Chemical equation

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A chemical equation is the symbolic representation of a

As an example, the equation for the reaction of hydrochloric acid with sodium can be denoted:

2HCl + 2Na → 2NaCl + H₂

Given the formulas are fairly simple, this equation could be read as "two H-C-L plus two N-A yields

Different variants of the arrow symbol are used to denote the type of a reaction:[1]

${\displaystyle \rightarrow }$	net forward reaction
${\displaystyle \rightleftarrows }$	reaction in both directions[c]
${\displaystyle \rightleftharpoons }$	equilibrium[d]
${\displaystyle =}$	stoichiometric relation
${\displaystyle \leftrightarrow }$	resonance (not a reaction)

To indicate

The standard notation for chemical equations only permits all reactants on one side, all products on the other, and all stoichiometric coefficients positive. For example, the usual form of the equation for

2 CH₃OH → CH₃OCH₃ + H₂O

Sometimes an extension is used, where some substances with their stoichiometric coefficients are moved above or below the arrow, preceded by a plus sign or nothing for a reactant, and by a minus sign for a product. Then the same equation can look like this:

${\displaystyle {\ce {2CH3OH->[{\overset {}{\ce {-H2O}}}]CH3OCH3}}}$

Such notation serves to hide less important substances from the sides of the equation, to make the type of reaction at hand more obvious, and to facilitate chaining of chemical equations. This is very useful in illustrating multi-step

Another extension used in reaction mechanisms moves some substances to branches of the arrow. Both extensions are used in the example illustration of a mechanism.

Use of negative stoichiometric coefficients at either side of the equation (like in the example below) is not widely adopted and is often discouraged.^[5][6]

${\displaystyle {\ce {2 CH3OH \;-\; H2O -> CH3OCH3}}}$

Because no nuclear reactions take place in a chemical reaction, the chemical elements pass through the reaction unchanged. Thus, each side of the chemical equation must represent the same number of atoms of any particular element (or nuclide, if different isotopes are taken into account). The same holds for the total electric charge, as stated by the charge conservation law. An equation adhering to these requirements is said to be balanced.

A chemical equation is balanced by assigning suitable values to the stoichiometric coefficients. Simple equations can be balanced by inspection, that is, by trial and error. Another technique involves solving a system of linear equations.

Balanced equations are usually written with smallest natural-number coefficients. Yet sometimes it may be advantageous to accept a fractional coefficient, if it simplifies the other coefficients. The introductory example can thus be rewritten as

${\displaystyle {\ce {HCl + Na -> NaCl + 1/2 H2}}}$

In some circumstances the fractional coefficients are even inevitable. For example, the reaction corresponding to the standard enthalpy of formation must be written such that one molecule of a single product is formed. This will often require that some reactant coefficients be fractional, as is the case with the formation of lithium fluoride:

${\displaystyle {\ce {Li(s) + 1/2F2(g) -> LiF(s)}}}$

The method of inspection can be outlined as setting the most complex substance's stoichiometric coefficient to 1 and assigning values to other coefficients step by step such that both sides of the equation end up with the same number of atoms for each element. If any

Balancing of the chemical equation for the complete combustion of methane

${\displaystyle {\ce {{\mathord {?}}\,{CH4}+{\mathord {?}}\,{O2}->{\mathord {?}}\,{CO2}+{\mathord {?}}\,{H2O}}}}$

is achieved as follows:

1. A coefficient of 1 is placed in front of the most complex formula (CH₄):

   ${\displaystyle {\ce {1{CH4}+{\mathord {?}}\,{O2}->{\mathord {?}}\,{CO2}+{\mathord {?}}\,{H2O}}}}$

2. The left-hand side has 1 carbon atom, so 1 molecule of CO₂ will balance it. The left-hand side also has 4 hydrogen atoms, which will be balanced by 2 molecules of H₂O:

   ${\displaystyle {\ce {1{CH4}+{\mathord {?}}\,{O2}->1{CO2}+2H2O}}}$

3. Balancing the 4
   O₂
   yields the equation

   ${\displaystyle {\ce {1 CH4 + 2 O2 -> 1 CO2 + 2 H2O}}}$

4. The coefficients equal to 1 are omitted, as they do not need to be specified explicitly:

   ${\displaystyle {\ce {CH4 + 2 O2 -> CO2 + 2 H2O}}}$

5. It is wise to check that the final equation is balanced, i.e. that for each element there is the same number of atoms on the left- and right-hand side: 1 carbon, 4 hydrogen, and 4 oxygen.

For each chemical element (or nuclide or unchanged

${\displaystyle \sum _{j\,\in \,{\text{reactants}}}\!\!\!\!\!a_{ij}s_{j}\ =\!\!\!\!\!\sum _{j\,\in \,{\text{products}}}\!\!\!\!\!a_{ij}s_{j}}$

where

a_ij is the number of atoms of element i in a molecule of substance j (per formula in the chemical equation), and

s_j is the stoichiometric coefficient for the substance j.

