Quiz 135Name:Class :.

Chapter : Rate of Reaction1Diagram 1 shows a graph of the volume of gas produced against time for the reaction between zinc granules and hydrochloric acid.Rajah 1 menunjukkan graf isipadu gas yang dihasilkan melawan masa bagi tindak balas antara ketulan zink dan asid hidroklorik.

Volume of gas, cm3Isipadu gas, cm3

Time, sMasa, s

Diagram 1Rajah 1The gradient of the graph decreases with time becauseKecerunan graf berkurang dengan masa kerana

Acatalyst is not usedmangkin tidak digunakan

Bvolume of mixture decreasesisipadu campuran berkurang

Ctemperature of reaction decreasessuhu tindak balas berkurang

Dconcentration of hydrochloric acid decreaseskepekatan asid hidroklorik berkurang

2Magnesium reacts with acid to produce hydrogen gas, H2. Which solution would give the highest initial rate of reaction?Magnesium bertindak balas dengan asid untuk menghasilkan gas hidrogen, H2. Larutan manakah akan memberikan kadar awal tindak balas yang tertinggi?

A100 cm3 of 1.0 mol dm-3 of nitric acid, HNO3100 cm3 asid nitrik, HNO3 1.0 mol dm-3

B100 cm3 of 1.0 mol dm-3 of hydrochloric acid, HCl100 cm3 asid hidroklorik, HCl 1.0 mol dm-3

C100 cm3 of 1.0 mol dm-3 of sulphuric acid, H2SO4100 cm3 asid sulfurik, H2SO4 1.0 mol dm-3

D100 cm3 of 1.0 mol dm-3 of ethanoic acid, CH3COOH100 cm3 etanoik asid, CH3COOH 1.0 mol dm-3

3Table 1 shows the experiments carried out to study the rate of reaction between zinc carbonate and nitric acid. Jadual 1 menunjukkan eksperimen yang dijalankan bagi mengkaji tindak balas antara zink karbonat dengan asid nitrik. ExperimentEksperimenZinc carbonate, ZnCO3Zink karbonatNitric acid, HNO3Asid nitrik

Mass (g)JisimStateKeadaanVolume(cm3)IsipaduConcentration (mol dm-3)Kepekatan

I5GranuleKetulan500.1

II5PowderSerbuk250.2

Table 1Jadual 1

Which of the following graph represents the two experiments?Antara graf berikut yang manakah mewakili kedua-dua eksperimen itu?

AVolume of CO2/cm3Isipadu CO2/cm3

IIITime/sMasa/s

BVolume of CO2/cm3Isipadu CO2/cm3

IIITime/sMasa/s

CVolume of CO2/cm3Isipadu CO2/cm3

IIITime/sMasa/s

DVolume of CO2/cm3Isipadu CO2/cm3

Time/sMasa/s III

4Diagram 2 shows the graph of the volume of hydrogen gas against time for the reaction between granulated zinc and 50 cm3 1.0 mol dm-3 sulphuric acid. Rajah 2 menunjukkan graf isipadu gas hidrogen melawan masa untuk tindak balas antara ketulan zink dan 50 cm3 asid sulfurik 1.0 mol dm-3Volume/ cm3Isipadu/ cm3

Time / s Masa/ s

XY

Diagram 2Rajah 2Curve X is obtained when excess granulated zinc is reacted with 50 cm3 1.0 mol dm-3 sulphuric acid. Which of the following must be done to produce curve Y?Lengkung X diperoleh apabila ketulan zink yang berlebihan bertindak balas dengan 50 cm3 asid sulfurik 1.0 mol dm-3.Antara berikut yang manakah perlu dilakukan untuk menghasilkan lengkung Y?

AAdd distilled water to sulphuric acid Tambahkan air suling ke dalam asid sulfurikBReplace granulated zinc with zinc powderGantikan ketulan zink dengan serbuk zinkCAdd a few drops of copper (II) sulphate solutionTambahkan beberapa titis larutan kuprum (II) sulfatDAdd sulphuric acid to the mixtureTambahkan asid sulfurik ke dalam campuran

STRUCTURE QUESTION (11 MARKS)An experiment is carried out to study the effect of concentration on rate of reaction. 50.0 cm3 of 0.5 mol dm-3 hydrochloric acid and excess granulated calcium carbonate is put into a conical flask and is placed on an electronic balance as shown in Diagram 1.The mass of the conical flask and its contents is recorded at 60 seconds intervals. Satu eksperimen telah dijalankan untuk mengkaji kesan kepekatan ke atas kadar tindak balas. 50.0 cm3 asid hidroklorik 0.5 mol dm-3 dan ketulan kalsium karbonat berlebihan dimasukkan ke dalam kelalang kun diletakkan di atas neraca elektronik seperti dalam Rajah 1.Jisim kelalang kun dan kandungannya direkodkan pada sela masa 60 saat. DIAGRAM 1RAJAH 1

Table 1 shows the results.Jadual 1 menunjukkan keputusan

Time/sec.[Masa/saat]060120180240300360420480

Balance readings /g[Bacaan neraca/g]140.64140.48140.37140.31140.26140.22140.20140.20140.20

TABLE 1JADUAL 1(a) Give the definition of rate of reaction. Berikan maksud kadar tindak balas

........................................................................................................................................ [1 mark]

(b) Explain why the mass of the conical flask and its content decreases with time. Terangkan mengapa jisim kelalang kun dan kandungannya berkurang dengan masa

........................................................................................................................................

