Answer the following question on the basis of the curve given below for a zero order reaction : R → P Find the relation between slope of this line and rate constant.

Straight line equation of the given graph can be derived as : R = R ∘ - mt , where R is the reactant concentration, R ∘ (initial reactant concentration) is the intercept and m is the slope. Integrated rate equation of zero order can be given as : R ∘ - R = kt , where R is molar concentration of reactant at time, t . This equation can also be given as : R = R ∘ - kt . On comparing the above two equations, we get : m = k , where m is the slope and k is the rate constant.

Answer the following question on the basis of the curve given below for a zero order reaction : R → P If the slope is 2 × 10 4 mol L - 1 s - 1 , then, calculate the rate constant.

Straight line equation of the given graph can be derived as : R = R ∘ - mt , where R is the reactant concentration, R ∘ (initial reactant concentration) is the intercept and m is the slope. Integrated rate equation of zero order can be given as : R ∘ - R = kt , where R is molar concentration of reactant at time, t . This equation can also be given as : R = R ∘ - kt . On comparing the above two equations, we get : m = k = 2 × 10 4 mol L - 1 s - 1 , where m is the slope and k is the rate constant.

238 U 96 changes to P 96 206 by successive radioactive decay. A sample of uranium was analyzed and found to contain 1 . 0 g of U 238 and 0 . 1 g of Pb 206 had accumulated due to decay of 238 U , find out the age of ore. (Half-life of U 238 = 4 . 5 × 10 9 years).

All radioactive reactions are first order reactions. Half life of first order reactions can be given as : t 1 2 = 0 . 693 k . So, k = 0 . 693 4 . 5 × 10 9 = 1 . 54 × 10 - 10 years -1 . Integrated rate equation for first order reactions can be given as : k = 2 . 303 t log R ∘ R , where k is rate constant. Age of the ore, t = 2 . 303 1 . 54 × 10 - 10 log 1 g 0 . 1 g = 2 . 303 × 10 10 1 . 54 = 1 . 49 × 10 10 years.

For the hydrolysis of methyl acetate in the aqueous solution, following results were obtained : t / s 0 30 60 CH 3 OCH 3 / mol L - 1 0 . 60 0 . 30 0 . 15 Show that it follows pseudo first order reaction, as the concentration of water remains constant.

Pseudo first order reactions are bimolecular but first order reactions. Hydrolysis of methyl acetate is a pseudo first order reaction as the concentration of water has no effect on the rate of reaction. Methyl acetate hydrolysis is a bimolecular reaction. In first order reactions, the value of rate constant remains same at different time intervals and reactant concentrations. Rate constant of first order reaction can be expressed as : k = 2 . 303 t log R ∘ R , where R ∘ is the initial molar concentration of reactants. At t = 30 s , k = 2 . 303 30 log 0 . 6 0 . 3 . k = 2 . 303 30 log 2 . At t = 60 s , k = 2 . 303 60 log 0 . 6 0 . 15 k = 2 . 303 60 log 2 2 = 2 . 303 × 2 60 log 2 = 2 . 303 30 log 2 . Hence, methyl acetate hydrolysis is a pseudo first order reaction.

For the hydrolysis of methyl acetate in the aqueous solution, following results were obtained : t / s 0 30 60 CH 3 OCH 3 / mol L - 1 0 . 60 0 . 30 0 . 15 Calculate the average rate of reaction between the time interval 30 s to 60 s .

The rate of reaction over a certain period of time during the course of reaction is called average rate of reaction. Average rate of reaction = R 1 - R 2 t 2 - t 1 = 0 . 30 - 0 . 15 60 - 30 = 0 . 15 30 = 0 . 005 . Hence, the average rate of reaction for the given reaction is 5 × 10 - 3 mol L - 1 s - 1 .

Saleha Pervez and Vandana Solutions for Chapter: Chemical Kinetics, Exercise 5: Chapter Practice

Author:Saleha Pervez & Vandana

Saleha Pervez Chemistry Solutions for Exercise - Saleha Pervez and Vandana Solutions for Chapter: Chemical Kinetics, Exercise 5: Chapter Practice

Attempt the practice questions on Chapter 4: Chemical Kinetics, Exercise 5: Chapter Practice with hints and solutions to strengthen your understanding. All In One ISC Chemistry Class 12 solutions are prepared by Experienced Embibe Experts.

Questions from Saleha Pervez and Vandana Solutions for Chapter: Chemical Kinetics, Exercise 5: Chapter Practice with Hints & Solutions

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 43

Answer the following question on the basis of the curve given below for a zero order reaction : $R \to P$

Question Image

Find the relation between slope of this line and rate constant.

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 43

Answer the following question on the basis of the curve given below for a zero order reaction : $R \to P$

Question Image

If the slope is $2 \times 10^{4} mol L^{- 1} s^{- 1}$ , then, calculate the rate constant.

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 44

Consider the reaction, $A \to B$ . The change in concentration with time $t$ is shown in the graph. Predict the order of the reaction.

Question Image

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 44

Consider the reaction, $A \to B$ . The change in concentration with time $t$ is shown in the graph. Derive the expression for the time which is
required to complete the reaction.

Question Image

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 44

Consider the reaction, $A \to B$ . The change in concentration with time $t$ is shown in the graph. Predict the order of the reaction.

Question Image

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 45

$^{238} U_{96}$ changes to ${}^{206}P_{96}$ by successive radioactive decay. A sample of uranium was analyzed and found to contain $1.0 g$ of ${}^{^{238}}U$ and $0.1 g$ of ${}_{206}{Pb}$ had accumulated due to decay of $^{238} U$ , find out the age of ore. (Half-life of ${}^{^{238}}U = 4.5 \times 10^{9}$ years).

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 46

For the hydrolysis of methyl acetate in the aqueous solution, following results were obtained :

$t / s$	$0$	$30$	$60$
${CH}_{3} {OCH}_{3} / mol L^{- 1}$	$0.60$	$0.30$	$0.15$

Show that it follows pseudo first order reaction, as the concentration of water remains constant.

MEDIUM

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 4 - Chemical Kinetics>Chapter Practice>Q 46

For the hydrolysis of methyl acetate in the aqueous solution, following results were obtained :

$t / s$	$0$	$30$	$60$
${CH}_{3} {OCH}_{3} / mol L^{- 1}$	$0.60$	$0.30$	$0.15$

Calculate the average rate of reaction between the time interval $30 s$ to $60 s$ .

Saleha Pervez and Vandana Solutions for Chapter: Chemical Kinetics, Exercise 5: Chapter Practice

Saleha Pervez Chemistry Solutions for Exercise - Saleha Pervez and Vandana Solutions for Chapter: Chemical Kinetics, Exercise 5: Chapter Practice

Questions from Saleha Pervez and Vandana Solutions for Chapter: Chemical Kinetics, Exercise 5: Chapter Practice with Hints & Solutions

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