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Lesson 20. Review Chapter 6 Pages 60, 61, 62 SBT Chemistry 10 Connecting Knowledge to Life>


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Recognizing 20.1

For the following chemical reaction:

C(s) + O2(g) → CO2(g)

The following weakness No affect the above reaction rate?

A. Temperature.

B. Pressure O2.

C. Carbon content.

D. Carbon surface area.

Detailed explanation:

– Answer:

– Explain:

+ The carbon content only affects the amount of product produced, not the reaction rate

+ High carbon content but not enough contact area does not affect the reaction rate

Recognizing 20.2

Let Zn react with HCl to prepare hydrogen. Describe a way to speed up this reaction.

Solution method:

Based on the influence of factors on the reaction rate:

– As the concentration of reactants increases, the rate of the reaction increases

– As the temperature increases, the reaction rate increases

– When the pressure is increased (for a gaseous reaction), the rate of the reaction increases

– As the surface area of ​​the reactants is increased, the rate of the reaction increases

– Catalysts increase the rate of a chemical reaction (in particular, reduce the activation energy of the reaction) but still preserve the quality and quantity when the reaction ends

Detailed explanation:

– Method 1: Increase the temperature by heating the reaction vessel

– Method 2: Increase concentration (using concentrated HCl solution)

– Method 3: Increase the surface area of ​​the piece of zinc (zinc)

Recognizing 20.3

Oxygen gas is prepared in the laboratory by pyrolysis of potassium chlorate. To make the experiment successful and shorten the time to conduct, some measures can be used:

(1) Use manganese dioxide catalyst.

(2) Fired at high temperature.

(3) Use the water displacement method to collect oxygen.

(4) Crush potassium chlorate.

(5) Mix the potassium chlorate powder and catalyst well.

Number of measures used to speed up the reaction

A. 2.

B. 3.

C. 4.

D. 5.

Solution method:

Based on the influence of factors on the reaction rate:

– As the concentration of reactants increases, the rate of the reaction increases

– As the temperature increases, the reaction rate increases

– When the pressure is increased (for a gaseous reaction), the rate of the reaction increases

– As the surface area of ​​the reactants is increased, the rate of the reaction increases

– Catalysts increase the rate of a chemical reaction (in particular, reduce the activation energy of the reaction) but still preserve the quality and quantity when the reaction ends

Detailed explanation:

– Measures to increase the reaction rate are (1), (2), (4) and (5) → Answer: C

– Measure (3) only describes how to collect oxygen

Recognizing 20.4

Which of the following statements? No correct?

A. Fuel burns faster in high altitude than in low altitude.

B. Food stored at lower temperatures will keep longer.

C. Using yeast as a catalyst to convert sticky rice into alcohol.

D. If the pickle juice is not added when the pickles are salted, the pickles will still be sour but slower.

Solution method:

Based on the influence of factors on the reaction rate:

– As the concentration of reactants increases, the rate of the reaction increases

– As the temperature increases, the reaction rate increases

– When the pressure is increased (for a gaseous reaction), the rate of the reaction increases

– As the surface area of ​​the reactants is increased, the rate of the reaction increases

– Catalysts increase the rate of a chemical reaction (in particular, reduce the activation energy of the reaction) but still preserve the quality and quantity when the reaction ends

Detailed explanation:

– Answer: A

– Explanation: The temperature in the highlands is lower than the temperature in the lowlands → Fuel in the highlands burns more slowly

Recognizing 20.5

In the sulfuric acid production process, the following chemical reaction occurs:

Which of the following statements? No correct?

A. When increasing SO . gas pressure2 or O2 the reaction rate increases.

B. Increase the surface area of ​​the catalyst V2O5 will increase the reaction rate.

C. The catalyst will gradually transform into another substance but the mass remains unchanged.

D. It is necessary to heat the reaction vessel to accelerate the reaction rate.

Solution method:

Based on the influence of factors on the reaction rate:

– As the concentration of reactants increases, the rate of the reaction increases

– As the temperature increases, the reaction rate increases

– When the pressure is increased (for a gaseous reaction), the rate of the reaction increases

– As the surface area of ​​the reactants is increased, the rate of the reaction increases

– Catalysts increase the rate of a chemical reaction (in particular, reduce the activation energy of the reaction) but still preserve the quality and quantity when the reaction ends

Detailed explanation:

– Answer:

– Explanation: A catalyst increases the rate of a chemical reaction (in particular, reduces the activation energy of the reaction) but remains qualitative and quantitative at the end of the reaction.

