For the reaction, N 2 + 3 H 2 → 2 NH 3 , the rate of change of concentration for hydrogen is 0 . 3 × 10 - 4 Ms - 1 . What will be the rate of change of concentration of ammonia?

Given: ∆ [ H 2 ] ∆ t = 0 . 3 × 10 - 4 M s - 1 . Solution: For the given reaction, N 2 + 3 H 2 → 2 NH 3 , Rate of reaction ( r ) = - ∆ [ N 2 ] ∆ t = - 1 3 ∆ [ H 2 ] ∆ t = + 1 2 ∆ [ NH 3 ] ∆ t . The rate of change in concentration of NH 3 : ( r ) = - 1 3 ∆ [ H 2 ] ∆ t = + 1 2 ∆ NH 3 ∆ t ∆ NH 3 ∆ t = 2 3 × ∆ [ N 2 ] ∆ t = 2 3 × 0 . 3 × 10 - 4 = 0 . 2 × 10 - 4 M s - 1

For a reaction between A and B, the initial rate of reaction is measured for various initial concentrations of A and B . The data provided is : A B Initial rate mol L - 1 s - 1 0 . 020 0 . 010 2 . 40 × 10 - 4 0 . 020 0 . 030 2 . 16 × 10 - 3 0 . 040 0 . 030 4 . 32 × 10 - 3 Calculate the overall order of the reaction.

Rate law can be given as : Rate, r = k A x B y , where k is rate constant. For the given reactant concentrations, rate equations can be given as : 2 . 40 × 10 - 4 = k 0 . 020 x 0 . 010 y 2 . 16 × 10 - 3 = k 0 . 020 x 0 . 030 y 4 . 32 × 10 - 3 = k 0 . 040 x 0 . 030 y Dividing the first two rate equations : 2 . 40 × 10 - 4 2 . 16 × 10 - 3 = k 0 . 020 x 0 . 010 y k 0 . 020 x 0 . 030 y 3 y = 9 y = 2 Dividing the second and third rate equations : 4 . 32 × 10 - 4 2 . 16 × 10 - 3 = k 0 . 040 x 0 . 030 y k 0 . 020 x 0 . 030 y 2 x = 2 x = 1 The over order of reaction = 2 + 1 = 3 .

A first order reaction has a rate constant 1 . 15 × 10 - 3 s - 1 . How long will 5 g of this reactant take to reduce to 3 g ?

Given : k = 1 . 15 × 10 - 3 s - 1 Integrated rate equation for first order reactions : t = 2 . 303 k log A ∘ A t . So, t = 2 . 303 1 . 15 × 10 - 3 s - 1 log 5 3 t = 2 × 10 3 × log 5 - log 3 t = 2 × 10 3 × 0 . 698 - 0 . 477 t = 2 × 10 3 × 0 . 221 t = 442 s

E M B I B E

All In One ISC Chemistry Class 12>Chapter 19 - Sample Question Paper 4>SAMPLE QUESTION PAPER 4

Saleha Pervez and Vandana Solutions for Chapter: Sample Question Paper 4, Exercise 4: SAMPLE QUESTION PAPER 4

Author:Saleha Pervez & Vandana

Saleha Pervez Chemistry Solutions for Exercise - Saleha Pervez and Vandana Solutions for Chapter: Sample Question Paper 4, Exercise 4: SAMPLE QUESTION PAPER 4

Attempt the practice questions on Chapter 19: Sample Question Paper 4, Exercise 4: SAMPLE QUESTION PAPER 4 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: Sample Question Paper 4, Exercise 4: SAMPLE QUESTION PAPER 4 with Hints & Solutions

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 19 - Sample Question Paper 4>SAMPLE QUESTION PAPER 4>Q 7

Calculate the emf of the cell in which the given reaction takes place: $Ni (s) + 2 {Ag}^{+} (0.002 M) \to {Ni}^{2 +} (0.160 M) + 2 Ag$ . Given that $E °_{cell} = 1.05 V$ .

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 19 - Sample Question Paper 4>SAMPLE QUESTION PAPER 4>Q 7

Can you store copper sulphate solution in a zinc vessel?

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 19 - Sample Question Paper 4>SAMPLE QUESTION PAPER 4>Q 16

Consider the reaction, $A \to P$ . The change in the concentration of A with time is shown in the following plot.

Question Image

Predict the order of reaction.

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 19 - Sample Question Paper 4>SAMPLE QUESTION PAPER 4>Q 16

For the reaction, $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ , the rate of change of concentration for hydrogen is $0.3 \times 10^{- 4} {Ms}^{- 1}$ . What will be the rate of change of concentration of ammonia?

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 19 - Sample Question Paper 4>SAMPLE QUESTION PAPER 4>Q 16

For a reaction between A and B, the initial rate of reaction is measured for various initial concentrations of $A$ and $B$ . The data provided is :

$[A]$	$[B]$	Initial rate $(mol L^{- 1} s^{- 1})$
$0.020$	$0.010$	$2.40 \times 10^{- 4}$
$0.020$	$0.030$	$2.16 \times 10^{- 3}$
$0.040$	$0.030$	$4.32 \times 10^{- 3}$