Chapter 12, Problem 12.116PAE

(a)

Interpretation Introduction

Interpretation:

The balanced chemical equation for the reaction of sodium sulfite and hydrochloric acid should be written.

Concept introduction:

A balanced chemical equation is defined as reaction in which there are same number of constituent atoms on both side of the reaction arrow, that is reactant and product side.

Answer to Problem 12.116PAE

Solution: Balanced chemical equation.

${\text{NaHSO}}_{\text{3}}\left(s\right)+\text{HCl}\left(g\right)\stackrel{}{\rightleftharpoons }\text{NaCl}\left(s\right)+{\text{SO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

Here, sulfur dioxide, sodium chloride and water are formed when sodium hydrogen sulfite is reacted with hydrochloric acid. The balanced reaction is as follows:${\text{NaHSO}}_{\text{3}}\left(s\right)+\text{HCl}\left(g\right)\stackrel{}{\rightleftharpoons }\text{NaCl}\left(s\right)+{\text{SO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(b)

Interpretation Introduction

Interpretation: Mass of ${\mathrm{SO}}_2$ produced if $1.9\text{g}$ of sodium hydrogen sulfite is reacted with excess $\text{HCl}$ should be identified.

Concept introduction:

For a reaction, if any reactant is present in excess, the mass of product obtained depends on the limiting reactant. The ratio of number of moles remains the same for any reaction.

Number of mole is related to mass and molar mass as follows:$\text{n}=\frac{m}{M}$

where, n is the number of moles

m is mass

M is molar mass

Answer to Problem 12.116PAE

Solution: Amount of sulfur dioxide produced is $1.2\text{g}$

Explanation of Solution

The balanced chemical reaction of sodium hydrogen sulfite and hydrochloride acid is as follows:${\text{NaHSO}}_{\text{3}}\left(s\right)+\text{HCl}\left(g\right)\stackrel{}{\rightleftharpoons }\text{NaCl}\left(s\right)+{\text{SO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Here, HCl is present in excess, thus, ${\text{NaHSO}}_{\text{3}}$ is a limiting reactant. 1 mole of ${\text{NaHSO}}_{\text{3}}$ produce 1 mole of sulfur dioxide ${\text{SO}}_{\text{2}}$.

Calculate the number of moles of ${\text{NaHSO}}_{\text{3}}$ obtained from 1.9 g of ${\text{NaHSO}}_{\text{3}}$ as follows:$\text{n}=\frac{m}{M}$

Molecular mass of ${\text{NaHSO}}_{\text{3}}=104\text{g/mol}$

Thus,$\begin{array}{c}{\text{n}}_{{\text{NaHSO}}_{\text{3}}}=\frac{1.9\text{ g}}{104\text{ g/mol}}\\ =0.0183\text{mol}\end{array}$

Since, $1$ mole of ${\text{NaHSO}}_{\text{3}}$ produces $1$ mole of ${\text{SO}}_{\text{2}}$ thus, $0.0183$ mole of ${\text{NaHSO}}_3$ produces $\text{ 0}\text{.0183 mole}$ of ${\text{SO}}_2$

Now, calculate mass of ${\text{SO}}_2$ produced as follows:$\text{m}=\text{n}\times \text{M}$

Molar mass of ${\text{SO}}_2=64\text{ g/mol}$

Thus,$\begin{array}{c}{\text{m}}_{{\text{SO}}_{\text{2}}}\text{=}0.0183\text{ mol}\times 64\text{g/mol}\\ =1.2\text{g}\end{array}$

Therefore, 1.2 g of SO₂ is produced from 1.9 g of ${\text{NaHSO}}_{\text{3}}$.

(c)

Interpretation Introduction

Interpretation: Equilibrium concentration of each species in the given reaction should be identified.${\text{SO}}_{\text{2}}+\frac{1}{2}{\text{O}}_{\text{2}}\stackrel{}{\rightleftharpoons }{\text{SO}}_{\text{3}}\text{K}=2.6\times {10}^{12}$

Concept introduction: For a general reaction, $\text{aA}+\text{bB}\stackrel{}{\rightleftharpoons }\text{cC}+\text{dD}$

Expression for the equilibrium constant (K) is as follows:$\text{K}=\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

Also, concentration or molarity of a solution is defined as number of moles in 1 L of solution. The unit of concentration or molarity is mol/L or M.

Answer to Problem 12.116PAE

Solution: Equilibrium concentration of ${\text{SO}}_{\text{2}}\left[{\text{SO}}_{\text{2}}\right]=\text{0}\text{M}$

Equilibrium concentration of ${\text{SO}}_{\text{3}}\left[{\text{SO}}_{\text{3}}\right]=0.183\text{M}$

Equilibrium concentration of ${\text{O}}_{\text{2}}\left[{\text{O}}_{\text{2}}\right]=0.41\text{M}$

Explanation of Solution

Given reaction is ${\text{SO}}_{\text{2}}\text{+}\frac{1}{2}{\text{O}}_2\stackrel{}{\rightleftharpoons }{\text{SO}}_{\text{3}}$

Equilibrium constant for the reaction will be:$\text{k=}\frac{\left[{\text{SO}}_{\text{3}}\right]}{\left[{\text{SO}}_{\text{2}}\right]{\left[{\text{O}}_{\text{2}}\right]}^{\frac{\text{1}}{\text{2}}}}$

The ICE table for the reaction will be:

	${\text{SO}}_{\text{2}}$	$\frac{1}{2}{\text{O}}_2$	${\text{SO}}_{\text{3}}$
Initial	$0.0183$	$0.05$	$0$
Change	$-x$	$-\frac{x}{2}$	$x$
equilibrium	$0.0183-x$	$0.05-\frac{x}{2}$	$x$

Since, the value of equilibrium constant is very high thus, SO₂ completely react with oxygen to form SO₃. The number of moles of SO₂ remain will be zero.

Thus,$\begin{array}{c}0.0183-x=0\\ x=0.0183\end{array}$

Calculate number of moles of O₂ remain as follows:

Moles of O₂ remain,$\begin{array}{c}{\text{n}}_{{\text{O}}_{\text{2}}}=\left(0.05-\frac{0.0183}{2}\right)\\ =0.041\text{mole}\end{array}$

And, ${\text{n}}_{{\text{SO}}_{\text{3}}}=0.0183\text{mole}$

Concentration can be calculated as follows:$\text{C}=\frac{\text{n}}{\text{V}}$

Since, volume of container is 100 mL or 0.1 L, equilibrium concentration of all species can be calculated.$\begin{array}{c}\left[{\text{SO}}_{\text{2}}\right]\text{=}\frac{{\text{n}}_{{\text{SO}}_{\text{2}}}}{\text{V}}\\ \text{=}\frac{\text{0}}{\text{0}\text{.1}\text{L}}\\ \text{=0}\\ \left[{\text{O}}_{\text{2}}\right]\text{=}\frac{\text{0}\text{.041}}{\text{0}\text{.1}}\\ \text{=0}\text{.41}\text{M}\\ \left[{\text{SO}}_{\text{3}}\right]\text{=}\frac{\text{0}\text{.0183}}{\text{0}\text{.1}}\\ \text{=0}\text{.183}\text{M}\end{array}$

Therefore, equilibrium concentration of SO₂, O₂ and SO₃ is 0 M, 0.41 M and 0.183 M respectively.