Understanding 20.6

When kept at 30 °C, an apple is spoiled after 3 days. When stored at 0 °C (in the refrigerator), jujubes spoil after 24 days.

a) Calculate the temperature coefficient of the reaction that occurs when the apple is damaged.

b) If stored at 20 °C, how many days will the apple go bad?

Solution method:

Based on the expression, calculate the temperature coefficient of the reaction

\(\gamma = \frac{{{V_{T + 10}}}}{{{V_T}}}\)or\({\gamma ^{\frac{{\Delta T}}{{10}} }} = \frac{{{V_1}}}{{{V_2}}} = \frac{{{t_2}}}{{{t_1}}}\)

Inside:

+ DRAWBILLION is the reaction rate at temperature T

+ DRAWT + 10 is the reaction rate at T + 10

– Reaction speed is inversely proportional to time

Detailed explanation:

a) Since the rate of reaction is inversely proportional to time

→ Reaction speed increases \(\frac{{24}}{3} = 8\) times → \({\gamma ^{\frac{{30 – 0}}{{10}}}} = \ frac{{24}}{3} = 8\)→ \(\gamma = 2\)

b) Yes \({2^{\frac{{30 – 20}}{{10}}}} = \frac{{{t_2}}}{3}\)→ \({t_2} = 2.3 = 6 \)day

Understanding 20.7

Indicate whether the following statements are true or false. Explain.

(1) For a chemical reaction to take place, the particles (molecules, atoms, ions) of the reactants must collide with each other.

(2) When the CO gas pressure increases, the reaction rate 4CO + Fe3O4 → 4CO2 + 3Fe increased.

(3) When the temperature is increased to 10 oC, the rate of chemical reactions is doubled.

(4) If the collision energy between two reactant molecules is less than the activation energy, a chemical reaction will occur.

(5) The lower the activation energy, the faster the reaction occurs.

Detailed explanation:

(1) False because the particles (molecules, atoms, ions) of the reactant must effectively collide with each other for the reaction to occur

(2) True because the gaseous reaction is CO

(3) Wrong because how many times the reaction rate increases or decreases depends on the temperature coefficient

(4) Wrong because the collision energy between two reactant molecules must be higher than the activation energy to cause a chemical reaction.

(5) Right

Understanding 20.8

At 225 °C, NO . gas2 enter2 have the following reaction:

2NO + O2 → 2NO2

The reaction rate expression has the form: \(v = k.C_{NO}^2.C_{{O_2}}^{}\)

Indicate how the reaction rate will change if:

(i) Increase NO concentration by 2 times.

(ii) Decrease in O . concentration2 go 3 times.

(iii) Increased NO . concentration2 up 2 times.

Detailed explanation:

Based on the instantaneous speed expression we have: \(v = k.C_{NO}^2.C_{{O_2}}^{}\)

– When NO concentration increases 2 times, we have: \(v’ = k. {(2.C_{NO}^{})^2}.C_{{O_2}}^{}\) → \(v’ = 4v\)

→ Reaction speed increased by 4 times

– When the concentration of O2 reduced by 3 times we have: \(v’ = kC{_{NO}^2^{}}. {(\frac{1}{3}.C_{{O_2}}^{})^1}\) → \(v’ = \frac{1}{3}v\)

→ Reaction speed reduced by 3 times

– When the concentration of NO2 2x increase → Reaction speed unchanged

Application 20.9

The decomposition reaction of ethyl iodide in the gas phase takes place as follows:

OLD2H5I → OLD2H4 + HIGHLIGHTS

At 127 °C, the rate constant of the reaction is 1.60.10-7S-first; at 227°C is 4.25.10-4S-first

a) Calculate the temperature coefficient of the above reaction.

b) Calculate the rate constant of the reaction at 167 °C.