........................................................................................................................................ [1 mark]

(c) Based on Table 1, plot a graph of the mass of the conical flask and its content versus time on the graph paper below. Berdasarkan Jadual 1, plotkan graf jisim kelalang kun dan kandungannya melawan masa pada kertas graf di bawah.Graph of mass of the conical flask and its content versus timeGraf jisim kelalang kun dan kandungannya melawan masa

[3 marks](d) Based on the graph, calculate the instantaneous rate of reaction at 120 seconds. Berdasarkan graf, hitungkan kadar tindak balas pada saat ke 120

[3 marks](e) (i) Calculate the mass of carbon dioxide gas produced. Hitungkan jisim gas karbon dioksida yang dihasilkan

[1 mark]

(ii) Calculate the expected volume of gas produced at room condition.[1 mol of gas occupies 24 dm3 at room condition; relative molecular mass of CO2 is 44 ]Hitungkan isipadu gas yang dijangka akan terhasil pada keadaan bilik. 1 mol gas menempati 24 dm3 pada suhu bilik ; jisim molekul relatif CO2 ialah 44

[2 marks]ESSAY QUESTION (20 MARKS)(a)A student carried out an experiment to investigate the effect of catalyst on the rate of reaction. Diagram 2 shows the decomposition of hydrogen peroxide in the presence of manganese(IV) oxide. Seorang pelajar menjalankan satu eksperimen untuk mengkaji kesan mangkin ke atas kadar tindak balas. Rajah 8 menunjukkan penguraian hidrogen peroksida dengan kehadiran mangan(IV) oksida.

Diagram 2Rajah 2Using Collision Theory, explain the effect of manganese(IV) oxide on the decomposition of hydrogen peroxide.Dengan menggunakan Teori Perlanggaran, terangkan kesan mangan(IV) oksida ke atas penguraian hidrogen peroksida. [5 marks][5 markah]

(b) A student carried out two experiments to determine the time taken to collect a maximum volume of 50 cm3 of gas. Table 2 shows the result of the experiments.Seorang pelajar menjalankan dua eksperimen untuk menentukan masa yang diambil bagi mengumpul isipadu maksima gas sebanyak 50 cm3.Jadual 2 menunjukkan keputusan eksperimen tersebut.

ExperimentEksperimenReactantBahan tindak balasTime, sMasa, s

I50 cm3 of 1-volume hydrogen peroxide solution and 1 g manganese(IV) oxide50 cm3 larutan hidrogen peroksida 1-isipadu dan 1g mangan(IV) oksida40

II25 cm3 of 2-volume hydrogen peroxide solution and 1 g manganese(IV) oxide25 cm3 larutan hidrogen peroksida 2-isipadu dan 1g mangan(IV) oksida 20

Table 2Jadual 2

(i) Draw a labelled diagram of the apparatus set up for the experiment. Lukiskan gambar rajah berlabel bagi susunan radas eksperimen tersebut.[2 marks][2 markah](ii)Sketch the graph for the gas liberated against time for Experiment I and Experiment II on the same axes.Lakarkan graf isipadu gas terbebas melawan masa pada paksi yang sama bagi Eksperimen I dan Eksperimen II. [2 marks][2 markah]

(iii)Calculate the average rate of reaction for both experiments.Tentukan kadar tindak balas purata bagi kedua-dua eksperimen. [2 marks][2 markah]

(iv)Compare the rate of reaction between Experiment I and Experiment II. Explain the differences based on Collision Theory.Bandingkan kadar tindak balas antara Eksperimen I dan Eksperimen II. Terangkan perbezaan tersebut berdasarkan Teori Perlanggaran. [5 marks][5 markah]

(v)Calculate the concentration in mol dm-3 hydrogen peroxide used in Experiment I in order to produce 50 cm3 of oxygen. Hitung kepekatan hidrogen peroksida dalam mol dm-3 yang digunakan dalam Eksperimen I untuk menghasilkan 50 cm3 oksigen.

[Molar volume of gas = 24 dm3 mol-1 at room conditions] [ Isipadu molar gas = 24 dm3 mol-1 pada keadaan bilik][4 marks][4 markah]

10QUIZ 1_CHEM/F5