31 Year NEET Previous Questions: Chemical Kinetics - 2 Free MCQ Practice

From data 1 and 3, it is clear that keeping (B) const, When [A] is doubled, rate remains unaffected. Hence rate is independent of [A]. from 1 and 4, keeping [A] constant, when [B] is doubled, rate become 8 times. Hence [rate ∝ [ B]³].

 E_a(Forward) + ΔH = E_{a(back word)}
For Exothermic reaction, ΔH = –ve and
∴  activation energy of reactant is less than the energy of activation of products.

A catalyst affects equally both forward and backward reactions, therefore it does not affect equilibrium constant of reaction.

Given initial concentration (a) = 2.00 m; Time taken (t) = 200 min and final concentration (a – x)    = 0.15 m. For a first order reation rate constant,

Further

The rate of a reaction is the speed at which the reactants are converted into products.
It depends upon the concentration of reactants. e.g for the reaction

Thus energy of activation for reverse reaction depend upon whether reaction is exothermic or endothermic If reaction is exothermic,  ΔH = +ve E_a(b) > E_a(f) If reaction is endothermic  ΔH =+ve E_a(b) < E_{a(f )}

We know that the activation energy of chemical reaction is given by formula

where k₁ is the rate constant at temperature T₁ and k₂ is the rate constant at temperature T₂ and E_a is the activation energy.  Therefore activation energy of chemical reaction is determined by evaluating rate constant at two different temperatures.

t_1/2 = 4 s T = 16

where Å = initial concentration &  A = concentration left after time t

Rate of reaction

The presence of enzyme (catalyst) increases the speed of reaction by lowering the energy barrier, i.e. a new path is followed with lower activation energy.

Here E_T is the threshold energy.
E_a and E_a1 is energy of activation of reaction in absence and presence of catalyst respectively.

from the unit of rate constant it is clear that the reaction follow first order kinetics. Hence by rate law equation,    
 r = k [N₂O₅] where r = 1.02 × 10^–4,
k = 3.4 × 10^–5 1.02 × 10^–4 = 3.4 × 10^–5 [N₂O₅]
[N₂O₅] = 3M

Rate of appearance of B is equal to rate of disappearance of A.

For reaction If it is zero order reaction r = k [A]⁰, i.e the rate remains same at any concentration of  'A'. i.e independent upon concentration of A.

k = Ae^{-E_a / Rt}

Comparing the above equation with y = mx + c

Thus A plot of log10k vs 1/T should be a straight line, with slope equal to – E_a/2.303 RT and intercept equal to log A

or  E_a = –2.303R x Slope

∵ r = k[A]ⁿ
if n = 0
r = k[A]⁰ or  r = k thus for zero order reactions rate is equal to the rate constant.

The activation energy of reverse reaction will depend upon whether the forward reaction is exothermic or endothermic.
As ΔH = E_{a (forward reaction)} – E_{a(backward reaction)}
For exothermic reaction

for endothermic reaction

A → B For a first order reaction given a = 0.8 mol,    (a – x) = 0.8 – 0.6 = 0.2

again a = 0.9,   a – x = 0.9 – 0.675 = 0.225

Hence t = 1 hour

For a first order reaction, A → products

= 3 × 10^–2

Further, 

Given [A] = 0.01 M
Rate = 2.0 × 10^–5 mol L^–1 S^–1
For a first order reaction Rate = k[A]

RateI  = k [A]^x[B]^y ... (1)

or Rate1 = 4k[A]^x[2B]^y

From (1) and (2) we get

or    y = –2.

In the given options  will not represent the reaction rate. It should not have –ve sign as it is product. since  show the rate of formation ofproduct C which will be positive.

If we write rate of reaction in terms of concentration of NH₃ and H₂,then Rate of reaction

So, 

As the slowest step is the rate determining step thus the mechanism B will be more consistent with the given information also because it involve one molecule of H₂ and one molecule of  ICl it can expressed as r = k [H₂][ICl] Which shows that the reaction is first order w.r.t. both H₂ & ICl.

For a first order reaction

when t = t_½

For a first order reaction

when t = 60 and x = 60%

Now,

 =  45.31 min.

Given, 

when k₁ and k₂ are equal at any temperature T, we have

or 2.303 ×1×T=1000           [∴ log 10= 1]

Rewriting the given data for the reaction

Actually this reaction is autocatalyzed and involves complex calculation for concentration terms.
We can look at the above results in a simple way to find the dependence of reaction rate (i.e. rate of disappearance of Br₂).
From data (1) and (2) in which concentration of CH₃COCH₃ and H⁺ remain unchanged and only the concentration of Br₂ is doubled, there is no change in rate of reaction. It means the rate of reaction is independent of concentration of Br₂.
Again from (2) and (3) in which (CH₃CO CH₃) and (Br₂) remain constant but H⁺ increases from 0.05 M to 0.10 i.e. doubled, the rate of reaction changes from 5.7×10^–5 to 1.2 × 10^–4 (or 12 × 10^–5), thus it also becomes almost doubled. It shows that rate of reaction is directly proportional to [H⁺].
From (3) and (4), the rate should have doubled due to increase in conc of [H⁺] from 0.10 M to 0.20 M but the rate has changed from 1.2× 10^–4 to 3.1×10^–4. This is due to change in concentration of CH₃ CO CH₃ from 0.30 M to 0.40 M. Thus the rate is directly proportional to [CH₃ COCH₃].
We now get rate = k [CH₃COCH₃]¹[Br₂]⁰[H⁺]1        
= k [CH₃COCH₃][H⁺].

Rate of disappearance of H₂ = rate of formation of NH₃.

= 3 ×10 ^–4 mol L^–1s ^–1

Rate of disappearance of Br^–                
= rate of appearance of Br₂