Chapter 14, Problem 14.1P

Interpretation Introduction

Interpretation:

The equilibrium constant for the combustion of Propane in the given balanced chemical equation is to be stated.

Concept introduction: The relation between the amount of reactant and the product that is present at equilibrium in a reversible reaction at a specified temperature is known as equilibrium constant.

To determine: The equilibrium constant for the combustion of Propane in the given balanced chemical equation.

Answer to Problem 14.1P

Solution:

The value of equilibrium constant for the given reaction is given as,

  $K=\frac{{\left[{\text{CO}}_2\right]}^3{\left[{\text{H}}_2\text{O}\right]}^4}{\left[{\text{C}}_{\text{3}}{\text{H}}_8\right]{\left[{\text{O}}_2\right]}^5}$

Explanation of Solution

Given information:

The chemical equation for the combustion of Propane is given as,

  ${\text{C}}_3{\text{H}}_8\left(g\right)+5{\text{O}}_2\left(g\right)\to {\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(g\right)$

The value of equilibrium constant is given by the ratio of concentration of products to the concentration of reactants each raised to the power of their respective stoichiometric coefficient.
In the given reaction the products are ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ in their gaseous form and the reactants are ${\text{C}}_{\text{3}}{\text{H}}_8$ and ${\text{O}}_2$ in their gaseous form. Therefore, the value of equilibrium constant is given as,

  $K=\frac{{\left[{\text{CO}}_2\right]}^3{\left[{\text{H}}_2\text{O}\right]}^4}{\left[{\text{C}}_{\text{3}}{\text{H}}_8\right]{\left[{\text{O}}_2\right]}^5}$

Conclusion

The value of equilibrium constant for the given reaction is given as,

  $K=\frac{{\left[{\text{CO}}_2\right]}^3{\left[{\text{H}}_2\text{O}\right]}^4}{\left[{\text{C}}_{\text{3}}{\text{H}}_8\right]{\left[{\text{O}}_2\right]}^5}$