# 1.4 Chemical Reactions

## What are chemical reactions? 

Understanding how chemical reactions work is very fundamental in geology and environmental sciences.  It is particularly useful for determining how minerals and rocks form from basic elements, how chemical weathering reactions breakdown minerals and rocks, how water dissolves salts and minerals, how contaminants move in aqueous environments, how ocean acidification is impacting corals, nuclear reactions occur, and so on.  

The main principles that underlie chemical equations are the conservation of mass and the transformation of chemical entities or species. These chemical equations can contain significant information regarding the chemical system depicted by the chemical equations.  In all chemical equations, chemical species are separated by an arrow and indicate the phrase *to produce*} or, in other contexts, the direction of the reaction.  This arrow can point in one direction or both directions (e.g., $\ce{->, <-, <=>}$).  Chemical species that appear to the left of the arrow are called reactants, and species on the right side of the arrow are products.  The "+" sign indicates the word *and* whether it appears on the left or right side of the arrow.  The states of matter are also important in chemical equations. They must be explicitly shown to deal with ambiguous situations that are unclear if the chemical in the equation is solid, liquid, or gas. The standard symbols used are $s$, $l$, $g$, and $aq$ for solid, liquid, gas, and aqueous (dissolved in water).

```{dropdown} Example: Dissolution of $\ce{CO2}$ in $\ce{H2O}$

Consider how liquid water is acidified by atmospheric $\ce{CO2 (g)}$.

$$
\begin{align*}
\ce{		
CO2 (g) &<=>[H2O]  CO2 (aq)\\
CO2 (aq) + H2O (l) &<=> H2CO3 (aq)
}
\end{align*}
$$

>The first equation is a representation of the chemical reaction where gaseous $\ce{CO2}$ (or $\ce{CO2 (g)}$) dissolves in water to form aqueous $\ce{CO2}$ (or $\ce{CO2 (aq)}$). Because $\ce{CO2 (aq)}$ does not explicitly form a chemical bond with water, $\ce{H2O}$ is shown on top of the arrow to indicate its implicit involvement in the reaction.  The $\ce{<=>}$ arrow indicates that the reaction is reversible.  The second equation represents a chemical reaction where $\ce{CO2 (aq)}$ explicitly reacts with $\ce{H2O (l)}$ to produce $\ce{H2CO3 (aq)}$ (carbonic acid.)  

```

Different types of chemical reactions will be explored over the rest of the course, including precipitation, dissolution, acid-base, redox, and nuclear reactions.


## Balancing Chemical Reactions

Consider the following chemical reaction where $\ce{H2(g)}$ reacts explosively with $\ce{O2(g)}$ to form $\ce{H2O(l)}$.

$$\ce{H2(g) + O2(g) -> H2O(l)}$$

According to this reaction, two atoms of $\ce{H}$ and two atoms of $\ce{O}$ react to form two atoms of $\ce{H}$ and one atom of $\ce{O}$.  This is a violation of the conservation of mass. 

```{admonition} The three hypotheses that make up Dalton's atomic theory

1. Matter is composed of tiny, indivisible particles called atoms, and all atoms of a given element are identical.
2. Compounds comprise specific combinations of atoms of two or more elements.
3. Chemical reactions cause the rearrangement of atoms but do not cause either the creation or the destruction of atoms.
```


This equation must be balanced so that the same number of atoms appear on both sides of the reaction arrow.  We write appropriate stoichiometric coefficients (or just *coefficients*) to the left of each chemical species until the same number of atoms on both sides of the equation is *balanced*.  In this particular case, let's tackle $\ce{O}$ first.  Let's add a coefficient of 2 before $\ce{H2O}$ to balance $\ce{O}$. There are 4 $\ce{H}$ atoms on the right.  If we add a coefficient of 2 before $\ce{H2}$, there are the same number of $\ce{H}$ atoms on both sides.  The balanced reaction can be shown as follows:

