Class 12 Chemistry Sample Paper With Solutions Set A

Please refer to Class 12 Chemistry Sample Paper With Solutions Set A below. These Class 12 Chemistry Sample Papers will help you to get more understanding of the type of questions expected in the upcoming exams. All sample guess papers for Chemistry Class 12 have been designed as per the latest examination pattern issued by CBSE. Please practice all CBSE Sample Papers for Chemistry in Standard 12.

Sample Paper Class 12 Chemistry With Solutions Set A

Topic-1
Rate of a Chemical Reaction and Factors Affecting Rate of Reactions

Very Short Answer-Objective Type Questions

A. Multiple choice Questions: 

Question. Rate law for the reaction A+2B- → C, is found to be Rate =k[A][B]. If the concentration of reactant B is doubled keeping the concentration of A constant, the value of rate constant will be :
(a) the same.
(b) doubled.
(c) quadrupled.
(d) halved.

Question. In the presence of a catalyst, heat evolved or absorbed during reaction :
(a) increases.
(b) decreases.
(c) remains unchanged.
(d) may increase or decrease.

Question. Which of the following expressions is correct for the rate of reaction given below?
5Br⁻ (aq) + BrO₃– (aq) + 6H⁺ (aq) → 3Br₂ (aq) + 3H₂O (l)

B. Match the following :

Question. Match the items of Column I and Column II.
Column I                                     Column II
(i) Diamond                               (a) short interval of time
(ii) Instantaneous rate               (b) ordinarily rate of
                                                    conversion is imperceptible
(iii) Average rate                       (c) long duration of time
Answer. Correct option :
(ii) → (a)
(i) → (b)
(iii) → (c)

C. Answer the following:

Question. Express the rate of the following reaction in terms of the formation of ammonia :
N₂(g) + 3H₂(g) → 2 NH₃(g)
Answer. Rate of the reaction.

Question. Define rate constant (k).
Answer. Rate constant is the rate of reaction when the concentration of reactant is unity.

Short Answer Type Questions 

Question. For the reaction
2N₂O₅ (g) → 4NO₂ (g) + O₂ (g), the rate of formation of NO₂ (g) is 2.8 × 10^–3 M s^–1.
Calculate the rate of disappearance of N₂O₅ (g).
Answer.

Question. What is meant by rate of a reaction ? Differentiate between average rate and instantaneous rate of a reaction.
Answer. Rate of a reaction is defined as the change in concentration of reactant or product in a chemical reaction at particular time interval.
Rate of reaction = Change in concentration / Timetaken

Topic-2
Order of a Reaction, Integrated Rate Equations and Half-life of a Reaction

Very Short Answer-Objective Type Questions 

A. Multiple choice Questions: 

Question. A first order reaction is 50% completed in 1.26 × 1014 s. How much time would it take for 100% completion?
(a) 1.26 × 10¹⁵ s
(b) 2.52 × 10¹⁴ s
(c) 2.52 × 10²⁸ s
(d) Infinite

Question. The value of rate constant of a pseudo first order reaction :
(a) depends on the concentration of reactants present in small amount.
(b) depends on the concentration of reactants present in excess.
(c) is independent of the concentration of reactants.
(d) depends only on temperature.

Question. Consider a first order gas phase decomposition reaction given below :
A(g) → B(g) + C(g)
The initial pressure of the system before decomposition of A was ‘Pi.’ After lapse of time ‘t’,total pressure of the system increased by x units and became ‘Pt’. The rate constant k for the reaction is given as ___________.

B. Answer the following:

Question. Write the unit of rate constant for a zero order reaction. 
Answer. Unit is mol L^–1 s^–1. 

Question. If the rate constant of a reaction is k = 3 × 10^–4 s^–1, then identify the order of the reaction.
Answer. On the basis of unit of rate constant (s^–1), the order of reaction is first order.

Question. For the reaction A → B, the rate of reaction becomes three times when the concentration of A is increased by nine times. What is the order of reaction?
Answer. Order of reaction = 1/2 

Question. For a reaction R → P, half-life (t_1/2) is observed to be independent of the initial concentration of reactants . What is the order of reaction?
Answer. First order.

Question. For a reaction, A + B → Product, the rate law is given by r = k [A]^1/2 [B]². What is the order of the reaction ? 
Answer.

Question. For a chemical reaction R → P, the variation in the
concentration (R) vs. time (t) plot is given as :

(i) Predict the order of the reaction.
(ii) What is the slope of the curve ?
Answer. (i) The reaction is a zero order reaction.
(ii) The slope of curve is (–k) i.e., negative of rate constant. 

