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Q9.8(ii):
Write chemical equations for combustion reaction of the following hydrocarbons: (ii) Pentene

Solution :

Step 1: Determining the Molecular Formula of Pentene

Pentene is an alkene with the general formula $C_nH_{2n}$. For pentene, $n = 5$.

Calculation: $C_5H_{2(5)} = C_5H_{10}$.

Thus, the molecular formula for pentene is $C_5H_{10}$. [Per the General Formula for Alkenes]

Step 2: Defining the General Combustion Reaction

The complete combustion of a hydrocarbon involves reacting it with excess oxygen ($O_2$) to produce carbon dioxide ($CO_2$) and water ($H_2O$) as the only products. The skeletal chemical equation is:

$C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + (\frac{y}{2})H_2O$

Step 3: Balancing the Equation for Pentene ($C_5H_{10}$)

Substituting $x = 5$ and $y = 10$ into the stoichiometric coefficients:

  • Carbon atoms ($C$): $5$ atoms on the left require $5$ molecules of $CO_2$ on the right.
  • Hydrogen atoms ($H$): $10$ atoms on the left require $5$ molecules of $H_2O$ on the right ($10/2 = 5$).
  • Oxygen atoms ($O$):
    • Products side: $(5 \times 2) + (5 \times 1) = 10 + 5 = 15$ oxygen atoms.
    • Reactants side: To balance $15$ atoms, we require $\frac{15}{2} = 7.5$ molecules of $O_2$.

Step 4: Finalizing the Balanced Equation

The balanced chemical equation is:

$C_5H_{10}(g) + \frac{15}{2}O_2(g) \rightarrow 5CO_2(g) + 5H_2O(g)$

To eliminate the fraction, multiply the entire equation by $2$:

$2C_5H_{10}(g) + 15O_2(g) \rightarrow 10CO_2(g) + 10H_2O(g)$

Visual Representation of Combustion Stoichiometry

2 C₅H₁₀ + 15 O₂ → (Combustion) 10 CO₂ + 10 H₂O

Final Solution: The balanced combustion reaction for pentene is $2C_5H_{10}(g) + 15O_2(g) \rightarrow 10CO_2(g) + 10H_2O(g)$.


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