In the reaction, BrO 3 - ( aq ) + 5 Br - ( aq ) + 6 H + → 3Br 2 (l)+3H 2 O(l) The rate of appearance of bromine (Br 2 ) is related to rate of disappearance of bromide ions as following

d [ B r 2 ] d t = - 3 5 d [ B r - ] d t

d [ B r 2 ] d t = - 5 3 d [ B r ] d t

d [ B r 2 ] d t = 5 3 d [ B r - ] d t

d [ B r 2 ] d t = 3 5 d [ B r - ] d t

The plot of log k vs 1 T helps to calculate :

Energy of activation as well as the frequency factor.

Rate constant K = 1.2×10 3 mol –1 L s –1 and E a = 2.0×10 2 kJ mol –1 . When T → ∞ , A is equal to

2.4 × 10 3 kJ mol –1 L s –1

Chemistry - Chemical Kinetics - Quiz-7

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

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ΔG = 0
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ΔH = 0
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ΔH = ΔG = ΔS = 0
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ΔS = 0

In a reaction, A + B Product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as

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$R a t e = k [A] [B]^{2}$
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$R a t e = k [A]^{2} [B]^{2}$
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$R a t e = k [A] [B]$
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$R a t e = k [A]^{2} [B]$

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

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256 times
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512 times
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64 times
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128 times

The incorrect statement regarding order of reaction is:

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Order is not influenced by the stoichiometric coefficient of the reactants.
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Order of reaction is the sum of power to the concentration terms of reactants to express the rate of reaction.
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Order of reaction is always whole number.
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Order can be determined by experiments only.

During the kinetic study of the reaction, 2A + B$ \rightarrow$C + D, following results were obtained

Run	[A)/ mol L $^{- 1}$	[B)/ mol L $^{- 1}$	Initial rate of formation of D/mol L $^{- 1}$
I	0.1	0.1	6.0 $\times 10^{- 3}$
II	0.3	0.2	7.2 $\times 10^{- 2}$
III	0.3	0.4	2.88 $\times 10^{- 1}$
IV	0.4	0.1	2.40 $\times 10^{- 2}$

Based on the above data which one of the following is correct?

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rate= k[A]²[B]
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rate= k[A][B]
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rate= k[A]²[B]²
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rate= k[A][B]²

Half-life period of a first order reaction is 1386 s. The specific rate constant of the reaction is

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$5.0 \times 10^{- 3} s^{- 1}$
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$0.5 \times 10^{- 2} s^{- 1}$
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$0.5 \times 10^{- 3} s^{- 1}$
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$5.0 \times 10^{- 2} s^{- 1}$

For the reaction, N₂ + 3H₂ $\to$ 2NH₃, if $\frac{d [N H_{3}]}{d t}$ =2 x 10^-4 mol L^-1s^-1, the value of $\frac{- d [H_{2}]}{d t}$ would be :

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$3 \times 10^{- 4} m o l L^{- 1} s^{- 1}$
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$4 \times 10^{- 4} m o l L^{- 1} s^{- 1}$
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$6 \times 10^{- 4} m o l L^{- 1} s^{- 1}$
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$1 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

In the reaction,
${BrO}_{3}^{-} (aq) + 5 {Br}^{-} (aq) + 6 H^{+} \to$ 3Br₂(l)+3H₂O(l)
The rate of appearance of bromine (Br₂) is related to rate of disappearance of bromide ions as following

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$\frac{d [B r_{2}]}{d t} = - \frac{3}{5} \frac{d [B r^{-}]}{d t}$
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$\frac{d [B r_{2}]}{d t} = - \frac{5}{3} \frac{d [B r]}{d t}$
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$\frac{d [B r_{2}]}{d t} = \frac{5}{3} \frac{d [B r^{-}]}{d t}$
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$\frac{d [B r_{2}]}{d t} = \frac{3}{5} \frac{d [B r^{-}]}{d t}$

