversed the the equilibrium constant is the reciprocal of the original equilibrium constant.

In the second case consider the equation,

The equilibrium constant expression for this reaction is

The equilibrium constant

K''_c = (K_c)^1/2

K'_c = (2 x 10⁷)^1/2 = 4.5 x 10³

Sample problems;

a) 2SO_2(g) + O_2(g) 2SO_3(g) K = 5.6 x 10⁴ at 350 degrees Celsius. Calculate K for the reaction SO_2(g) + 1/2O_2(g) SO₃_(g)

b) 1/2N_2(g) + 3/2H_2(g) NH_3(g) K = 1.09 at 375 degrees Celsius. Calculate K for the reaction N_2(g) + 3H_2(g) 2NH₃_(g)

We can use both of these properties to determine equilibrium constants for new reactions. Consider the following example;

Given the following information

Chemical Equation	Equilibrium Constant (K)
C_(s) + CO_2(g) 2CO_(g)	1.3 x 10¹⁴
CO_(g) + Cl_2(g) COCl_2(g)	6.0 x 10^-3

Calculate K for the reaction,

C_(s) + CO_2(g) + Cl_2(g) 2COCl_2(g)

To solve this problem we can use the Hess' Law approach that allows addition of chemical equations. If we add the equations together, after multiplying the second equation by 2 we will obtain the new equation.

C_(s) + CO_2(g) 2CO_(g)

2CO_(g) + 2Cl_2(g) 2COCl_2(g)

Adding (note the 2CO_(g) cancel in each equation),

C_(s) + CO_2(g) + 2Cl_2(g) 2COCl_2(g)

How do we handle the equilibrium constants? First, we need to look at the equilibrium constant expressions for all three reactions,

The only wat to obtain the equilibrium expression for the last equation is to multiply the equilibrium expressions of the first two equations,

Therefore, the equilibrium constant for the reaction

C_(s) + CO_2(g) + 2Cl_2(g) 2COCl_2(g)

is,

K''c = Kc x K'c = 1.3 x 10¹⁴ x (6.0 x 10^-3)2 K''c = 4.68 x 10⁹

When doing these types of problems when we add chemical equations we must multiply the equilibrium constants.

Try this sample problem,

Given the following information

Chemical Equation	Equilibrium Constant (K)
S_(s) + O_2(g) SO_2(g)	4.2 x 10⁵²
2S_(s) + 3O_2(g) 2SO_3(g)	9.8 x 10¹²⁸

Calculate K for the reaction,

2SO_2(g) + O_2(g) 2SO_3(g)

Equilibrium Lecture Notes

In general we can write the equilibrium expression for a reaction. Consider the following general chemical equation;

aA(g) + bB(g) cC(g) + dD(g)

The equilibrium expression for the reaction is,

K measures the extent of a reaction.

Consider the following reactions;

H2(g) + Cl2(g) 2HCl(g)

The equilibrium constant expression for this reaction is

For the reaction

3H2(g) + N2(g) 2NH3(g)

The equilibrium constant expression for this reaction is

To summarize;

large K:

position of equilibrium lies on the products side;

concentration of the products is greater than the concentration of the reactants;

small K:

position of equilibrium lies on the reactants side;

concentration of the reactants is greater than the concentration of the products;

Sample Problems;

Write the equilibrium constant expressions for the following reactions;

a) PCl5(g) PCl3(g) + Cl2(g)

b) 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)

For each of the following reactions which side (reactants or products) will have the higher concentrations at equilibrium?

a) C(s) + CO2(g) + 2Cl2(g) 2COCl2(g) K = 4.68 x 109

b) H2O(l) H+(aq) + OH-(aq) K = 1.0 x 10-14

There are some additional 'properties' of the equilibrium constant and the equilibrium constant expression that are important.

1) reversing the reaction;

2) multiplying the chemical equation by a number.

In the first case consider the reaction;

H2(g) + Cl2(g) 2HCl(g)

The equilibrium constant expression for this reaction is

If the reaction is reversed

2HCl(g) H2(g) + Cl2(g)

The equilibrium constant expression for this reaction is

The equilibrium constant

K'c = 1/Kc

K'c = 1/(2 x 107) = 5 x 10-8

When the chemical equation is reversed the the equilibrium constant is the reciprocal of the original equilibrium constant.