This results in a

All solutions to this system of linear equations are of the following form, where r is any real number:

${\displaystyle {\begin{aligned}s_{1}&=r\\s_{2}&=2r\\s_{3}&=r\\s_{4}&=2r\end{aligned}}}$

The choice of r = 1 yields the preferred solution,

${\displaystyle {\begin{aligned}s_{1}&=1\\s_{2}&=2\\s_{3}&=1\\s_{4}&=2\end{aligned}}}$

which corresponds to the balanced chemical equation:

${\displaystyle {\ce {CH4 + 2 O2 -> CO2 + 2 H2O}}}$

The system of linear equations introduced in the previous section can also be written using an efficient matrix formalism. First, to unify the reactant and product stoichiometric coefficients s_j, let us introduce the quantity

${\displaystyle \nu _{j}={\begin{cases}-s_{j}&{\text{for a reactant}}\\+s_{j}&{\text{for a product}}\end{cases}}}$

called

Techniques have been developed[7]^[8] to quickly calculate a set of J_N independent solutions to the balancing problem, which are superior to the inspection and algebraic method^{[citation needed]} in that they are determinative and yield all solutions to the balancing problem.

Example

Using the same chemical equation again, write the corresponding matrix equation:

${\displaystyle {\ce {{\mathit {s}}_{1}{CH4}+{\mathit {s}}_{2}{O2}->{\mathit {s}}_{3}{CO2}+{\mathit {s}}_{4}{H2O}}}}$

${\displaystyle {\begin{matrix}{\text{C:}}\\{\text{H:}}\\{\text{O:}}\end{matrix}}\quad {\begin{bmatrix}1&0&1&0\\4&0&0&2\\0&2&2&1\end{bmatrix}}{\begin{bmatrix}\nu _{1}\\\nu _{2}\\\nu _{3}\\\nu _{4}\end{bmatrix}}=\mathbf {0} }$

Its solutions are of the following form, where r is any real number:

${\displaystyle {\begin{bmatrix}\nu _{1}\\\nu _{2}\\\nu _{3}\\\nu _{4}\end{bmatrix}}={\begin{bmatrix}-s_{1}\\-s_{2}\\s_{3}\\s_{4}\end{bmatrix}}=r{\begin{bmatrix}-1\\-2\\1\\2\end{bmatrix}}}$

The choice of r = 1 and a sign-flip of the first two rows yields the preferred solution to the balancing problem:

${\displaystyle {\begin{bmatrix}-\nu _{1}\\-\nu _{2}\\\nu _{3}\\\nu _{4}\end{bmatrix}}={\begin{bmatrix}s_{1}\\s_{2}\\s_{3}\\s_{4}\end{bmatrix}}={\begin{bmatrix}1\\2\\1\\2\end{bmatrix}}}$

An ionic equation is a chemical equation in which

In this reaction, the Ca²⁺ and the NO₃⁻ ions remain in solution and are not part of the reaction. That is, these ions are identical on both the reactant and product side of the chemical equation. Because such ions do not participate in the reaction, they are called spectator ions. A net ionic equation is the full ionic equation from which the spectator ions have been removed.^[9] The net ionic equation of the proceeding reactions is:

${\displaystyle {\ce {2Cl^- + 2Ag+ -> 2AgCl(v)}}}$

or, in reduced balanced form,

${\displaystyle {\ce {Ag+ + Cl^- -> AgCl(v)}}}$

In a neutralization or acid/base reaction, the net ionic equation will usually be:

${\displaystyle {\ce {H+ (aq) + OH^{-}(aq) -> H2O(l)}}}$

There are a few acid/base reactions that produce a precipitate in addition to the water molecule shown above. An example is the reaction of barium hydroxide with phosphoric acid, which produces not only water but also the insoluble salt barium phosphate. In this reaction, there are no spectator ions, so the net ionic equation is the same as the full ionic equation.

${\displaystyle {\ce {3Ba(OH)2 + 2H3PO4 -> 6H2O + Ba3(PO4)2(v)}}}$

${\displaystyle {\ce {{3Ba^{2}+}+{6OH^{-}}+{6H+}}}+\underbrace {\ce {2PO4^{3}-}} _{\ce {phosphate}}{\ce {->{6H2O}+\underbrace {Ba3(PO4)2(v)} _{barium~phosphate}}}}$

Double displacement reactions that feature a carbonate reacting with an acid have the net ionic equation:

${\displaystyle {\ce {2H+}}+\underbrace {{\ce {CO3^2-}}} _{{\ce {carbonate}}}{\ce {-> H2O + CO2 (^)}}}$

If every ion is a "spectator ion" then there was no reaction, and the net ionic equation is null.

Generally, if z_j is the multiple of elementary charge on the j-th molecule, charge neutrality may be written as:

${\displaystyle \sum _{j=1}^{J}z_{j}\nu _{j}=0}$

where the ν_j are the stoichiometric coefficients described above. The z_j may be incorporated^[7][8] as an additional row in the a_ij matrix described above, and a properly balanced ionic equation will then also obey:

${\displaystyle \sum _{j=1}^{J}a_{ij}\nu _{j}=0}$

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