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Chemical Kinetics

Short Answer Type

211. The rate of reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction that it does not change with temperature.

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212. The decomposition of NH₃ on platinum surface is a zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 x 10^–4 mol^–1 L s^–1 ?

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Long Answer Type

213. What is the effect of temperature on the rate constant of reaction? How can this effect of temperature on rate constant be respresented quantitatively?

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214.

For a certain chemical reaction variation in the concentration in [R] vs time(s) plot is given below:
For the reaction write/draw
(i) What is the order of the reaction?
(ii) What are the units of rate constant K?

(iii) Give the relationship between K and t_1/2 (half-life period).
(iv) What does the slope of the above line indicate?
(v) Draw the plot Vs. time(s).
Or
For a certain chemical reaction:
$A + 2 B ⇋ 2 C + D$
The experimentally obtained information is tabulated below:

The experimentally obtained information is tabulated below :

Exp.

[A]₀

[B]₀

Initial rate of reaction

0.30

0.60

0.30

0.60

0.30

0.60

0.096

0.384

0.192

0.768

For this reaction:
(i) Derive the order of reaction w.r.t. both the reactants A and B.
(ii) Write the rate law.
(iii) Calculate the value of rate constant K.
(iv) Write the expression for the rate of reaction in terms of A and C.

308 Views

215. What do you mean by zero order reaction? How the value of rate constant is determined? What is the relation between rate constant and half-life period?

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216. For the reaction A → B, deduce the integrated form of rate law. What will be the nature of the curve when concentration is plotted against time for such reaction?

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217.

Answer the following questions on the above curve for a first order reaction A → P.
(a) What is the relation between slope of this line and rate constant.
(b) (i) Calculate the rate constant of the above reaction if the slope is 2 x 10^–4 s^–1.
(ii) Derive the relationship between half life of a first order and its rate constant.

(a) Slope = – k

or k = – 2 x 10^–4 mol^–1Ls^–1.

(ii) Consider the following first order reaction
A → P

at t = 0 a 0

at t = t (a – x) x

Suppose a is the initial concentration of reactant. After time t, x gm mol lit^–1 is changed to P. According to law of mass action the rate of reaction at time t is directly proportional to the concentration of A at that instant i.e., (a – x)

Hence $\frac{- dx}{dt} α (a - x) or \frac{- dx}{dt} = x (a - x) where$

is the rate and K is the rate constant, after $\frac{dx}{dt}$ rearranging the above equation, we get

$\frac{- dx}{(a - x)} = k dt$ ...(i)
On integrating the above equation

$\int_{}^{} - \frac{dx}{(a - x)} = \int_{}^{} K dt$

We get
$- In (a - x) = kt + constant [∴ \int_{}^{} \frac{dx}{a - x} = \ln (a - x)] . . . (ii)$

When t = 0, x = 0 hence – In a = constant substituting this value of constant in eqn. (ii) we get

$- In (a - x) = kt - In a or In a - In (a - x) = kt or In \frac{a}{a - x} = kt [∵ In A - In B = In \frac{A}{B}] or k = \frac{1}{t} In \frac{a}{x - x} or k = \frac{2.303}{t} \log \frac{a}{a - x} .$