The reaction of hydrogen and iodine monochloride is given as :
H₂(g) + 2ICl(g) → 2HCl(g) + I₂(g)
This reaction is of first order with respect to H₂(g) and ICl(g), following mechanisms were proposed :
Mechanism A :
H₂(g) + 2ICl(g) → 2HCl(g) + I₂(g)
Mechanism B :
H₂(g) + ICl(g) →HCl(g) + HI(g); slow
HI(g) + ICl(g) →HCl(g) + I₂(g); fast

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B Only
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A and B both
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Neither A nor B
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A only

For the reaction ,

2A + B → 3C + D

Incorrect expression for rate of reaction is:

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$- \frac{d [C]}{3 d t}$
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$- \frac{d [B]}{d t}$
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$\frac{d [D]}{d t}$
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$- \frac{d [A]}{2 d t}$

Consider the reaction
N₂(g) + 3H₂(g) → 2NH₃(g)
The equality relationship between $\frac{{d}\left[{{NH}_{3}}\right]}{dt}$ and ${-}\frac{{d}\left[{{H}_{2}}\right]}{dt}$ is :

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$\frac{d [N H_{3}]}{d t} = - \frac{1}{3} \frac{d [H_{2}]}{d t}$
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$+ \frac{d [N H_{3}]}{d t} = - \frac{2}{3} \frac{d [H_{2}]}{d t}$
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$+ \frac{d [N H_{3}]}{d t} = - \frac{3}{2} \frac{d [H_{2}]}{d t}$
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$+ \frac{d [N H_{3}]}{d t} = - \frac{d [H_{2}]}{d t}$

If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:

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t=2.303/k
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t=0.693/k
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t=6.909/k
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t=4.606/k

For the chemical reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$ the correct option is:

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$3 \frac{d [H_{2}]}{dt} = 2 \frac{d [{NH}_{3}]}{dt}$
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$- \frac{1}{3} \frac{d [H_{2}]}{dt} = - \frac{1}{2} \frac{d [{NH}_{3}]}{dt}$
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$- \frac{d [N_{2}]}{dt} = 2 \frac{d [{NH}_{3}]}{dt}$
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$- \frac{d [N_{2}]}{dt} = \frac{1}{2} \frac{d [{NH}_{3}]}{dt}$

Assertion: The enthalpy of reaction remains constant in the presence of a catalyst.

Reason: A catalyst participating in the reaction forms a different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same.

5. The assertion is incorrect, but the reason is correct.

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Both assertion and reason are correct and the reason is the correct explanation of assertion.
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Both assertion and reason are correct, but reason does not explain the assertion
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Assertion is correct, but the reason is incorrect.
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Both assertion and reason are incorrect.

A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as -

$k_{1} = 1.26 \times 10^{- 4} s^{- 1} and k_{2} = 3.8 \times 10^{- 5} s^{- 2}$

The percentage distributions of B and C respectively are –

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80% and 20%
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76.83% and 23.17%
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90% and 10%
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60% and 40%

A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in presence of a catalyst at the same rate with same frequency factor, the temperature required is 400 K the activation energy of the reaction, if the catalyst lowers the activation energy barrier by 16 kJ/mol is:

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100 kJ/mol
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80 kJ/mol
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60 kJ/mol
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None of th above

$Consider the reaction .$

$The rate constant for two parallel reactions were found to be 10^{- 2} {dm}^{3} {mol}^{- 1} s^{- 1} and 4 x 10^{- 2} {dm}^{3} {mol}^{- 1} s^{- 1} . If the corresponding energies of activation of the parallel reaction are 100 and 120 K J / mol respecctively, the net energy of activation (E_{a}) of A is$

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100 kJ/mol
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120 kJ/mol
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116 kJ/mol
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220 kJ/mol

$A reaction takes place in various steps . The rate constant for first, second, third, and fifth step are k_{1}, k_{2}, k_{3} and k_{5} respectively . The overall rate constant is given by k = \frac{k_{2}}{k_{3}} {(\frac{k_{1}}{k 5})}^{1 / 2} If activation energy are 40, 60, 50, and 10 kJ / mol respectively, the overall energy of activation (kJ / mol) is :$

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10
0%

20
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25
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None of the above

