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QUESTION: 1

The rate of a reaction is expressed in different ways as follows :

=

The reaction is :

Solution:

B + 3D = 2C + 4A

QUESTION: 2

The rate constant for the forward reaction A(g) 2B(g) is 1.5 × 10^{-3} s^{-1} at 100 K. If 10^{-5 }moles of A and 100 moles of B are present in a 10 litre vessel at equilibrium then rate constant for the backward reaction at this temperature is

Solution:

K_{f} = = 1.5 × 10^{–3} s^{–1}

A_{t} eq.^{m}, R_{f} = R_{b}

K_{f} [A] = K_{b}[B]^{2}

On solving : K_{b} = 1.5 × 10^{-11}

QUESTION: 3

Reaction A + B → C + D follow's following rate law : rate = k[A]^{+1/2}[B]^{1/2}. Starting with initial conc. of 1 M of A and B each, what is the time taken for concentration of A of become 0.25 M.

Given : k = 2.303 × 10^{-3} sec^{-1}.

Solution:

Rate= K[A]^{1/2}[B]^{1/2}

Overall order of reaction is = 1/2+1/2=1

∴ It is a 1^{st} order reaction.

For concentration of A to become 0.5M:-t^{1/2}=0.693/k

It will require 2 half times for the concentration of A to become .0.25M.

∴ Time taken= 2×t_{1/2 }sec

=2×300

=600sec

QUESTION: 4

Consider the following first order competing reactions :

X A+ B and y C+D

if 50% of the reaction of X was completed when 96% of the reaction of Y was completed, the ratio of their rate constants is

Solution:

QUESTION: 5

Units of rate constant for first and zero order reactions in terms of molarity (M) are respectively.

Solution:

sec^{–1}, Msec^{–1}

QUESTION: 6

The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the reaction increases by about

Solution:

For every 10°C rise of temperature, the rate is doubled. Thus, the temperature coefficient of the reaction = 2

When temperature is increased by 50°, rate becomes

=2(50/10) = 25 times = 32 times

QUESTION: 7

For a reaction pA + qB → products, the rate law expression is r = k[A]^{l}[B]^{m}, then :

Solution:

pA + qB → Products r = k[A]^{l}[B]^{m}

(p + q) may or may not be equal to (l + m) (order may or may not be equal to stoichiometric coefficient for a complex reaction.

QUESTION: 8

In the reaction : A + 2B → 3C + D, which of the following expression does not describe changes in the conc. of various species as a function of time :

Solution:

QUESTION: 9

A first order reaction is 87.5% complete in an hour. The rate constant of the reaction is

Solution:

The correct answer is Option A.

KT = ln (a_{0} / a_{t})

⇒ K*60 = ln (a_{0} / a_{t})

87.5 % completed in 1 hour.

Therefore, a_{t} = a_{0} - 0.875 a_{0}

a_{t} = 0.125 a_{0}

K * 60 = ln (a_{0} / 0.125 a_{0})

⇒ K * 60 = ln 8

⇒ K = 2.079 / 60

⇒ K = 0.0346 min^{-1}

QUESTION: 10

Half-life period of a second order reaction is

Solution:

The correct answer is Option B.

Relation between half-life period of a reaction and initial concentration is as follows:

For first order reaction (Half life α a^{0} )

For second order reaction (Half life ∝

1/a)

For third order reaction (Half life ∝ 1a^{2})

QUESTION: 11

A first order reaction is 50% completed in 20 minutes at 27°C and in 5 min at 47°C. The energy of activation of the reaction is

Solution:

t_{1/2} = 20 min at 300 K

t’_{1/2} = 5 min at 320 K

On solving, Ea = 55303.12 J

= 55.3 KJ

QUESTION: 12

For the first order reaction A → B + C, carried out at 27°C if 3.8 × 10^{-16}% of the reactant molecules exists in the activated state, the E_{a} (activation energy) of the reaction is

Solution:

A → B + C

Ea = 100 KJ/mol

QUESTION: 13

The rate constant, the activation energy and the Arrhenius parameter (A) of a chemical reaction at 25°C are 3.0 × 10^{-4} s^{-1}, 104.4 kJ mol^{-1} and 6.0 × 10^{-4}s^{-1}respectively. The value of the rate constant at T → ∞ is

Solution:

At temperature = ∞

Rate constant = Arrhenious constant.

QUESTION: 14

The following mechanism has been proposed for the exothermic catalyzed complex reaction.

If k_{1} is much smaller than k_{2}. The most suitable qualitative plot of potential energy (P.E.) versus reaction coordinate for the above reaction.

