Chapter 5, Problem 91QAP

Interpretation Introduction

(a)

Interpretation:

By burning sulfur with oxygen sulfur trioxide can be prepared:

${\text{2S(s) + 3O}}_{\text{2}}\text{(g) }\to {\text{ 2SO}}_{\text{3}}\text{(g)}$

A 5.00 L flask having 5 g of oxygen and sulfur at a temperature of 25 ⁰C and pressure of 995 mm Hg is heated. The temperature in the flask increases to 138 ⁰C after the completion of reaction. The pressure of SO₃ in the flask needs to be calculated.

Concept introduction:

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. n, temperature i.e. T and Universal gas constant i.e. R are interrelated as below:

$\text{PV = nRT}$

The molarity of the solution is calculated as follows:

$M=\frac{n}{V\left(\text{L}\right)}$

Here, n is number of moles of solute and V is volume of solution in L.

Answer to Problem 91QAP

1.05 atm

Explanation of Solution

By burning sulfur with oxygen sulfur trioxide can be prepared

${\text{2S(s) + 3O}}_{\text{2}}\text{(g) }\to {\text{ 2SO}}_{\text{3}}\text{(g)}$

In the 5 g of sulfur the mole of sulfur present is:

$\begin{array}{l}{\text{n}}_s=\frac{m_s}{{\left(MM\right)}_s}\\ \text{ =}\frac{5\text{ g}}{32.0\text{ g/mol}}\\ =0.15\text{6 mol}\end{array}$

Moles of sulfur reacted is 0.156 mol.

Similarly, number of moles of oxygen can be calculated using the ideal gas equation as follows:

$n=\frac{PV}{RT}$

The given pressure can be converted into atm as follows:

$1\text{ mm Hg}=0.001316\text{ atm}$

Thus,

$995\text{ mm Hg}=1.31\text{ atm}$

The given temperature can be converted into Kelvin as follows:

$0{}^{\circ }\text{C}=273.15\text{ K}$

Thus,

$25{}^{\circ }\text{C}=298.15\text{ K}$

Putting the values,

$n=\frac{\left(1.31\text{ atm}\right)\left(5\text{ L}\right)}{\left(0.082{\text{ atm L K}}^{-1}{\text{ mol}}^{-1}\right)\left(298.15\text{ K}\right)}=0.268\text{mol}$

From the balanced chemical reaction, 2 mol of S must react with 3 mol of ${\text{O}}_{\text{2}}$ thus, 0.156 mol of S must react with $\frac{3}{2}\times 0.156=0.233\text{ mol}$.

The number of moles of oxygen gas given is 0.268 mol thus, it is present in excess and the sulfur is limiting reactant.

The number of moles of SO₃ produced is same as that of sulfur.

After completion of reaction, the volume is 5.00 L, temperature is 138 ⁰C and the pressure in the flask due to SO₃ is calculated as below:

$\begin{array}{l}{\text{P}}_{S{O}_3}V={\text{n}}_{S{O}_3}RT\\ {\text{P}}_{S{O}_3}\left(5.0\text{0 L}\right)=\left(0.156\text{ mol}\right)\left(0.08205{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(138{}^0\text{C}\right)\\ {\text{ P}}_{S{O}_3}=\frac{\left(0.156\text{ mol}\right)\left(0.08205{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(138+273\right)\text{K}}{5.00\text{ L}}\\ =\text{ 1}\text{.05 atm}\end{array}$

Therefore, the pressure of SO₃ is 1.05 atm.

Interpretation Introduction

(b)

Interpretation:

The total pressure in the flask needs to be determined.

Concept introduction:

The ideal gas equation is as follows:

$PV=nRT$

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

Answer to Problem 91QAP

1.27 atm.

Explanation of Solution

The molar ratio of O₂ :S is 3:2. The moles of oxygen used to react with 0.156 mol of sulfur are calculated as below:

$\begin{array}{l}{\left(n_{{\text{O}}_{\text{2}}}\right)}_{\text{reacted}}=n_s\left({\text{Molar ratio of O}}_{\text{2}}\text{ and S}\right)\\ \text{ = 0}\text{.156 mol}\left(\frac{3}{2}\right)\\ \text{ =}0.234\text{ mol}\end{array}$

Therefore, the moles of O₂ used in 0.156 mol of sulfur are 0.234 mol.

