Chapter 17.3, Problem 18PP

(a)

Interpretation Introduction

Interpretation:

The concentration of the $\left[\text{CO}\right]$ is to be calculated.

Concept introduction:

The concentration is the amount of the reactant $\left[\text{CO}\right]$ which is used in the reaction. The equilibrium constant is the ratio of concentration of the products to the concentration of reactants.

${\text{K}}_{\text{eq}}\text{=}\frac{\left[{\text{CH}}_{\text{3}}\text{OH}\right]}{\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}}$

$\left[{\text{CH}}_{\text{3}}\text{OH }\right]$ = concentration of the product

$\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{2 }}$ = concentration of the reactants

Answer to Problem 18PP

The concentration of $\left[\text{CO}\right]$ is $\text{0}\text{.144}$ mol/L.

Explanation of Solution

Given information is: $\text{Keq = 10}\text{.5}$

Concentration of $\left[{\text{CH}}_{\text{3}}\text{OH}\right]\text{ = 1}\text{.32}$

Concentration of $\left[{\text{H}}_{\text{2}}\right]\text{ = 0}\text{.933}$

${\text{K}}_{\text{eq}}\text{=}\frac{\left[{\text{CH}}_{\text{3}}\text{OH}\right]}{\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}}$

$\left[\text{CO}\right]\text{ =}\frac{\left[{\text{CH}}_{\text{3}}\text{OH}\right]\begin{array}{l}\hfill \end{array}}{\text{Keq × }{\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}}$

$\text{ = }\frac{\text{ 1}\text{.32 }}{\text{ 10}\text{.5×}{\left(\text{0}\text{.933}\right)}^{\text{2}}}$

$\text{ = 0}\text{.144 mol/L}$

(b)

Interpretation Introduction

Interpretation:

The concentration of the $\left[{\text{H}}_{\text{2}}\right]$ is to be calculated.

Concept introduction:

The concentration is the amount of the reactant $\left[{\text{H}}_{\text{2}}\right]$ which is used in the reaction. The equilibrium constant is the ratio of concentration of the products to the concentration of reactants.

${\text{K}}_{\text{eq}}\text{=}\frac{\left[{\text{CH}}_{\text{3}}\text{OH}\right]}{\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}}$

${\text{[CH}}_{\text{3}}\text{OH ] =}$ Concentration of the product

$\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{2 }}\text{=}$ Concentration of the reactants

Answer to Problem 18PP

The concentration of $\left[{\text{H}}_{\text{2}}\right]$ is $\text{0}\text{.1685}$ mol/L.

Explanation of Solution

Given in information is $\text{Keq = 10}\text{.5}$

Concentration of $\left[{\text{CH}}_{\text{3}}\text{OH}\right]\text{ = 0}\text{.325}$

Concentration of $\left[\text{CO}\right]\text{ = 1}\text{.09}$

${\text{K}}_{\text{eq}}\text{=}\frac{\left[{\text{CH}}_{\text{3}}\text{OH}\right]}{\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}}$

${\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}\text{=}\frac{{\left[{\text{CH}}_{\text{3}}\text{OH}\right]}^{}}{\text{Keq × }\left[\text{CO}\right]}$

$\text{ = }\frac{\text{0}\text{.325}}{\text{10}\text{.5×}\left(\text{1}\text{.09}\right)}$

$\text{ = 0}\text{.1685}$ mol/L

(c)

Interpretation Introduction

Interpretation:

The concentration of the $\left[{\text{CH}}_{\text{3}}\text{OH}\right]$ is to be calculated.

Concept introduction:

The concentration is the amount of the Product $\left[{\text{CH}}_{\text{3}}\text{OH}\right]$ which is produced in the reaction. The equilibrium constant is the ratio of concentration of the products to the concentration of reactants.

${\text{K}}_{\text{eq}}\text{=}\frac{\left[{\text{CH}}_{\text{3}}\text{OH}\right]}{\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}}$

$\left[{\text{CH}}_{\text{3}}\text{OH}\right]$ = concentration of the product

$\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{2 }}$ = concentration of the reactants

Answer to Problem 18PP

The concentration of $\left[{\text{CH}}_{\text{3}}\text{OH}\right]$is $\text{0}\text{.176}$ mol/L

Explanation of Solution

Given information is $\text{Keq = 10}\text{.5}$

Concentration of $\left[{\text{CH}}_{\text{3}}\text{OH}\right]\text{ = ?}$

Concentration of $\left[\text{CO}\right]\text{ = 3}\text{.85}$

Concentration of $\left[{\text{H}}_{\text{2}}\right]\text{ = 0}\text{.0661}$

${\text{K}}_{\text{eq}}\text{=}\frac{\left[{\text{CH}}_{\text{3}}\text{OH}\right]}{\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}}$

${\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}{\text{K}}_{\text{eq}}\text{× }{\left[\text{CO}\right]}^{}\text{=}{\left[{\text{CH}}_{\text{3}}\text{OH}\right]}^{}$

${}^{}\left[{\text{CH}}_{\text{3}}\text{OH}\right]\text{ = 10}\text{.5× }{\left(\text{0}\text{.0661}\right)}^{\text{2}}\text{×3}\text{.85}$

$\text{=0}\text{.1766}$ mol/L