Question

By using Gaussian elimination method, balance the chemical reaction equation:

\[\ce{C2H + O2 -> H2O + CO2}\]

Answer 1

We are searching for positive integers x₁, x₂, x₃ and x₄ 

\[\ce{x1 C2H6 + x2 O2 -> x3HO + x4 CO2}\]  ........(1)

The number of carbon atoms on the LHS of (1) should be equal to the number of carbon atoms on the RHS of (1)

So we get a linear homogeneous equation.

2x₁ x₄ = 2x₁ – x₄ = 0   ........(2)

6x₁ = 2x₃ 

= 6x₁ – 2x₃ = 0

÷ 2 ⇒ 3x₁ – x₃ = 0  ........(3)

2x₂ = x₃ + 2x₄ 

⇒ 2x₂ – x₃ – 2x₄ = 0  .........(4)

Equation (2), (3) and (4) constitute a homogeneous system of linear equations in four unknowns.

Augmented matrix

[A|B] = `[(2, 0, 0, -1, |, 0),(3, 0, -1, 0, |, 0),(0, 2, -1, -2, |, 0)]`

By Gaussian elimination method, we get

`{:("R"_2 -> 2"R"_2 - 3"R"_1),(->):} [(2, 0, 0, -1, 0),(0, 0, -2, 3, 0),(0, 2, -1, -2, 0)]`

`{:("R"_2 ↔ "R"_3),(->):} [(2, 0, 0, -1, |, 0),(0, 2, -1, -2, |, 0),(0, 0, -2, 3, |, 0)]`

P(A) = P(A|B) = 3 < 4

The system is consistent and has an infinite number of solutions.

Writing the equations using the echelon form we get

– 2x₃ + 3x₄ = 0  ........(1)

2x₂ – x₃ – 2x₄ = 0  ........(2)

2x₁ – x₄ = 0  ........(3)

Put x₄ = t

(3) ⇒ 2x₁ – t = 0

x₁ = `"t"/2`

(1) ⇒ – 2x₃ + 3x₄ = 0

– 2x₃ = – 3t

x₃ = `3/2 "t"`

(2) ⇒ 2x₂ – x₃ – 2x₄ = 0

`2x_2 - 3/2 "t" - 2"t"` = 0

2x₂ = `3/2 "t" + 2"t"` 

x₂ = `(7"t")/2`

(x₁, x₂, x₃, x₄) = `("t"/2. (7"t")/4, 3/2 "t", "t")` ∀ t ∈ R

Since x₁, x₂, x₃ and x₄ are positive integers.

Let us choose t = 4

x₁ =`4/2` = 2

x₂ = `(7"t")/4 = (7(4))/4` = 7

x₃ = `3/2 (4)` = 6

x₄ = t = 4

x₁ = 2, x₂ = 7, x₃ = 6, x₄ = 4

So the balanced equation is

\[\ce{2C2H6 + 7O2 -> 6H2O + 4CO2}\]