The pressure in the flask prior the reaction is 995 mm Hg. The moles of O₂ present initially a 25 ⁰C are calculated as below:

$\begin{array}{l}{\text{P}}_{O_2}V={\left({\text{n}}_{O_2}\right)}_{excess}RT\\ \left(995\text{ mm Hg}\right)\left(5.0\text{0 L}\right)={\left({\text{n}}_{O_2}\right)}_{excess}\left(0.08205{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(25{}^0\text{C}\right)\\ {\left({\text{n}}_{O_2}\right)}_{excess}=\frac{\left(995\text{ mm Hg}\right)\left(5.0\text{0 L}\right)}{\left(0.08205{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(\text{25+273}\right)\text{K}}\left(\frac{0.001315}{1\text{ mmHg}}\right)\\ =\text{ 0}\text{.268 mol}\end{array}$

Therefore, mole of oxygen present in the mixture is 0.268 mol.

The oxygen which is left unreacted is calculated as below:

$\begin{array}{l}{\left({\text{n}}_{O_2}\right)}_{unreacted}={\left({\text{n}}_{O_2}\right)}_{excess}-{\left({\text{n}}_{O_2}\right)}_{reacted}\\ \text{ = 0}\text{.268 mol }-\text{ 0}\text{.234 mol}\\ \text{ = 0}\text{.034 mol}\end{array}$

The total number of moles in the flask =

$\begin{array}{l}{n}_{tot}=n_{S{O}_3}-{\left({\text{n}}_{O_2}\right)}_{reacted}\\ \text{ = 0}\text{.156 mol }+\text{ 0}\text{.034 mol}\\ \text{ = 0}\text{.190 mol}\end{array}$

The total number of moles in the flask is 0.190 mol.

The total pressure of gas in the flask due to SO₃ and unreacted oxygen is calculated as

$\begin{array}{l}{\text{P}}_{tot}V=n_{tot}RT\\ {\text{P}}_{tot}\left(5.0\text{0 L}\right)=\left(0.190\text{ mol}\right)\left(0.08205{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(138{}^0\text{C}\right)\\ {\text{ P}}_{tot}=\frac{\left(0.190\text{ mol}\right)\left(0.08205{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(\text{138+273}\right)\text{K}}{5.00\text{ L}}\\ =\text{ 1}\text{.27 atm}\end{array}$

Therefore, the total pressure in the flask is 1.27 atm.

Interpretation Introduction

(c)

Interpretation:

The molarity of H₂ SO₄ formed if 250.0 mL is added into the flask needs to be determined.

Concept introduction:

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. n, temperature i.e. T and Universal gas constant i.e. R are interrelated as below:

$\text{PV = nRT}$

The molarity of the solution is calculated as follows:

$M=\frac{n}{V\left(\text{L}\right)}$

Here, n is number of moles of solute and V is volume of solution in L.

Answer to Problem 91QAP

0.03 M.

Explanation of Solution

The volume of the flask containing SO₃ is 5.00 L, if 250.0 mL of water is added, the total volume of solution becomes:

$V_{\text{Solution}}=5.00\text{ L+}\frac{250}{1000}\text{ L}=5.25\text{ L}$

The molarity of solution is calculated as follows:

${\text{Molarity of H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{ = }\frac{{\text{moles of solute (SO}}_{\text{3}}\text{)}}{{\text{volume of solution (H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{)}}$

The moles of SO₃ is 0.156 mol. The molarity of H₂ SO₄ is calculated as below:

$\begin{array}{l}{\text{Molarity of H}}_{\text{2}}{\text{SO}}_{\text{4}}=\frac{0.156\text{ mol}}{5.25\text{ L}}\\ =\text{ 0}\text{.03 M}\end{array}$

Therefore, the molarity of H₂ SO₄ is 0.03 M.