Solution method:

Based on the expression, calculate the temperature coefficient of the reaction

\(\gamma = \frac{{{V_{T + 10}}}}{{{V_T}}}\)or\({\gamma ^{\frac{{\Delta T}}{{10}} }} = \frac{{{V_1}}}{{{V_2}}} = \frac{{{t_2}}}{{{t_1}}}\)

Inside:

+ DRAWBILLION is the reaction rate at temperature T

+ DRAWT + 10 is the reaction rate at T + 10

Detailed explanation:

a) The temperature constant is \({\gamma ^{\frac{{227 – 127}}{{10}}}} = \frac{{4,{{25.10}^{ – 4}}}}{ – 4}}}}{) {1,{{60.10}^{ – 7}}}}\) → \(\gamma = 2,2\)

b) At a temperature of 167 °C there is: \(2,{2^{\frac{{167 – 127}}{{10}}}} = \frac{{{V_1}}}{{1,{{ 60.10}^{ – 7}}}}\)

→ The rate constant of the reaction at 167 °C is \({V_1} = 2,{2^{\frac{{167 – 127}}{{10}}}}.1,{60.10^{ – 7 }} = 3,{75.10^{ – 6}}\)

Use 20.10

In the lowlands (altitudes near sea level), water boils at 100 °C. On the top of Mount Fansipan (3200 m above sea level), the water boils at 90 °C. When boiling a piece of meat in boiling water, it takes 3.2 minutes in the plains, while it takes 3.8 minutes on the top of Fansipan.

a) Calculate the temperature coefficient of the reaction that cooks the meat above.

b) If the meat is boiled on a higher mountain, where the water boils at 80 °C, how long will it take to cook the meat?

Solution method:

: Based on the expression, calculate the temperature coefficient of the reaction

\(\gamma = \frac{{{V_{T + 10}}}}{{{V_T}}}\)or\({\gamma ^{\frac{{\Delta T}}{{10}} }} = \frac{{{V_1}}}{{{V_2}}} = \frac{{{t_2}}}{{{t_1}}}\)

Inside:

+ DRAWBILLION is the reaction rate at temperature T

+ DRAWT + 10 is the reaction rate at T + 10

Detailed explanation:

a) The temperature constant is \({\gamma ^{\frac{{100 – 90}}{{10}}}} = \frac{{3,8}}{{3,2}}\) → \(\gamma = 1.1875\)

b) At 80 °C there is \(1,{1875^{\frac{{90 – 80}}{{10}}}} = \frac{{{V_1}}}{{3,8} }\)

→ The time it takes to cook a piece of meat at 80 °C is \({V_1} = 1,{1875^{\frac{{90 – 80}}{{10}}}}.3.8 = 4.5125\ )(min)

Application 20.11

Agent Orange dioxin causes extremely serious harm to the environment and human health. It decomposes very slowly in the soil. Research shows that it takes eight years for dioxin levels in soil to be halved. If a piece of soil contains 0.128 mg of dioxin, how long will it take for the remaining amount of dioxin to be 10 .?-6g dioxin?

Detailed explanation:

– From 0.128.10-3g of dioxin decomposes to 10-6g dioxin has been reduced: \(\frac{{0,{{128.10}^{ – 3}}}}{{{{10}^{ – 6}}}} = 128 = {2^7}\) time

→ Time required to decompose 0.128.10-3g of dioxin decomposes to 10-6g dioxin is:

8.7 = 56 years

Application 20.12

The decomposition reaction of an active antibiotic has a temperature coefficient of 2.5. At 27 °C, after 10 hours, the amount of active ingredient is halved.

a) When put into the human body (37 °C), how long does it take for the amount of active ingredient to be halved?

b) After how long will this antibiotic active ingredient in the human body remain 12.5% ​​compared to the original?

Solution method:

Based on the expression, calculate the temperature coefficient of the reaction

\(\gamma = \frac{{{V_{T + 10}}}}{{{V_T}}}\)or\({\gamma ^{\frac{{\Delta T}}{{10}} }} = \frac{{{V_1}}}{{{V_2}}} = \frac{{{t_2}}}{{{t_1}}}\)

Inside:

+ DRAWBILLION is the reaction rate at temperature T

+ DRAWT + 10 is the reaction rate at T + 10

Detailed explanation:

a) – Yes \(2,{5^{\frac{{37 – 27}}{{10}}}} = \frac{{10}}{{{t_1}}}\)

→ At 37 oC, the amount of active ingredient is halved after: \({t_1} = \frac{{10}}{{2,5}} = 4\)hours

b) The remaining antibiotic is 12.5% ​​ie reduced: \(\frac{{100}}{{12,5}} = 8 = {2^3}\) times

→ The time required for the antibiotic to remain 12.5% ​​in the human body is:

4.3 = 12 years



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