# Exothermic vs. Endothermic and K

An exothermic reaction occurs when the temperature of an isolated system, or one that interacts with its surroundings, increases due to the evolution of heat. This heat is then released into the surrounding resulting in an overall negative quantity for the heat of reaction (qrxn< 0). Conversely, an endothermic reaction occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system gains heat. Endothermic reactions result in an overall positive heat of reaction (qrxn> 0).

### Exothermic Reactions

Exothermic and endothermic reactions cause energy level differences and thus differences in enthalpy (ΔH), the sum of all potential and kinetic energies. ΔH is determined by the system, not the surrounding environment in a reaction.

• A system that releases heat to the surroundings, an exothermic reaction, creates a negative ΔH because the enthalpy of the products is lower than the enthalpy of the reactants of the system (Figure 1).
• A system that absorbs heat from the surroundings, an endothermic reaction, creates a positive ΔH, because the enthalpy of the products is higher than the enthalpy of the reactants of the system.

### Exothermic Reactions

C(s) + O2(g) → CO2(g); ΔH = –393.5 kJ

H2(g) + 1/2 O2(g) → H2O(l); ΔH = –285.8 kJ

Since the enthalpies of these reactions are less than zero, they are exothermic reactions.

### Endothermic Reactions

N2(g) + O2(g) → 2NO(g); ΔH = +180.5 kJ > 0

C(s) + 2S(s) → CS2(l); ΔH = +92.0 kJ > 0

Since the enthalpies of these reactions are greater than zero, they are endothermic reactions.

### The Equilibrium Constant K

The equilibrium constant, represented by K, defines the relationship amongst the concentrations of chemical substances involved in a reaction at the time of equilibrium. The Le Châtelier's Principle states that if a stress, such as changing temperature, pressure, or concentration, is added on an equilibrium reaction, the reaction will either shift to restore the equilibrium. When using exothermic and endothermic reactions, this added stress is a change in temperature. The equilibrium constant is represented by K and shows how far the reaction will progress at a specific temperature by determining the ratio of products to reactions using equilibrium concentrations.

In an equilibrium expression: $$aA + bB \rightleftharpoons cC + dD$$

$K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$

K=equilibrium constant, [A], [B], [C], [D] are concentrations and a, b, c, d are the stoichiometric coefficients of the balanced equation.

#### Exothermic Reactions

• If K decreases with an increase in temperature, the reaction moves to the left.
• If K increases with a decreases in temperature, the reaction to moves to the right.

#### Endothermic Reactions

• If K increases with an increase in temperature, the reaction to moves to the right.
• If K decreases with a decrease in temperature, the reaction to move to the left.

### K Values

If the products dominate in a reaction, the value for K >1. (The larger the K value, the more the reaction will go to completion and thus to the right).

• If K=1, neither the reactants nor the products are favored. Note that this is not the same as both being favored.
• If the reactants dominate in a reaction, the value for K<1. (The smaller the K value, the more the reaction will go to the left.

Example

Lets say that the following reaction is at equilibrium and that the concentration of N2 is 2 M, the concentration of H3 is 4 M and that concentration of NH3 is 3 M. What is the value of K?

$N_2 + 2H_3 \rightleftharpoons 2NH_3$

Since we know the formula for Kc, we just plug in the coefficients and the concentrations into the equation to get the value.

$K_c = \dfrac{[NH_3]^2}{[N_2]^1[H3]^2}$

$K_c = \dfrac{[3]^2}{[2]^1[4]^2}$

We get that the value of $$K_c = \frac{9}{32}$$.

### Practice Problems

1. Determine the equilibrium constant, K, for the following chemical reaction at equilibrium if the molar concentrations of the molecules are 0.20 M H2, 0.10 M NO, 0.20 M H2O, and 0.10M N2

2H2 (g) + 2NO (g) → 2H2O (g) + N2 (g)

2. In the previous equation, in what way will the equilibrium shift to?

3. In the following reaction, the temperature is increased and the K value decreases from 0.75 to 0.55. What kind of reaction is this, an exothermic or an endothermic reaction?

N2 (g) + 3H2 ↔ 2NH3 (g)

4. In the following reaction, in what direction will the equilibrium shift to if there is an increase in temperature and the enthalpy of reaction is given such that ΔH = -92.5 kJ:

PCl3(g) + Cl2(g) ↔ PCl5(g)    ΔH = -92.5 kJ

5. In the following reaction, in what direction will the equilibrium shift to if there is an increase in temperature and the enthalpy of reaction is given such that ΔH= +92.0 kJ:

C(s) + 2S(s) → CS2(l); ΔH = +92.0 kJ

### Solutions

1. Using the equilibrium constant equation and plugging in the concentration values of each molecule:

$K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$

$K_c= 0.2$

2. Since the Kc = 0.2, which is less than zero, the reaction will shift to the left because the energy of the reactants will be greater than the energy of the products.

3. Since the K value decreases with an increase in temperature, the reaction is an exothermic reaction.

4. In the initial reaction, the energy given off is negative and thus is an exothermic reaction. However, an increase in temperature will allow the system to absorb energy and thus favor an endothermic reaction and shift to the left.

5. In the initial reaction, the energy given off is positive and thus is an endothermic reaction. Moreover, an increase in temperature will allow the system to absorb more energy and keep favoring an endothermic reaction and shift to the right.

### References

1. Petrucci, Harwood, Herring, Madura. General Chemistry Principles & Modern Applications. Prentice Hall. New Jersey, 2007

### Contributors

• Alyson Salmon, Nikita Patel (UCD), Deepak Nallur (UCD)

Viewing 1 of 1 comments: view all
RE Practice Problem 5 (above)
I thought that in an endothermic reaction we could consider heat as a “reactant”. Increasing temperature adds heat to the reactant side. Then, it seems, the reaction should shift right; forming more product. But your answer says shift left.
Posted 11:57, 20 Oct 2013
Viewing 1 of 1 comments: view all
19:04, 18 Jan 2014

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