Chapter 15, Problem 15.63SP

Interpretation Introduction

(a)

Interpretation:

The value of the equilibrium constant $K_p$ for the reaction $2{\text{AsH}}_{3(g)}+{\text{P}}_{2(g)}\rightleftharpoons 2{\text{PH}}_{3(g)}+{\text{As}}_{2(g)}$ at $873K$ is to be calculated.

Concept introduction:

The term equilibrium constant expresses the relationship between the reactants and the products of a reaction that is in the equilibrium state. This relationship using the partial pressures of reactants and products to calculate the equilibrium constant, it is denoted as $K_p$.

If the equilibrium constant for the reverse reaction of a given chemical reaction needs to be determined, the given equilibrium constant can be reversed to obtain the new equilibrium constant, there one takes $1/\text{equilibrium constant}$.

Interpretation Introduction

(b)

Interpretation:

The value of the equilibrium constant $K_p$ for the reaction $6{\text{PH}}_{3(g)}+3{\text{As}}_{2(g)}\rightleftharpoons 6{\text{AsH}}_{3(g)}+3{\text{P}}_{2(g)}$ at $873K$ is to be calculated.

Concept introduction:

The term equilibrium constant expresses the relationship between the reactants and the products of a reaction that is in the equilibrium state. This relationship using the partial pressures of reactants and products to calculate the equilibrium constant, it is denoted as $K_p$.

But if the reaction is multiplied by a factor, then the new equilibrium constant for the reaction can be obtained by taking the equilibrium constant raised to power of that factor. For example, if the reaction is multiplied by $"n"$ then the new equilibrium constant will be:

$K_p{}^{\text{'}}=K_p{}^n$

Interpretation Introduction

(c)

Interpretation:

The value of the equilibrium constant $K_p$ for the reaction $2{\text{P}}_{2(g)}+4{\text{AsH}}_{3(g)}\rightleftharpoons 2{\text{As}}_{2(g)}+4{\text{PH}}_{3(g)}$ at $873K$ is to be calculated.

Concept introduction:

The term equilibrium constant expresses the relationship between the reactants and the products of a reaction that is in the equilibrium state. This relationship using the partial pressures of reactants and products to calculate the equilibrium constant, it is denoted as $K_p$.

But if the reaction is multiplied by a factor, then the new equilibrium constant for the reaction can be obtained by taking the equilibrium constant raised to power of that factor. For example, if the reaction is multiplied by $"n"$ then the new equilibrium constant will be:

$K_p{}^{\text{'}}=K_p{}^n$