In the second case consider the equation,

The equilibrium constant expression for this reaction is

The equilibrium constant

K''c = (Kc)1/2

K'c = (2 x 107)1/2 = 4.5 x 103

Sample problems;

a) 2SO2(g) + O2(g) 2SO3(g) K = 5.6 x 104 at 350 degrees Celsius. Calculate K for the reaction SO2(g) + 1/2O2(g) SO3(g)

b) 1/2N2(g) + 3/2H2(g) NH3(g) K = 1.09 at 375 degrees Celsius. Calculate K for the reaction N2(g) + 3H2(g) 2NH3(g)

We can use both of these properties to determine equilibrium constants for new reactions. Consider the following example;

Given the following information

Chemical Equation

Equilibrium Constant (K)

C(s) + CO2(g) 2CO(g)

1.3 x 1014

CO(g) + Cl2(g) COCl2(g)

6.0 x 10-3

Calculate K for the reaction,

C(s) + CO2(g) + Cl2(g) 2COCl2(g)

To solve this problem we can use the Hess' Law approach that allows addition of chemical equations. If we add the equations together, after multiplying the second equation by 2 we will obtain the new equation.

C(s) + CO2(g) 2CO(g)

2CO(g) + 2Cl2(g) 2COCl2(g)

Adding (note the 2CO(g) cancel in each equation),

C(s) + CO2(g) + 2Cl2(g) 2COCl2(g)

How do we handle the equilibrium constants? First, we need to look at the equilibrium constant expressions for all three reactions,

The only wat to obtain the equilibrium expression for the last equation is to multiply the equilibrium expressions of the first two equations,

Therefore, the equilibrium constant for the reaction

C(s) + CO2(g) + 2Cl2(g) 2COCl2(g)

is,

K''c = Kc x K'c = 1.3 x 1014 x (6.0 x 10-3)2 K''c = 4.68 x 109

When doing these types of problems when we add chemical equations we must multiply the equilibrium constants.

Try this sample problem,

Given the following information

Chemical Equation

Equilibrium Constant (K)

S(s) + O2(g) SO2(g)

4.2 x 1052

2S(s) + 3O2(g) 2SO3(g)

9.8 x 10128

Calculate K for the reaction,

2SO2(g) + O2(g) 2SO3(g)

H_2(g) + Cl_2(g) 2HCl_(g)

3H_2(g) + N_2(g) 2NH_3(g)

a) PCl_5(g) PCl_3(g) + Cl_2(g)

b) 4NH_3(g) + 3O_2(g) 2N_2(g) + 6H₂O_(g)

a) C_(s) + CO_2(g) + 2Cl_2(g) 2COCl₂_(g) K = 4.68 x 10⁹

b) H₂O_(l) H⁺_(aq) + OH^-_(aq) K = 1.0 x 10^-14

H_2(g) + Cl_2(g) 2HCl_(g)

2HCl_(g) H_2(g) + Cl_2(g)

K'_c = 1/K_c

K'_c = 1/(2 x 10⁷) = 5 x 10^-8

K''_c = (K_c)^1/2

K'_c = (2 x 10⁷)^1/2 = 4.5 x 10³

a) 2SO_2(g) + O_2(g) 2SO_3(g) K = 5.6 x 10⁴ at 350 degrees Celsius. Calculate K for the reaction SO_2(g) + 1/2O_2(g) SO₃_(g)

b) 1/2N_2(g) + 3/2H_2(g) NH_3(g) K = 1.09 at 375 degrees Celsius. Calculate K for the reaction N_2(g) + 3H_2(g) 2NH₃_(g)

C_(s) + CO_2(g) 2CO_(g)

1.3 x 10¹⁴

CO_(g) + Cl_2(g) COCl_2(g)

6.0 x 10^-3

C_(s) + CO_2(g) + Cl_2(g) 2COCl_2(g)

C_(s) + CO_2(g) 2CO_(g)

2CO_(g) + 2Cl_2(g) 2COCl_2(g)

Adding (note the 2CO_(g) cancel in each equation),

C_(s) + CO_2(g) + 2Cl_2(g) 2COCl_2(g)

C_(s) + CO_2(g) + 2Cl_2(g) 2COCl_2(g)

K''c = Kc x K'c = 1.3 x 10¹⁴ x (6.0 x 10^-3)2 K''c = 4.68 x 10⁹

S_(s) + O_2(g) SO_2(g)

4.2 x 10⁵²

2S_(s) + 3O_2(g) 2SO_3(g)

9.8 x 10¹²⁸

2SO_2(g) + O_2(g) 2SO_3(g)