$$
\ce{2 H2 (g) + O2 (g) -> 2 H2O (l)}
$$
Balancing a chemical reaction requires a trial-and-error approach.  Sometimes, you have to try several iterations of the coefficients of a reactant or product until conservation of mass is achieved.  Below are a couple of approaches that are commonly applied.

```{admonition} Rules for balancing chemical reactions

1. Change the coefficients of compounds (e.g., $\ce{CO2}$) before changing the coefficients of elements (e.g., $\ce{O2}$).
2. Treat polyatomic ions that appear on both sides of the equation (e.g., $\ce{CO_3^2-}$, $\ce{OH-}$, $\ce{SO_4^2-}$) as units, rather than counting their constituent atoms individually.
3. Count atoms and/or polyatomic ions carefully and track their numbers each time you change a coefficient.
```


```{dropdown} Example: Balancing chemical reactions - Approach 1

Balance the reaction that represents the combustion of butane ($\ce{C4H10 (g)}$):

$$
\ce{C4H10(g) + O2(g) ->	CO2(g) + H2O(l)	}
$$

>Let's inventory all atoms on both sides of the arrow.
>
> | 4 	| C 	| 1 	|
> | --- | --- | --- |
> | 10 	| H 	| 2 	|
> | 2 	| O 	| 3 	|
>
>Now, let's fix $\ce{C}$ atoms first by changing the coefficient of $\ce{CO2 (g)}$.
>
>$$
\ce{C4H10(g) + O2(g) -> 4 CO2(g) + H2O(l)	}
$$
>
> | 4 	| C 	| $\cancel{1}$ 4 	|
> | --- | --- | --- |
> | 10 	| H 	| 2 	|
> | 2 	| O 	| $\cancel{3}$ 9 	|
>
>Now, let's fix $\ce{H}$ atoms by changing coefficient of $\ce{H2O (l)}$.
>
>$$
\ce{C4H10(g) + O2(g) -> 4 CO2(g) + 5 H2O(l)	}
$$	
>
> | 4 	| C 	| 4 	|
> | --- | --- | --- |
> | 10 	| H 	| $\cancel{2}$ 10 	|
> | 2 	| O 	| $\cancel{9}$  13	|
>
>Lastly, let's fix $\ce{O}$ atoms by changing coefficient of $\ce{O2 (g)}$.
>
>$$
\ce{C4H10(g) + 13/2 O2(g) -> 4 CO2(g) + 5 H2O(l)	}
$$	
>
> | 4 	| C 	| 4 	|
> | --- | --- | --- |
> | 10 	| H 	| 10 	|
> | $\cancel{2}$ 13 	| O 	| 13 	|
>
> Multiply the above equation by 2 to get the whole number of coefficients.
>
>$$
\ce{2 C4H10(g) + 13 O2(g) -> 8 CO2(g) + 10 H2O(l)	}
$$
>The equation is still balanced!
```

The example below shows another approach to balancing chemical equations. 
 
```{dropdown} Example: Balancing chemical reactions - Approach 2

Kaolinite ($\ce{Al2Si2O5(OH)4}$), a significant component of soils, is chemically weathered (by $\ce{H+}$) into $\ce{Al^3+}$, silicic acid ($\ce{H4SiO4}$), and $\ce{H2O}$. The basic chemical reaction that represents this chemical process is shown below:

$$
	\ce{Al2Si2O5(OH)4 + H+ -> Al^3+ + H4SiO4 + H2O}
$$


>1. Write all reactants and products in the correct locations of the chemical equation:
>
>$$
	\ce{Al2Si2O5(OH)4 + H+ -> Al^3+ + H4SiO4 + H2O}
$$
>2.  Balance the main elements ($\ce{Al}$ and $\ce{Si}$) first
>
>$$
	\ce{Al2Si2O5(OH)4 + H+ -> 2 Al^3+ + 2 H4SiO4 + H2O}
$$
>3. Balance $\ce{O}$ next and adjust $\ce{H2O}$ as needed
>
>$$
	\ce{Al2Si2O5(OH)4 + H+ -> 2 Al^3+ + 2 H4SiO4 + H2O}
$$
>4. Balance single elements and ions ($\ce{H+}$) last
>
>$$
	\ce{Al2Si2O5(OH)4 + 6 H+ -> 2 Al^3+ + 2 H4SiO4 + H2O}
$$
>5.  Check to see if there is mass balance for all elements.
```