Lesson 7: Chemistry on Fire and Explosion Reactions – Chemistry Learning Topic 10 Creative Horizons>

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Chapter 44 Introduction:

The phenomena of fire and explosion occur mostly due to chemical reactions, which give off a lot of heat and have a high reaction rate. Therefore, chemistry plays a very important role in studying the causes, giving prevention measures as well as handling when a fire occurs in the most effective and safest way. How is the heat of a fire or explosion reaction determined? What factors affect the intensity of a fire or explosion reaction?

Detailed explanation:

– Calculation of enthalpy change of combustion and explosion reaction according to binding energy

$${\Delta _r}H_{298}^o = \sum {{E_b}(cd) – \sum {{E_b}(sp)} }$$

– Calculation of enthaloy changes of combustion and explosion reactions according to the standard heat of formation of substances

$${\Delta _r}H_{298}^o = \sum {{\Delta _f}H_{298}^o(sp) – \sum {{\Delta _f}H_{298}^o(cd)} }$$

Factors affecting the rate of combustion reaction:

+ Flammable.

Oxidant (oxygen)

+ Heat source

Chapter 44 Discussion 1:

Based on the data in Tables 7.1 and 7.2, calculate the enthalpy change of the reaction of burning 1 mole of ethanol and 1 mole of gas.

Detailed explanation:

Ethanol combustion reaction:

The reaction of burning 1 mole of propane:

The reaction of burning 1 mole of butane:

The enthalpy change for the reaction of burning 1 mole of a gas containing propane (40%) and butane (60%) is:

$${\Delta _r}H_{298}^0 = ( – 1718).0.4 + ( – 2222).0.6 = 2020,4kJ$$

Chapter 45 Discussion 2:

Calculate the enthalpy change of the reaction of burning 1 mole of octane (C8H18, which is found in gasoline) and 1 mole of methane (the main component of natural gas). Predict the intensity of these reactions.

Detailed explanation:

The reaction of burning 1 mole of octane:

$$\begin{array}{l}{\Delta _r}H_{298}^o = {E_b}({C_8}{H_{18}}) + \frac{{25}}{2}. {E_b }({O_2}) – 8. {E_b}(C{O_2}) – 9. {E_b}({H_2}O)\\{\Delta _r}H_{298}^o = (7. {E_{ C – C}} + 18. {E_{C – H}}) + \frac{{25}}{2} {E_{O = O}} – 8.2. {E_{C = O}} – 9.2 {E_{O – H}}\\{\Delta _r}H_{298}^o = (7,347 + 18,413) + \frac{{25}}{2}.498 – 8.2.745 – 9.2.467 = – 4238kJ\end{array}$$

Burn 1 mol C8H18(g) gives off 4238 kJ of heat

The reaction burns 1 mole of methane

$$\begin{array}{l}{\Delta _r}H_{298}^o = {E_b}(C{H_4}) + 2. {E_b}({O_2}) – {E_b}(C{O_2 }) – 2{E_b}({H_2}O)\\{\Delta _r}H_{298}^o = 4. {E_{C – H}} + 2. {E_{O = O}} – 2 {E_{C = O}} – 2.2. {E_{O – H}}\\{\Delta _r}H_{298}^o = 4.413 + 2.498 – 2.745 – 2.2.467 = – 710kJ\end{array }$$

Burn 1 mole ONLY4(g) gives off 710 kJ of heat

Thus, the heat released when burning octane is much larger than burning methane. Or octane combustion is more intense than methane combustion.

Chapter 45 Discussion 3:

Under normal conditions (298K), oxygen makes up about 20.9% by volume in air, equivalent to a pressure of 0.209 atm. Calculate the molarity of oxygen in the air.

Detailed explanation:

Take 100 L of air with 20.9 L of oxygen

$$\begin{array}{l}{n_{{O_2}}} = \frac{{{P_{{O_2}}} {V_{{O_2}}}}{{RT}} = \frac {{0,209.20.9}}{{0,082,298}} = 0.179mol\\{C_{{O_2}}} = \frac{{{n_{{O_2}}}}}{{{V_{{O_2}}} }} = \frac{{0,179}}{{20.9}} = 8,{56.10^{ – 3}}mol/L\end{array}$$

Chapter 45 Discussion 4:

When the volume of oxygen is reduced to 15% of the volume of air, what is the mole/L concentration of oxygen?

Detailed explanation:

$$\begin{array}{l}\frac{{{C_{{O_2}(20.9\% )}}}}{{{C_{{O_2}(15\% )}}}} = \ frac{{20.9}}{{15}}\\ = > \frac{{8,{{56.10}^{ – 3}}}}{{{C_{{O_2}(15\% )}} }} = \frac{{20.9}}{{15}}\\ = > {C_{{O_2}(15\% )}} = 6,{14.10^{ – 3}}mol/L\end {array}$$

Chapter 45 Discussion 5:

Indicate how many times the rate of combustion of coal increases or decreases when the percentage by volume of oxygen in the air decreases from 20.9% to 15%.

Detailed explanation:

From the expression to calculate the burning reaction rate of the ice bar: v = k × CO2 shows that if the oxygen concentration is reduced by how many times, the rate of reaction decreases by many times.

The rate of combustion is directly proportional to the oxygen concentration.

\(\frac{{{v_{15\% }}}}{{{v_{20.9\% }}}} = \frac{{15}}{{20.9}} = 0.72\ ) time

Thus, the combustion reaction rate of coal is reduced to only 0.72 times compared to the original.

