We often hear of combustion reactions producing carbon dioxide and water. This is true when the reactants contain all three elements (hydrogen, oxygen, and carbon) needed to make carbon dioxide and water. The trick here is to understand that although combustion is just an oxidation reaction (ie, creating more carbon-oxygen bonds), all reactions must still be balanced. One answer choice fails to balance hydrogen in the reaction.

Although we do see an oxidation reaction with the formation of carbon dioxide, the reaction is NOT balanced. This reaction cannot possibly take place as written.

An oxidizing agent is one that oxidizes another element and, becomes reduced in the process. Oxidizing and reducing agents are found in the reactants of the equation, eliminating Al³⁺ and Cr(s) as answer choices. Cr³⁺ gains three electrons to become Cr(s), which has an oxidation number of zero. Cr³⁺ becomes reduced, while Al(s) is oxidized, therefore, Cr³⁺ is the oxidizing agent. Remember "OIL RIG": Oxidation Is Loss of electrons and Reduction Is Gain of electrons.

There are a couple options to solve this problem. First, we could simply remember that anything which combines with oxygen in a redox reaction is oxidized, and the associated oxygen is reduced.

Another option is to think about the oxidation numbers of each element (an element whose oxidation number increases is oxidized, and one whose oxidation number decreases is reduced). If we choose this method, recall that free elements (Fe and O₂ in this case) have oxidation number 0. Also, oxygen in compounds has oxidation number -2. Oxygen's oxidation number decreases from 0 to -2, meaning that oxygen is reduced. Since oxidation and reduction occur at the same time in different elements of the redox reaction, iron's oxidation number must increase, so iron is oxidized.

This problem requires us to determine the oxidation number of carbon as a reactant and as a product. The oxidation number of carbon as a reactant is –4, because it is attached to four hydrogens each with a charge of +1. The overall charge on the molecule is zero, the carbon must cancel out the charges contributed by hydrogen.

Carbon as a product has an oxidation number of +4, because it is attached to two oxygens, each with an oxidation number of –2. Again, we know that the molecule is neutral, and carbon must balance the charges from oxygen

Since the oxidation number of carbon went from –4 to +4, we conclude that carbon has been oxidized in the reaction. Any atom that loses electrons is oxidized, while any atom to gain electrons is reduced.

K_eq only includes the concentrations of gases and aqueous solutions. Pure solid and liquid concentrations are left out of the equation.

K_eq is given by the equation below, where the concentrations expressed are the equilibrium concentrations.

K_eq is a property of a given reaction at a given temperature. It is unaffected by catalysts, which only affect rate and activation energy. As the value of K_eq increases, the equilibrium concentration of products must also increase, based on the equation.

Because Q is now greater than K_eq, we know that we need to run the reaction in reverse to come back to equilibrium, where Q = K_eq.  A larger Q value indicates that [products] must be decreased in order to equilibrate at K_eq.

The first activation energy we have to overcome in the conversion of products to reactants is the difference between the energy of the products (point 5) and the first transition state (point 4) relative to the products.

The law of mass action is used to compare the chemical equation to the equilibrium constant. In the equation, the product concentration are on the top, and the reactant concentrations are on the bottom. Coefficients in the balanced equation become the exponents seen in the equilibrium equation. While pure solids and liquids can be excluded from the equation, pure gases must still be included.

If the reaction is at equilibrium, we know that the law of mass action will equal the equilibrium constant given in the above information. When the reaction contains only gases, partial pressure values can be substituted for concentrations. As a result, we simply need to add the values into the equation and solve for the partial pressure of carbon monoxide (CO).

When a reaction reaches equilibrium, the forward and reverse reaction rates are equal. The arrival of a reaction at equilibrium does not speak to the concentrations. Arrival at equilibrium also does not change the inherent energy properties of the reactants and products.

If the reaction quotient is larger than the equilibrium constant, then there is a relative abundance of products compared to their equilibrium concentration. By proxy, there must be a deficiency of reactants with respect to the equilibrium concentrations. The reactants will need to increase in concentration until the reaction reaches equilibrium.