For a chemical reaction, variation in concentration In [ R ] versus time ( min ) plot is given below : What are the units of rate constant k for the reaction?

The given graph is of first order reactions. The rate law for first order reaction can be given as : Rate = k R So, mol L - 1 s - 1 = k mol L - 1 k = s - 1 Hence, the unit of rate constant of a first order reaction is s - 1 .

For a chemical reaction, variation in concentration In [ R ] versus time ( min ) plot is given below : If the initial concentration of the reactant is half of the original concentration, how will t 1 2 change?

From the given graph, it is clear that the graph is of first order reactions. Half life period of first order reactions can be as : t 1 2 = 0 . 693 k , where k is rate constant. From the above equation, it is clear that t 1 2 is independent of the concentration of reactants. Hence, if the initial concentration of the reactant is half of the original concentration, half life period remains unchanged.

Half-life of a first order reaction is 5 × 10 4 s . Then, what percentage of the initial reactant will react in 2 h ?

Given : t 1 2 = 5 × 10 4 s Solution : Half life of first order reactions : t 1 2 = 0 . 693 k So, k = 0 . 693 5 × 10 4 s = 0 . 1386 × 10 - 4 s - 1 . Integrated rate equation for first order reactions : t = 2 . 303 k log A ∘ A t . So, 2 × 3600 = 2 . 303 0 . 1386 × 10 - 4 log A ∘ A t log A ∘ A t = 7200 × 0 . 06 × 10 - 4 = 0 . 0432 A ∘ A t = 1 . 104 A t = A ∘ 1 . 104 Percentage of reactant that will react in two hours = A ∘ - A t A ∘ × 100 = A ∘ - A ∘ 1 . 104 A ∘ = 0 . 104 1 . 104 × 100 = 9 . 4 %

Why is use of aspartame limited to cold foods and drinks?

Aspartame is an artificial sweetening agent which is 100 times as sweet as sucrose. The use of aspartame limited to cold foods and drinks because aspartame is unstable at cooking temperature. Note : The chemical formula of aspartame can be given as : C 14 H 18 N 2 O 5 .

HARD

12th ICSE

IMPORTANT

Earn 100

The rate constant of a reaction is $0.01439 \min^{- 1}$ at $25 ° C$ and its activation energy is $70, 000 J {mol}^{- 1}$ . What is the value of the rate constant at $40 ° C$ ? $(R = 8.314 J K^{- 1})$

Important Questions on Sample Question Paper 5

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 20 - Sample Question Paper 5>SAMPLE QUESTION PAPER 5>Q 17

Nitrogen pentoxide decomposes according to the following equation, $2 N_{2} O_{5} (g) \to 4 {NO}_{2} (g) + O_{2} (g)$ . This first order reaction was allowed to proceed at $40 ° C$ and the data given below:

$[N_{2} O_{5}] (M)$	Time $(\min)$
$0.400$	$0.00$
$0.289$	$20.0$
$0.209$	$40.0$
$0.151$	$60.0$
$0.109$	$80.0$

Calculate the rate constant.

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 20 - Sample Question Paper 5>SAMPLE QUESTION PAPER 6>Q 17

$[N_{2} O_{5}] (M)$	Time $(\min)$
$0.400$	$0.00$
$0.289$	$20.0$
$0.209$	$40.0$
$0.151$	$60.0$
$0.109$	$80.0$

What will be the concentration of $N_{2} O_{5}$ after $100 \min$ ?

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 20 - Sample Question Paper 5>SAMPLE QUESTION PAPER 7>Q 17

$[N_{2} O_{5}] (M)$	Time $(\min)$
$0.400$	$0.00$
$0.289$	$20.0$
$0.209$	$40.0$
$0.151$	$60.0$
$0.109$	$80.0$

Calculate the initial rate of reaction.

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 20 - Sample Question Paper 5>SAMPLE QUESTION PAPER 5>Q 17 or

For a chemical reaction, variation in concentration $In [R]$ versus time $(\min)$ plot is given below :

Question Image

What is the order of reaction?

MEDIUM

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 20 - Sample Question Paper 5>SAMPLE QUESTION PAPER 5>Q 17 or

For a chemical reaction, variation in concentration $In [R]$ versus time $(\min)$ plot is given below :

Question Image

What are the units of rate constant $k$ for the reaction?

MEDIUM

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 20 - Sample Question Paper 5>SAMPLE QUESTION PAPER 5>Q 17 or

For a chemical reaction, variation in concentration $In [R]$ versus time $(\min)$ plot is given below :

Question Image

If the initial concentration of the reactant is half of the original concentration, how will $t_{\frac{1}{2}}$ change?

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 20 - Sample Question Paper 5>SAMPLE QUESTION PAPER 5>Q 17 or

For a chemical reaction, variation in concentration $In [R]$ versus time $(\min)$ plot is given below :

Question Image

Draw the plot of log $[R_{\circ}] / [R]$ versus time $(s)$ .

HARD

12th ICSE

IMPORTANT

All In One ISC Chemistry Class 12>Chapter 20 - Sample Question Paper 5>SAMPLE QUESTION PAPER 5>Q 17 or

Half-life of a first order reaction is

5 \times 10^{4} s

. Then, what percentage of the initial reactant will react in

2 h

The rate constant of a reaction is 0.01439 min-1 at 25°C and its activation energy is 70,000 J mol-1. What is the value of the rate constant at 40°C? R=8.314 J K-1

Important Questions on Sample Question Paper 5

Learn and Prepare for any exam you want !

The rate constant of a reaction is $0.01439 \min^{- 1}$ at $25 ° C$ and its activation energy is $70, 000 J {mol}^{- 1}$ . What is the value of the rate constant at $40 ° C$ ? $(R = 8.314 J K^{- 1})$