Solution:

Since

K_{1} <<< K_{2} = most Imp. peack will be higher

QUESTION: 15

The activation energy of a reaction at a given temperature is found to be 2.303 RT J mol^{–1}. The ratio of rate constant to the Arrhenius factor is

Solution:

The correct answer is option A

Arrhenius equation is,

rate constant, k=Ae^{−Ea/RT}

k=Ae^{−2.303RT/RT} ⇒ k/A = e^{-2.303}

On solving, we get

k/A = 10^{-1 }or 0.1

QUESTION: 16

Consider + heat, If activation energy for forward reaction is 100 kJ/mole then activation energy for backward reaction and heat of reaction is :

Solution:

E_{af} = 100 KJ/mol

E_{ab} = ? heat of reaction = ?

Only one option because

E_{b} = E_{a} + ΔH.O. reaction

QUESTION: 17

In a reaction, the thershold energy is equal to :

Solution:

In a reaction, the threshold energy is equal to : Activation energy + Normal energy of reactants.

QUESTION: 18

The first order rate constant k is related to temperature as log k = 15.0 - (10^{6}/T). Which of the following pair of value is correct ?

Solution:

log A = 15

⇒ A = 10^{15}

E_{a} = 1.9 × 10^{4} KJ

QUESTION: 19

When a graph between log K and 1/T is drawn a straight line is obtained. The temperature at which line cuts y-axis and x-axis.

Solution:

When line cuts y axis

When it cut X -axis

log K = 0

QUESTION: 20

The rate constant, the activation energy and the frequency factor of a chemical reaction at 25°C are 3.0 × 10^{-2} s^{-1}, 104.4 KJ mol^{-1} and 6.0 × 10^{14} s^{-1} respectively. The value of the rate constant as T → ∞ is :

Solution:

K = 3 × 10^{–2} s^{–1}

Ea = 104.4 KJ/mol

A = 6 × 10^{14}

Value of rate constant at T = ∞ will be equal to frequency factor i.e. A = 6 × 10^{14} s^{-1}

QUESTION: 21

The rate data for the net reaction at 25°C for the reaction X + 2Y → 3Z are given below :

[X_{0}] [Y_{0}] Time required for [Z] to increase by 0.005 mol per litre.

0.01 0.01 72 sec

0.02 0.005 36 sec

0.02 0.01 18 sec

The intial rate (as given by Z) is :

Solution:

Since,

Rate becomes four times by doubling the concentration of A

⇒ Order w.r.t A is 2

Also,

by doubling the concentration of B, the rate becomes double

⇒ Order w.r.t. B is 1.

QUESTION: 22

The rate of production of NH_{3} in N_{2} + 3H_{2} → 2NH_{3} is 3.4 kg min^{-1}. The rate of consumption of H_{2} is :

Solution:

= 5.1 Kgmin^{-1}

QUESTION: 23

For a given reaction of first order it takes 20 min. for the conc. to drop from 1.0 M to 0.60 M. The time required for the conc. to drop from 0.60 M to 0.36 M will be :

Solution:

Also,

Since, Rate constant will be equal

On solving, t = 20 min

QUESTION: 24

For a first order reaction, the concentration of reactant :

Solution:

For a first order reaction, the concentration of the reactant varies exponentially with time (A = A_{0}e^{–kt})

QUESTION: 25

Radioactivity of a sample (z = 22) decreases 90% after 10 years. What will be the half-life of the sample?

Solution:

Reaction is or zero order hence, Option C will be correct.

QUESTION: 26

Mathematical representation for t_{1/4} life for first order reaction is over is given by :

Solution:

t_{1/4} = time taken for decomposition.

QUESTION: 27

For a reaction A → Products, the conc. of reactant C_{0}, aC_{0}, a^{2}C_{0}, a^{3}C_{0}............ after time interval 0, t, 2t ............ where 'a' is constant. Then :

Solution:

If we calculate K, it comes constant every time i.e. reaction is of first order.

QUESTION: 28

Half-life is independent of conc. of A. After 10 minutes volume N_{2} gas is 10 L and after complete reaction 50 L. Hence rate constant in min^{-1} :

Solution:

Half life is independent of concentration

⇒ reaction is of first order

QUESTION: 29

In a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become

Solution:

For rise in temperature, n=1

Therefore rate = 2^{n}

= 2^{1}

= 2

When temperature is increased from 10^{o} C to 100^{o}C

change in temperature = 100 − 10 = 90^{o}C

therefore n=9 therefore, rate = 2^{9}

= 512 times.

QUESTION: 30

In acidic medium the rate of reaction between (BrO_{3})^{-} & Br^{-} ions is given by the expression, –[d(BrO_{3}^{-}) /dt] = K[BrO_{3}^{-}][Br^{-}][H^{ +}]^{2 }It means :

Solution:

D option is correct.

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