# Πώς να εξισορροπήσετε τις χημικές εξισώσεις χρησιμοποιώντας γραμμική άλγεβρα: 8 βήματα

## Πίνακας περιεχομένων: Πώς να εξισορροπήσετε τις χημικές εξισώσεις χρησιμοποιώντας γραμμική άλγεβρα: 8 βήματα

## Βίντεο: Πώς να εξισορροπήσετε τις χημικές εξισώσεις χρησιμοποιώντας γραμμική άλγεβρα: 8 βήματα Βίντεο: Το καλύτερο Challenge να κάνεις με τους φίλους σου! 🤣 2023, Σεπτέμβριος

Η εξισορρόπηση των χημικών εξισώσεων γίνεται συνήθως με τον πρώτο εντοπισμό ασυνήθιστων στοιχείων στις ενώσεις και το δρόμο προς το υδρογόνο και το οξυγόνο. Υπάρχει επίσης μια πιο αργή αλλά συστηματική προσέγγιση χρησιμοποιώντας γραμμική άλγεβρα.

## Βήματα

### Βήμα 1. Προσδιορίστε την εξίσωση προς ισορροπία

• H3PO4+(NH4) 2MoO4+HNO3 → (NH4) 3PO4⋅12MoO3+NH4NO3+H2O { displaystyle { begin {straight} & / mathrm {H} _ {3} mathrm {PO} _ {4}+( mathrm {NH} _ {4}) _ {2} mathrm {MoO} _ {4}+\ mathrm {HNO} _ {3} & / to ( mathrm {NH} _ {4}) _ {3 } mathrm {PO} _ {4} cdot 12 \, / mathrm {MoO} _ {3}+\ mathrm {NH} _ {4} mathrm {NO} _ {3}+\ mathrm {H} _ {2} mathrm {O} end {στοιχισμένο}}}

### step 2. identify the elements

the number of elements present in the equation determines how many rows will be in the vectors and matrices that we are going to construct. below, the order we list corresponds to the order of the rows.

• h{displaystyle \mathrm {h} }

– hydrogen

• p{displaystyle \mathrm {p} }

– phosphorus

• o{displaystyle \mathrm {o} }

– oxygen

• n{displaystyle \mathrm {n} }

– nitrogen

• mo{displaystyle \mathrm {mo} }

– molybdenum

### step 3. set up the vector equation

the vector equation consists of column vectors corresponding to each compound in the equation. each vector has a corresponding coefficient, labeled x1{displaystyle x_{1}}

to x6, {displaystyle x_{6}, }

for which we are solving for. make sure you understand how to count the number of atoms in a molecule.

• x1(31400)+x2(80421)+x3(10310)=x4(12140312)+x5(40320)+x6(20100){displaystyle x_{1}{begin{pmatrix}3\\1\\4\\0\\0\end{pmatrix}}+x_{2}{begin{pmatrix}8\\0\\4\\2\\1\end{pmatrix}}+x_{3}{begin{pmatrix}1\\0\\3\\1\\0\end{pmatrix}}=x_{4}{begin{pmatrix}12\\1\\40\\3\\12\end{pmatrix}}+x_{5}{begin{pmatrix}4\\0\\3\\2\\0\end{pmatrix}}+x_{6}{begin{pmatrix}2\\0\\1\\0\\0\end{pmatrix}}}

### step 4. set the equation to 0 and obtain the augmented matrix

there are two major points to consider here. first, recognize that a vector equation like the one above has the same solution set as a linear system with its corresponding augmented matrix. this is a fundamental idea in linear algebra. second, when the augments are all 0, row-reduction does not change the augments. therefore, we need not write them at all – row-reducing the coefficient matrix is all that is necessary.

• note that moving everything to the left side causes the elements on the right side to negate.
• (381−12−4−2100−100443−40−3−1021−3−20010−1200){displaystyle {begin{pmatrix}3&8&1&-12&-4&-2\\1&0&0&-1&0&0\\4&4&3&-40&-3&-1\\0&2&1&-3&-2&0\\0&1&0&-12&0&0\end{pmatrix}}}

step 5. row-reduce to reduced row-echelon form.

for such a matrix, it is recommended that you use a calculator, although row-reducing by hand is always an option, albeit slower.

• (10000−1/1201000−100100−7/400010−1/1200001−7/4){displaystyle {begin{pmatrix}1&0&0&0&0&-1/12\\0&1&0&0&0&-1\\0&0&1&0&0&-7/4\\0&0&0&1&0&-1/12\\0&0&0&0&1&-7/4\end{pmatrix}}}
• it is clear that there is a free variable x6{displaystyle x_{6}}

here. those with sharp minds would've seen this coming, for there are more variables than equations, and hence more columns than rows. this free variable means that x6{displaystyle x_{6}}

can take on any value, and the resulting combination of x1{displaystyle x_{1}}

to x5{displaystyle x_{5}}

would be a valid solution (to our linear system, that is – the chemical equation results in further restrictions in this solution set).

### step 6. reparameterize the free variable and solve for the variables

let's set x6=t.{displaystyle x_{6}=t.}

since for positive values of t, {displaystyle t, }

none of the variables become negative, so we are on the right track.

• x1=t/12{displaystyle x_{1}=t/12}
• x2=t{displaystyle x_{2}=t}
• x3=7t/4{displaystyle x_{3}=7t/4}
• x4=t/12{displaystyle x_{4}=t/12}
• x5=7t/4{displaystyle x_{5}=7t/4}

• x6=t{displaystyle x_{6}=t}

step 7. substitute an appropriate value for t{displaystyle t}

remember that the coefficients in the chemical equation must be integers. therefore, set t=12, {displaystyle t=12, }

the least common multiple. from our solution set, it is clear that while there are an infinite number of solutions, as we would expect, it is nonetheless a countably infinite set.

• x1=1{displaystyle x_{1}=1}
• x2=12{displaystyle x_{2}=12}

• x3=21{displaystyle x_{3}=21}
• x4=1{displaystyle x_{4}=1}
• x5=21{displaystyle x_{5}=21}
• x6=12{displaystyle x_{6}=12}

### step 8. substitute the coefficients into the chemical equation

the equation is now balanced.

• h3po4+12(nh4)2moo4+21hno3→(nh4)3po4⋅12moo3+21nh4no3+12h2o{displaystyle {begin{aligned}&\mathrm {h} _{3}\mathrm {po} _{4}+12\, (mathrm {nh} _{4})_{2}\mathrm {moo} _{4}+21\, \mathrm {hno} _{3}\\&\to (mathrm {nh} _{4})_{3}\mathrm {po} _{4}\cdot 12\, \mathrm {moo} _{3}+21\, \mathrm {nh} _{4}\mathrm {no} _{3}+12\, \mathrm {h} _{2}\mathrm {o} end{aligned}}}