Chapter 7, Problem 7.90QP

Interpretation Introduction

Interpretation:

Equilibrium concentration of hydrogen sulfide has to be calculated for the given reaction if concentration of hydrogen is $1.0\times {10}^{-1}M$, concentration of ${\text{S}}_{\text{2}}$ is $1.0\times {10}^{-5}M$ and equilibrium constant is $1.09\times {10}^7$.

    ${\text{2H}}_{\text{2}}(g)+{\text{S}}_{\text{2}}(g)\rightleftharpoons 2{\text{H}}_{\text{2}}\text{S}(g)$

Concept Introduction:

Equilibrium constant is the one that relates the concentration of products and the reactants that is present in an equilibrium reaction.  In other words it can also be said that equilibrium constant is equal to the rate constant of forward reaction that is divided by the rate constant of reverse reaction.  Equilibrium constant is represented as $K_{eq}$.  General representation of equilibrium constant for a chemical reaction at equilibrium is,

    $\begin{array}{l}aA+bB\rightleftharpoons cC+dD\\ K_{eq}=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}\end{array}$

From the above equation, the equilibrium concentration of one of the substance can also be calculated if equilibrium constant is known.