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When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is

zero

first

second

more than zero but less than first

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Explanation

(b) For a zero order reaction t_1/2 is directly proportional to the initial concentration of the reactant [R]₀

t_1/2 $\propto$ [R]₀

For a first order reaction

k= 2.303/t. log[R]₀/[R] at t_1/2, [R] =[R]₀/2

So, the above equation becomes

K=2.303/t_1/2 .log[R]₀/([R]₀/2)

t_1/2 = 2.303/K = log2 = 2.303/K x .3010

t_1/2 = .693/K

i.e, half life period is independent of initial concentration of a reactant

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The rate constant of the reaction A $\to$ B is 0.6 x 10^-3 molar per second. If the concentration of A is 5 M then concentration of B after 20 min is

1.08 M

3.60 M

0.36 M

0.72 M

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Explanation

Key Concept: For a zero order reaction unit of rate constant is molar per second. Hence, we can easily calculate concentration of B after 20 min by the following formula,

x = Kt

X = Kt = 0.6 x 10^-3 x 20 x 60= 0.72M

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What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R=8.314 J mol^-1 K^-1)

342 kJ mol^-1

269 kJ mol^-1

34.7 kJ mol^-1

15.1 kJ mol^-1

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Explanation

(c) Given, initial temperature,
T₁=20+273=293K
Final temperature
T₂=35+273=308K

R=8.314 J mol^-1 K^-1

Since, rate becomes double on raising temperature,
∴ r₂=2r₁ or r₂/r₁=2

As rate constant, k∝r
∴ k₂/k₁=2

From Arrnhenius equation, we know that

log k₂/k₁= -E_a_{______} [T₁-T₂/T₁T₂]                 2.303R

log 2= -E_a_{______}[293-308/293x308]           2.303x8.314

0.3010= -E_a_{______}[-15/293x308]           2.303x8.314

∴ E_a=0.3010x2.303x8.314x293x308
15

=34673.48 J mol^-1=34.7 kJ mol^-1

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A reaction having equal energies of activation for forward and reverse reactions has

$∆$ S = 0

$∆$ G = 0

$∆$ H = 0

$∆$ H = $∆$ G = $∆$ S = 0

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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

256 times

512 times

64 times

128 times

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Explanation

(b) For 10° rise in temperature, n=1 so rate = 2ⁿ = 2¹ = 2

When temperature is increased from 10°C to 100°C, change in temperature = 100-10 = 90°C, i.e., n= 9
So, rate = 2⁹ 512 times

Alternate method With every 10° rise in temperature, rate becomes double, so
$\frac{r'}{r} = 2^{(\frac{100 - 10}{10})}$ = 2⁹ = 512 times.

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Which one of the following statements for the order of a reaction is incorrect?

Order is not influenced by stiochiometric coefficient of the reactants

Order of reaction is sum of power to the concentration terms of reactants to express the rate of reaction

Order of reaction is always whole number

Order can be determined only experimentally

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Explanation

Order of reaction may be zero, whole number or fraction number.

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For the reaction,

$N_{2} O_{5} (g) \to 2 N O_{2} (g) + \frac{1}{2} O_{2} (g)$

the value of rate of disappearance of $N_{2} O_{5}$ is given as $6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}$ . The rate of formation of $N O_{2} a n d O_{2}$ is given respectively as

$6.25 \times 10^{- 3} mol L^{- 1} s^{- 1} and 6.25 \times 10^{- 3} mol L^{- 1} s^{- 1}$

$1.25 \times 10^{- 2} mol L^{- 1} s^{- 1} and 3.125 \times 10^{- 3} mol L^{- 1} s^{- 1}$

$6.25 \times 10^{- 3} mol L^{- 1} s^{- 1} and 3.125 \times 10^{- 3} mol L^{- 1} s^{- 1}$

$1.25 \times 10^{- 2} mol L^{- 1} s^{- 1} and 6.25 \times 10^{- 3} mol L^{- 1} s^{- 1}$