$For reaction A \overset{}{\to} B, the rate constant k_{1} = A_{1} e^{- E_{a_{1}} / (RT)} and for the reaction X \to y, the rate constant k_{2} = A_{2} e^{- E_{a_{2}} / (RT)} . If A_{1} = 10^{8} A_{2} = 10^{10} a n d E_{a_{1}} = 600 cal / mol, E_{a_{2}} = 1800 cal / mol, then the temperature at which k_{1} = k_{2} is (given : R = 2 cal / K - mol)$

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$1600 K$
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$1200 x 4.606 K$
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$\frac{1200}{4.606} K$
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$\frac{600}{4.606} K$

$For first - order parallel reaction k_{1} and k_{2} are 4 and 2 \min^{- 1} respectively at 300 K . If the activation energies for the formation of B and C are respectively 30, 000 and 38, 314 J / mol respectively . The temperature at which B and C will be obtained in the equimolar ratio is :$

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757.48 K
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378.74 K
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600.91 K
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None of the above

The activation energies of two reactions are $E_{a_{1}}$ and $E_{a_{2}}$ with $E_{a_{1}}$ > $E_{a_{2}}$ . If the temperature of the reacting system is increased from T₁ to T₂ .The correct relation is:

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$\frac{K_{1}'}{K_{1}} = \frac{K_{2}'}{K_{2}}$
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$\frac{K_{1}'}{K_{1}} > \frac{K_{2}'}{K_{2}}$
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$\frac{K_{1}'}{K_{1}} < \frac{K_{2}'}{K_{2}}$
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$\frac{K_{1}'}{K_{1}} < 2 \frac{K_{2}'}{K_{2}}$

The units of rate constant and rate of reaction are same for :

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First order reaction
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Second order reaction
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Third order reaction
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Zero order reaction

The plot of log k vs $\frac{1}{T}$ helps to calculate :

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Energy of activation.
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Rate constant of reaction.
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Order of the reaction.
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Energy of activation as well as the frequency factor.

Rate constant K = 1.2×10³ mol^–1 L s^–1 and E_a = 2.0×10² kJ mol^–1. When T $\to \infty$ , A is equal
to

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2.0 × 10² kJ mol^–1
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1.2 × 10³ mol^–1L s^–1
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1.2 × 10³ mol L^–1 s^–1
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2.4 × 10³ kJ mol^–1L s^–1

The rate constant for a reaction 2×10^–2s^–1 at 300K and 8×10^–2s^–1 at 340K. The energy of
activation of the reaction is

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14.69 kJ mol^–1
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29.39 kJ mol^–1
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44.34 kJ mol^–1
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22.05 kJ mol^–1

The reaction 2A + B + C $\to$ D + E is found to be first order in A, second in B and zero order in C. On increasing concentration of A, B and C two times , the rate of reaction is:

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72 times
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8 times
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24 times
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36 times

Zieglar-Natta catalyst is

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$K [{PtCl}_{3} (C_{2} H_{4})]$
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${({Ph}_{3} P)}_{3} RhCl$
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${Al}_{2} {(C_{2} H_{5})}_{6} + {TiCl}_{4}$
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$Fe {(C_{5} H_{5})}_{2}$

A first order reaction has a rate constant of 2.303 $\times 10^{- 3} s^{- 1}$ . The time required for 40 g of this reactant to reduce to 10 g will be [Given that $\log_{10} 2 = 0.3010$ ]

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230.3 s
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301 s
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2000 s
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602 s

For a reaction, activation energy $E_{a} = 0$ and the rate constant at 200 K is 1.6 $X 10^{6} s^{- 1}$ . The rate constant at 400K will be [Given that gas constant, R=8.314 J $K^{- 1} m o l^{- 1}$ ]

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$3.2 x 10^{4} s^{- 1}$
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$1.6 x 10^{6} s^{- 1}$
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$1.6 x 10^{3} s^{- 1}$
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$3.2 x 10^{6} s^{- 1}$

If a reaction A + B → C is exothermic to the extent of 30 kJ mol⁻¹ and the forward reaction has an activation energy of 249 kJ mol⁻¹, the activation energy for reverse reaction in kJ mol is-

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324
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279
0%

40
0%

¹⁰⁰

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Practice Chemistry Quiz Questions and Answers

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Practice Chemistry Quiz Questions and Answers

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