Short Answer Type Questions

Question. (i) What is the order of the reaction whose rate constant has same units as the rate of reaction ?
(ii) For a reaction A + H₂O → B; Rate μ [A]. What is the order of this reaction?
Answer. (i) Zero Order 
(ii) Pseudo-first Order

Question. What do you understand by the rate law and rate constant of a reaction ? Identify the order of a reaction if the units of its rate constants are :
(i) L^–1 mol s^–1
(ii) L mol^–1 s^–1 
Answer. The representation of rate of reaction in terms of concentration of the reactants is known as rate law.
The rate constant is defined as the rate of reaction when the concentration of each reactant is unity.
(i) Zero order 
(ii) Second order

Question. (i) Explain why H₂ and O₂ do not react at room temperature.
(ii) Write the rate equation for the reaction A₂+3B₂ → 2C, if the overall order of the reaction is zero. A&E+A
OR
Derive integrated rate equation for rate constant of a first order reaction.
Answer. (i) Due to high activation energy 
(ii) Rate = k [A₂]⁰[B₂]⁰ 
OR

Question. Write units of rate constants for zero order and for the second order reactions if the concentration is expressed in mol L^–1 and time in second.
Answer. Zero order : mol L^–1s^–1
Second order : L mol^-1s^-1

Question. Define the following terms :
(i) Pseudo first order reaction
(ii) Half-life period of a reaction (t_1/2).
Answer. (i) Pseudo first order reaction : If a reaction is not truly of the first order but under certain conditions become reaction of first order is called pseudo first order reaction. 
(ii) Half-life period of a reaction (t_1/2) : Half-life of a reaction is the time in which the concentration of a reactant is reduced to half of its original value.

Question. For a reaction :

Rate = k
(i) Write the order and molecularity of this reaction,
(ii) Write the unit of k.
Answer. (i) It is a zero order reaction, molecularity = 2 (Bimolecular reaction)
(ii) Units of k = mol L^–1 s^–1. 1 × 2 = 2

Question. State a condition under which a bimolecular reaction is kinetically first order reaction.
Answer. Let us take a biomolecular reaction :
A + B → Product
Rate = k [A] [B]
When concentration of [B] is taken in excess then
rate law will become :
Rate = k [A]
where, k = constant
The rate depends only on one of the reactant as
there is negligible change in its concentration so it is
biomolecular but is of first order.

Question. Explain the following terms :
(i) Rate constant (k)
(ii) Half-life period of reaction (t_1/2).
Answer. (i) Rate constant (k) : Rate constant is rate of the reaction when the concentration of reactants is unity. 
(ii) Half-life period of a reaction (t_1/2) : Half-life of a reaction is the time in which the concentration of a reactant is reduced to half of its original value. 

Question. For a chemical reaction R → P, variation in log [R_o]/[R] vs time plot is given below:

For this reaction :
(i) Predict the order of reaction
(ii) What is the unit of rate constant (k)?
Answer. (i) First order. 
(ii) s^-1/time^-1

Question. Write two differences between ‘order of reaction’ and ‘molecularity of reaction’.
Answer. (i) Order of a reaction is meant for elementary as well as for complex reactions but molecularity is for elementary reactions. 
(ii) Order can be zero or fraction but molecularity cannot be zero or fraction. (or any other difference)

Question. For a certain chemical reaction variation in concentration [A] vs. time (s) plot is given below:
(i) Predict the order of the given reaction?

(ii) What does the slope of the line and intercept indicate?
(iii) What is the unit of rate constant k?
Answer. (i) Zero order reaction 
(ii) Slope represents –k ; Intercept represents [R]₀
(iii) mol L^–1 s^–1 

Question. For a chemical reaction R → P, variation in ln [R] vs time (t) plot is given below:

For this reaction :
(i) Predict the order of reaction
(ii) What is the unit of rate constant (k)?
Answer. (i) First order. 
(ii) s^-1/time^-1

Long Answer Type Questions-I

Question. For the first order thermal decomposition reaction,the following data were obtained :
C₂H₅Cl(g) → C₂H₄(g) + HCl(g)
Time / sec     Total pressure / atm
0                     0.30
300                 0.50
Calculate the rate constant.
(Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)
Answer. P₀ = 0.30 atm P_t = 0.50 atm t = 300 s

Question. Following data are obtained for the reaction:
N₂O₅ → 2NO₂ + 1/2 O₂

(i) Show that it follows first order reaction.
(ii) Calculate the half-life.
(Given : log 2 = 0.3010 log 4 = 0.6021)
Answer.

k is constant when using first order equation therefore it follows first order kinetics. 
or
In equal time interval, half of the reactant gets converted into product and the rate of reaction is independent of concentration of reactant, so it is a first order reaction.
(ii) t_1/2 = 0.693/k
= 0.693/2.31 ´ 10^-3
= 300 s

Question. A first order gas phase reaction :
A₂B₂(g) → 2A(g) + 2B(g)
at the temperature 400°C has the rate constant
k = 2.0 × 10^–4 sec^–1. What percentage of A₂B₂ is decomposed on heating for 900 seconds ? [Antilog 0.0781 = 1.197] 
Answer.