Chapter 46 Discussion 6:

Assume a room has a percentage by volume composition of oxygen in the air of 17%. How many times does the “respiratory response” rate of people in the room increase or decrease compared to outside the room? Know that oxygen makes up about 20.9% by volume in air.

Detailed explanation:

The rate of the “respiratory response” depends on the oxygen concentration according to the rate equation:

v = k × CO2

\(\frac{{{v_{17\% }}}}{{{v_{20.9\% }}}} = \frac{{17}}{{20.9}} = 0.81\ ) time

Thus, the rate of “respiratory response” of people in the room reduced to only 0.81 times compared to outside the room.

Chapter 45 Discussion 7:

List the factors that affect the rate of a combustion reaction. From there, let’s list some measures to put out a fire.

Detailed explanation:

Factors affecting the rate of combustion reaction:

+ Flammable.

Oxidant (oxygen)

+ Heat source

Some measures to extinguish a fire:

+ For fires that are solid substances such as wood, firewood, and straw, we can use water to reduce the temperature of the fire to below the burning temperature, diluting the burning gas.

+ For gasoline and oil fires, do not use water to extinguish, but use sand (for small fires) or use specialized fire extinguishing powder to extinguish.

Chapter 45 Discussion 8:

Give examples of some combustibles of each type of fire in Table 7.3.

Detailed explanation:

 Type of fire Flammables Eg Type A Fires of solids (usually organic) when burning are often accompanied by the creation of embers. Solids include wood, paper, cloth, garbage and other common materials. Type of cow Fires of liquefied liquids and solids Gasoline, oil, paint, etc. Size Type Fire of gases Natural gas, methane, hydrogen, etc. EASY Type Fire of metals Alkali metals (Na, K, Li), alkaline earth metals (Ca, Mg), aluminum (aluminium Al) Type F Fires of animal or vegetable oils and fats in cooking appliances Cooking oil, fat, etc.

Chapter 46 Discussion 9:

Why can’t water be used in some cases to fight fires (gasoline, oil, etc.)

Detailed explanation:

Do not use water to extinguish gasoline and oil fires because gasoline and oil are lighter than water and do not dissolve in water. Using water will cause gasoline and oil to spread, making the fire spread and more difficult to extinguish.

Chapter 47 Application:

Explain why the fire has active metals such as alkali metals, alkaline earth and aluminum, etc without using water, CO2sand (main component is SiO2), fire extinguishing foam (mixture of air, water and surfactant) to extinguish the fire?

Detailed explanation:

– Do not use water to extinguish fires with strong metals such as alkali metals, alkaline earth and aluminum, etc. Because these metals have the ability to react with water to release hydrogen, resulting in an explosion. Explosions of vapor scatter these metals everywhere, leading to a widespread fire. Furthermore, some metals when heated will split water into oxygen and hydrogen which can create a large hydrogen explosion.

– Do not use CO . gas2 to put out magnesium fires because when magnesium burns in the presence of CO2 The following reaction will occur:

2Mg + CO2 → 2MgO + C.

The above reaction is very exothermic and produces soot. This soot continues to burn and makes the fire even harder to control.

– Do not use sand to extinguish magnesium metal fires because magnesium can react with SiO .2 the main ingredient in sand, making the fire more difficult to control.

– Do not use fire extinguishing foams, as they will continue to burn due to the reaction with the air and water contained in the foam.

– Do not use fire extinguishing powder containing NaHCO3 because when high temperature will decompose into CO2 and continues to react with Mg causing the fire to spread further.

Chapter 47 Lesson 1:

The rate of the combustion reaction depends on the oxygen concentration. As oxygen concentration decreases, how does the rate of combustion change?

Detailed explanation:

As the oxygen concentration decreases, the combustion reaction rate decreases and vice versa.

Chapter 47 Lesson 2:

The rate of the respiratory reaction depends on the oxygen concentration. As oxygen concentration increases, how does the rate of the “respiratory response” change?

Detailed explanation:

When oxygen concentration decreases, the rate of “respiratory response” decreases and vice versa.

Chapter 47 Lesson 3:

The air at the top of the mountain is very thin. This can have an adverse effect on climbers. Therefore, climbers always equip oxygen tanks when they climb high mountain peaks. Let’s say the air on top of that mountain has 16% oxygen by volume. How many times does the rate of “respiratory response” increase or decrease compared to where the air is 20.9% oxygen by volume?

Detailed explanation:

The rate of the “respiratory response” depends on the oxygen concentration according to the rate equation:

V = kCO2

\(\frac{{{v_{16\% }}}}{{{v_{20.9\% }}}} = \frac{{16}}{{20.9}} = 0.77\ ) time

Thus, the rate of “respiratory response” of humans at the top of the mountain is reduced to only 0.77 times compared to where the air has 20.9% oxygen by volume.

Chapter 47 Lesson 4

Name some substances that can be used to extinguish a fire when a fire occurs in

a) lumberyard

b) petrol and oil stations.

Detailed explanation:

a) For Class A combustibles (wood, firewood) we can use chemicals such as water, carbon dioxide (CO2), foam extinguishing agent, dry powder (NaHCO3) to extinguish.

b) For Class B combustibles (gasoline, oil) we can use chemicals such as carbon dioxide (CO2), foam extinguishing agent, dry powder (NaHCO3) to extinguish. Absolutely do not use water.

Chapter 47 Lesson 5

In a gasoline or oil fire, a wet blanket or sand can be used to extinguish the fire. Explain why this is possible.

Detailed explanation:

Using a damp blanket or sand to extinguish the fire works to prevent the flame from coming into contact with oxygen, breaking up a flammable agent in the fire triangle.