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Explanation

Key Idea Rate of disappearance of reactant=rate of appearance of product

$or - \frac{1}{stoichiometric coefficient of reactant} \frac{d [reactant]}{dt} = + \frac{1}{stoichiometric coefficient of product} \frac{d [product]}{dt} For the reaction, N_{2} O_{2} (g) \to 2 {NO}_{2} (g) + \frac{1}{2} O_{2} (g) \frac{- d [N_{2} O_{5}]}{dt} = + \frac{1}{2} \frac{d [{NO}_{2}]}{dt} = + \frac{2 d [O_{2}]}{dt} ∴ \frac{d [{NO}_{2}]}{dt} = - 2 \frac{d [N_{2} O_{5}]}{dt} = 2 \times 6.25 \times 10^{- 3} mol L^{- 1} s^{- 1} = 12.5 \times 10^{- 3} mol L^{- 1} s^{- 1} = 1.25 \times 10^{- 2} mol L^{- 1} s^{- 1} \frac{d [O_{2}]}{dt} = - \frac{d [N_{2} O_{5}]}{dt} \times \frac{1}{2} = \frac{6.25 \times 10^{- 3} mol L^{- 1} s^{- 1}}{2} = 3.125 \times 10^{- 3} mol L^{- 1} s^{- 1}$

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For an endothermic reaction, energy of activation is E_a and enthalpy of reaction is $∆$ H (both of these in kJ/mol). Minimum value of E_a will be

less than $∆$ H

equal to $∆$ H

more than $∆$ H

equal to zero

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Explanation

For an endothermic reaction, the enthalpy change (Î”H) is positive. The activation energy (Ea) is always greater than or equal to Î”H. The minimum value of Ea is when it is equal to Î”H. Therefore, the minimum value of Ea will be more than Î”H.

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During the kinetic Study of the reaction, 2A+B→C+D, following results were obtained

Run [A]/mol L^-1 [B]/mol L^-1     Initial rate of formation of D/mol L^-1min^-1
I          0.1                 0.1                6.0x10^-3II         0.3                 0.2               7.2x10^-2
III        0.3                 0.4                2.88x10^-1
IV        0.4                0.1                2.40x10^-2Based on the above data which one of the following is correct ?

rate=k[A]²[B]

rate=k[A][B]

rate=k[A]²[B]²

rate=k[A][B]²

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Explanation

Let the order of reaction with respect to A is x and with respect to B is y. Thus, rate= k[A]^x[B]^y
For the given cases,
(I) rate= k(0.1)^x(0.1)^y=6.0x10^-3
(II) rate= k(0.3)^x(0.2)^y=7.2x10^-2
(III) rate= k(0.3)^x(0.40)^y=2.88x10^-1(IV) rate= k(0.4)^x(0.1)^y=2.40x10^-2
On dividing eq. (II) by (III), we get
(0.3/0.3)^x (0.2/0.4)^y = 7.2X10^-2/2.88X10^-1or (1/2)^y=1/4
or (1/2)^y=(1/2)²
y=2

Thus, rate law is, rate= k[A]¹[B]² or =k[A][B]²

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For the reaction A+B $\to$ products, it is observed that

On doubling the initial concentration of A only, the rate of reaction is also doubled and

On doubling the initial concentrations of both A and B, there is a change by a factor of 8 in that rate of the reaction

The rate of this reaction is, given by

(a) rate = k[A]²[B]

(b) rate = k[A][B]²

rate = k[A]²[B]²

rate = k[A][B]

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Explanation

For the reaction,

A+B $\to$ Products

On doubling the initial concentration of A only, the rate of reaction is also doubled, therefore

Rate $\propto$ [A]¹ ....(i)

Let initially rate law is

Rate = k[A][B]^y ...(ii)

If concentration of A and B both are doubled, the rate gets changed by a factor of 8.

8 x Rate = k[2A][2B]^y ...(iii) [._.. Rate $\propto$

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