Question. The rate constant for a first order reaction is 60 s^-1.How much time will it take to reduce 1 g of the reactant to 0.0625 g? 
Answer.

Question. A reaction is first order in A and second order in B
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times ?
(iii) How is the rate affected when the concentration of both A and B are doubled ?
Answer. (i) Rate = k[A][B]² 
(ii) Rate becomes 9 times 
(iii) Rate becomes 8 times

Question. The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume:
Experiment    Time (s)     Total pressure (atm)
1                   0                    0.4
2                   100                 0.7
Calculate the rate constant (k)
[Given: log 2 = 0.3010; log 4 = 0.6021]
Answer. P_A = 2P₀ – P_t 
= (2 x 0.4) – 0.7 = 0.1

Question. Why molecularity is applicable only for elementary reactions and order is applicable for elementary as well as complex reactions ?
Answer. A complex reaction occurs through several elementary reactions. Numbers of molecules involved in each elementary reaction may be different, that is, the molecularity of each step may be different. Therefore, the molecularity of overall complex reaction is meaningless. On the other hand, order of a complex reaction is experimentally determined by the slowest step in its mechanism and is therefore, applicable even in the case of complex reactions.

Question. The rate constant for the first order decomposition of H₂O₂ is given by the following equation :
log k = 14.2 – (1.0×10⁴/)T
Calculate E_a for this reaction and rate constant k if its half-life period be 200 minutes.
(Given : R = 8.314 JK^–1 mol^–1)
Answer.

Question. The following data were obtained during the first order thermal decomposition of N₂O₅(g) at a constant volume:
2N₂O₅(g) → 2N₂O₄(g) + O₂(g)

Calculate the rate constant.
Answer. 
2N₂O₅(g) → 2N₂O₄(g) + O₂(g)
At t = 0       0.5 atm        0 atm      0 atm
At time t    0.5–2x atm    2x atm    x atm
pt = pN₂O₅ + pN₂O₄ + pO₂
= (0.5–2x) + 2x + x = 0.5+x 
x = pt – 0.5
pN₂O₅ = 0.5 – 2x
= 0.5 – 2(pt – 0.5) 
= 1.5 – 2pt
At t=100 s; pt = 0.512 atm pN₂O₅
= 1.5–2 × 0.512
= 0.476 atm

k = 2.303 × 0.0216 × 10^–2
= 4.98 × 10^–4 s^–1

Question. Rate constant k for a first order reaction has been found to be 2.54 × 10^–3, sec^–1. Calculate its 3/4th life.
(log 4 = 0.6020) 
Answer. k = 2.54 × 10^–3 sec^–1

Question. A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.
(Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) 
Answer.

Question. Half-life for a first order reaction is 693 s. Calculate the time required for 90% completion of this reaction. 
Answer.

Topic-3
Concept of Collision Theory, Activation Energy and
Arrhenius Equation

Very Short Answer-Objective Type Questions 

A. Multiple choice Questions:

Question. Activation energy of a chemical reaction can be determined by
(a) determining the rate constant at standard temperature.
(b) determining the rate constants at two temperatures.
(c) determining probability of collision.
(d) using catalyst

Question. Consider the following figure and mark the correct option :

(a) Activation energy of forward reaction is E₁ + E₂ and product is less stable than reactant.
(b) Activation energy of forward reaction is E₁ + E₂ and product is more is stable than reactant.
(c) Activation energy of both forward and backward reaction is E₁ + E₂ and reactant is more stable than product.
(d) Activation energy of backward reaction is E₁ and product is more stable than reactant.

Question. Which of the following statement is not correct for the catalyst?
(a) It catalyses the forward and backward reactions to the same extent.
(b) It alters Gibbs energy (ΔG) of the reaction.
(c) It is a substance that does not change the equilibrium constant of a reaction.
(d) It provides an alternate mechanism by reducing activation energy between reactants and products.

Question. Mark the incorrect statements :
(a) Catalyst provides an alternative pathway to reaction mechanism.
(b) Catalyst raises the activation energy.
(c) Catalyst lowers the activation energy.
(d) Catalyst alters enthalpy change of the reaction.

Question. The role of a catalyst is to change ______________.
(a) Gibbs energy of reaction.
(b) enthalpy of reaction.
(c) activation energy of reaction.
(d) equilibrium constant.

