Chapter 17, Problem 17.86P

Interpretation Introduction

Interpretation:

The number of grams of ${\text{H}}_{\text{2}}$ present at equilibrium has to be calculated for the given reaction.

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

Given information is a chemist introduces $\text{2}{\text{.0 atm of H}}_{\text{2}}$ and $\text{2}{\text{.0 atm of CO}}_{\text{2}}$ into a $\text{1}\text{.00-L}$ container at $\text{25}{\text{.0}}^{\text{°}}\text{C}$ and then raises the temperature to ${\text{700}}^{\text{°}}\text{C}$ at which ${\text{K}}_{\text{c}}\text{= 0}\text{.534}$.

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by $\text{K}$.

For a reaction,

  $\text{xX + yY }\rightleftharpoons \text{ zZ}$

The expression of $\text{K}$ can be given as

  $\begin{array}{l}{\text{K}}_{\text{c}}\text{ = }\frac{{\text{[Z]}}^{\text{z}}}{{\text{[X]}}^{\text{x}}{\text{[Y]}}^{\text{y}}}\\ \\ \text{where,}\text{[X] = equilibrium concentration of X}\\ \\ \text{[Y] = equilibrium concentration of Y}\\ \\ \text{[Z] = equilibrium concentration of Z}\end{array}$

Moles can be calculated using ideal gas equation.

  $\begin{array}{l}\text{PV =nRT}\\ \\ \text{P =}\left(\frac{\text{n}}{\text{V}}\right)\text{RT}\frac{\text{n}}{\text{V}}\text{=M}\\ \\ \text{P =MRT}\\ \\ \text{M =}\frac{\text{P}}{\text{RT}}\\ \\ \text{where,}\text{P = Pressure}\\ \\ \text{R = universal gas constant}\\ \\ \text{T = temperature}\\ \\ \text{M = number of moles}\end{array}$