Chapter 14, Problem 14.143CP

Interpretation Introduction

Interpretation:

In the following reaction of formation of fumarate.

$L-aspartate\rightleftharpoons fumarate\text{ + }ammoniumion.$

An initial concentration of L-aspartate is $8.32\times {10}^{-3}M$ at 37 ⁰C. Calculate the equilibrium concentration of L-aspartate, fumarate and ammonium ion.

Concept introduction:

Equilibrium is the process in which temperature, pressure, concentration of reactant and product do not show any change with respect to time.
There are two types of equilibrium.

1. Physical equilibrium: In this physical state of reactant and product does not change when dynamic equilibrium occurred.
2. Chemical equilibrium: In this chemical composition of reactant and product does not change when dynamic equilibrium occurred. There are two types of reaction occurred in this.

(a) Reversible reactions: In this reaction two ways reaction can occurs that is reactant can convert to product as well as product can convert back to reactant.
(b) Irreversible reaction: In this reaction two ways reaction cannot occurs that is reactant can convert to product but product cannot convert back to reactant.

Equilibrium constant $K_c$ is defined as the ratio of the concentrations of the products raised to the power of their stoichiometric coefficients and reactants raised to the power of their stoichiometric coefficients.
Reaction quotient Q is defined as the ratio at any point of the reaction of the concentration of the product raised to the power of their stoichiometric coefficients and reactant raised to the power of their stoichiometric coefficients.

$\begin{array}{l}aA+bB\to cC+dD\\ K_c=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b},equilibrium\\ Q=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}\end{array}$

To calculate:

The equilibrium concentration of L-aspartate, fumarate and ammonium ion.