B. Answer the following:

Question. In the Arrhenius equation, what does the factor e-Ea/RT corresponds to? 
Answer. e-Ea/RT corresponds to the fraction of molecules that have kinetic energy greater than E_a

Question. What is the effect of catalyst on
(i) Gibbs energy (ΔG) and
(ii) activation energy of a reaction?
Answer. (i) No effect. 
(ii) Decreases

Question. On increasing temperature, activation energy of a reaction decreases, why ? 
Answer. Because temperature and activation energy are inversely proportional to each other. 

Question. In some cases, it is found that a large number of colliding molecules have energy more than threshold energy yet the reaction is slow why ?
Answer. It is due to improper orientation. Because energy more than threshold energy and proper orientation, are the two main factors which are responsible for a reaction to occur. 

Short Answer Type Questions

Question. The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K.
Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature.
[R = 8.314 JK^–1 mol^–1, log 4 = 0.6021]
Answer. Given if rate at 293 K is R thus at 313 K rate becomes 4R

Question. With the help of a diagram, explain the physical significance of energy of activation (Ea) in chemical reactions. 
Answer. The excess energy which must be supplied to the reactants to undergo chemical reactions is called activation energy E_a. It is equal to the difference between the threshold energy E_P, needed for the reaction and the average of all the reacting molecules, E_R.
Activation energy = Threshold energy – Average
kinetic energy of the reacting molecule.

Long Answer Type Questions-I 

Question. A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K.
Calculate the activation energy of the reaction.
(Given : log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^–1 mol^–1)
Answer.

Question. The rate of most reactions becomes double when their temperature is raised from 298 K to 308 K.
Calculate their activation energy.
[Given R = 8.314 J mol^–1 K^–1]
Answer.

E_a = 52905 J mol^–1 or 52.905 kJ mol^–1

Question. Two reactions of the same order have equal pre exponential factors but their activation energies differ by 24.9 kJ mol^–1. Calculate the ratio between the rate constants of these reactions at 27°C. (Gas constant R = 8.314 J K^–1 mol^–1)
Answer.

Long Answer Type Questions-II

Question. (i) The decomposition of A into products has a value of k as 4.5 × 10³ s^–1 at 10°C and energy of activation 60 kJ mol^–1. At what temperature would k be 1.5 × 104 s^–1 ?
(ii) (a) If half life period of a first order reaction is x and 3/4^th life period of the same reaction is y, how are x and y related to each other ?
(b) In some cases it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why ?
Answer. (i) We know that

(ii) (a) y = 2x. This is because for 3/4th of the reaction to complete, time required = two half lives. 
(b) This is because of improper orientation of the colliding molecules at the time of collision.

Question. All energetically effective collisions do not result in a chemical change. Explain with the help of an example. 
Answer. Effective collision leads to the formation of products implying that during collisions molecules collide with sufficient kinetic energy called threshold energy (Threshold Energy = Activation energy + Energy possessed by reacting species) and proper orientation leads to a chemical change because it enhance the old bond breaking between reactant molecules and formation of the new ones in products. For example : Formation of methanol from bromoethane depends upon the orientation of the reactant molecules. Proper orientation of reactant molecules leads to bond formation while improper orientation does not allow bond formation and hence no products are formed. To estimate effective collisions, another factor P which is called as probability or steric factor is introduced.

Question. (i) For the reaction A→B, the rate of reaction becomes twenty seven times when the concentration of A is increased three times. What is the order of the reaction ?
(ii) The activation energy of a reaction is 75.2 kJ mol^–1 in the absence of a catalyst and it lowers to 50.14 kJmol^–1 with a catalyst. How many times will the rate of reaction grow in the presence of a catalyst if the reaction proceeds at 25°C ? 
Answer.

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

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
OR
(i) For a reaction A + B → P, the rate is given by Rate = k[A] [B]²
(a) How is the rate of reaction affected if the concentration of B is doubled ?
(b) What is the overall order of reaction if A is present in large excess ?
(ii) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reactions. (log 2 = 0.3010)
Answer. (i) For the first order reaction—

(a) When concentration of B is doubled it means concentration of B becomes 2 times.
Thus, Rate = k[A]¹ [2B]²
= k[A] [4B²]
So, the rate becomes 4 times. 
(b) Order of reaction is the no. of molecules whose concentration alters after the reaction.
If A is present in excess i.e., its concentration is unaffected.
So, rate depends only on the concentration of B.
As
k = [B]²
Thus, the reaction is of second order. 
(ii) For the 1st order reaction:

Question. (i) A first order reaction takes 100 minutes for completion of 60% of the reaction. Find the time when 90% of the reaction will be completed.
(ii) With the help of diagram explain the role of activated complex in a reaction.
Answer.(i) (a) For the reaction of first order,

(ii) It is believed that when the reactant molecules absorb energy, their bonds are loosened and new loose bonds are formed between them (A and B both). The intermediate thus formed is called an activated complex. It is unstable and immaediately dissociated to form the stable products.