Chemical Kinetics MCQs for NEET — Chemistry Questions with Answers

(A) $t_{1/2}\propto \frac{1}{{\left(a\right)}^{n-1}}$ where a is initial conc. or presuure or volume

$\frac{{\left(t_{1/2}\right)}_1}{{\left(t_{1/2}\right)}_2}={\left(\frac{a_2}{a_1}\right)}^{n-1}$

from first & second data

$\left(\frac{136}{112.5}\right)={\left(\frac{300}{250}\right)}^{n-1};\left(1.2\right)={\left(1.2\right)}^{n-1}\Rightarrow n-1=1$ 

n=2, Hence reaction is of second order.

(B) For a first order reaction K has the dimensions of reciprocal time viz., $se{c}^{-1}i.e.,$a frequency.

(A) For a second order reaction, $\frac{dx}{dt}=K\left(a-x\right)\left(b-x\right)$

where ‘a’ and ‘b’ are the initial concentrations of the reactants. On integration we get,

$$\begin{gathered} K=\frac{2.303}{t(a-b)}\log \frac{b\left(a-x\right)}{a\left(b-x\right)} \\ \therefore \frac{Kt\left(a-b\right)}{2.303}=\log \left[\frac{\left(a-x\right)}{\left(b-x\right)}\right]-\log \left[\frac{a}{b}\right] \\ \therefore A plot of \log \left[\frac{\left(a-x\right)}{\left(b-x\right)}\right] vs time, t would give a straight \end{gathered}$$

line with a positive slope= $\frac{K\left(a-b\right)}{2.303}$

(A) $$\begin{gathered} K=\frac{0.6932}{120} ....\left(1\right) \\ K=\frac{2.303}{t}\log \frac{a}{0.10a}=\frac{2.303}{t}\log 10 ......\left(2\right) \\ Equating \left(1\right) and \left(2\right)\frac{0.6932}{120}=\frac{2.303}{t} \\ t=399 minutes \end{gathered}$$

(C) $$\begin{gathered} \frac{d\mathcal{l}nK}{dT}=\frac{E_a}{R{T}^2}\therefore \frac{dK}{K}=\frac{E_a}{R{T}^2}dT \\ \frac{dK}{K}=\frac{11.11}{100}; dT=1°C; T=27°C=300 K \\ \therefore 0.1111=\frac{E_a}{1.987\times 300\times 300} \\ \therefore E_a=0.1111\times 1.987\times {\left[300\right]}^2 \\ At 127°C \\ \frac{dK}{K}=\frac{E_a}{1.987\times {\left(400\right)}^2}\frac{0.1111\times 1.987\times {\left(300\right)}^2}{1.987\times {\left(400\right)}^2} \\ =0.1111\times {\left(\frac{3}{4}\right)}^2=0.06249 \\ \therefore \%change=6.25 \end{gathered}$$

(B) $t_{1/2}=5730 y$

If a is the initial ${}^{14}C$ activity in a living tree, then

Activity in the dead wood,$a_t = \frac{80}{100} \times a$

The nuclear / radioactive decays follow the first order

kinetics. So, $t=\frac{2.303}{0.693/t_{1/2}}\log \frac{a}{a_t}and k=\frac{0.693}{t_{1/2}}$

Therefore, $t=\frac{2.303}{0.693/t_{1/2}}\log \frac{a}{{\displaystyle \frac{80}{100}a}}$

                 $=\frac{2.303\times 5730y}{0.693}\log \frac{100}{80}$

$So, t=1845.4 y$

(A) Amount of A left in $n_1$ halves $= {\left(\frac{1}{2}\right)}^{n1} \left[A_0\right]$

Amount of B left in $n^2$ halves $= {\left(\frac{1}{2}\right)}^{n2} \left[B_0\right]$

At the end, according to the question

$$\begin{gathered} \frac{\left[A^0\right]}{2^{n1}}=\frac{{\displaystyle \left[B^0\right]}}{{\displaystyle 2^{n2}}}\Rightarrow \frac{4}{2^{n1}}=\frac{{\displaystyle 1}}{{\displaystyle 2^{n2}}},\left[\left[A_0\right]=4\left[B_0\right]\right] \\ \therefore \frac{2^{n1}}{2^{n2}}=4\Rightarrow 2^{n1-n2}={\left(2\right)}^2\therefore n_1-n_2=2 \\ \therefore n_2=\left(n_1-2\right) .....\left(1\right) \\ Also t=n_1\times t_{1/2\left(A\right)}; t=n_2\times t_{1/2\left(B\right)} \end{gathered}$$

(Let concentration of both become equal after time t)

$\therefore \frac{n_1\times t_{1/2\left(A\right)}}{n_2\times t_{1/2\left(B\right)}}=1\Rightarrow \frac{n_1\times 5}{n_2\times 15}=1\Rightarrow \frac{n_1}{n_1}=3 .....\left(2\right)$

For equation (1) and (2)

$$\begin{gathered} n_1=3, n_2=1 \\ t=3\times 5=15 minutes \end{gathered}$$

(A). A lump of coal burns at a moderate rate in air while coal dust burns explosively because coal dust contains very fine particles of carbon. So, for a certain mass, the surface area a coal dust is very large as compared to that of the lump coal. The increases surface area permits a close contact with the oxidiser oxygen gas (in the air) and therefore coal dust burns at a very high rate leading to an explosion. In the case of lump coal, the surface available for reaction is small. So, it burns slowly at moderate rate in air.

(C). Liquid bromine reacts slowly as compared to bromine vapour because in liquid bromine, the bromine molecules are held together by a force which is much stronger than the force existing between the two molecules of bromine in the vapour phase. So, additional energy is required to make the bromine molecules free.

(B). The reaction,$N_{2 }\left(10atm\right) 3H_2 \left(10atm\right)\to 2N{H}_3 \left(g\right) is faster because$ of higher partial pressures of the reactants.