Chemistry

This is a Long Answer Type Questions as classified in NCERT Exemplar

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}}dx\\ Put{x}^2=tforthepurposeofpartialfraction.\\ Weget\frac{t}{\left(t+a^2\right)\left(t+b^2\right)}\\ Put\frac{t}{\left(t+a^2\right)\left(t+b^2\right)}=\frac{A}{t+a^2}+\frac{B}{t+b^2}\left[whereAandBarearbitrar\mathrm{yconstants}.\right]\\ \frac{t}{\left(t+a^2\right)\left(t+b^2\right)}=\frac{A\left(t+b^2\right)+B\left(t+a^2\right)}{\left(t+a^2\right)\left(t+b^2\right)}\\ \Rightarrow t=At+A{b}^2+Bt+B{a}^2\\ Comparingtheliketerms,weget\\ A+B=1andA{b}^2+B{a}^2=0\\ \Rightarrow A=\frac{-a^2}{b^2}B\\ \therefore \frac{-a^2}{b^2}B+B=1\\ B\left(\frac{-a^2}{b^2}+1\right)=1\Rightarrow B\left(\frac{-a^2+b^2}{b^2}\right)=1\\ \Rightarrow B=\frac{b^2}{b^2-a^2}andA=\frac{-a^2}{b^2}*\frac{b^2}{b^2-a^2}=\frac{a^2}{a^2-b^2}\\ So,A=\frac{a^2}{a^2-b^2}andB=\frac{-b^2}{a^2-b^2}\\ \therefore {\displaystyle \int \frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}}dx=\frac{a^2}{a^2-b^2}{\displaystyle \int \frac{1}{x^2+a^2}dx-}\frac{b^2}{a^2-b^2}{\displaystyle \int \frac{1}{x^2+b^2}dx}\\ =\frac{a^2}{a^2-b^2}*\frac{1}{a}ta{n}^{-1}\frac{x}{a}-\frac{b^2}{a^2-b^2}.\frac{1}{b}.ta{n}^{-1}\frac{x}{b}\\ =\frac{a}{a^2-b^2}ta{n}^{-1}\frac{x}{a}-\frac{b}{a^2-b^2}ta{n}^{-1}\frac{x}{b}+C\\ Hence,I=\frac{1}{a^2-b^2}\left[ata{n}^{-1}\frac{x}{a}-bta{n}^{-1}\frac{x}{b}\right]+C.\end{array}$

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option B

If A→B then the concentration of both reactants and the products vary exponentially with time. But, in option B graph the reactant concentration decreases exponentially and the product concentration increases.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option C

Let's start with what a pseudo-first-order response is.

Although the pseudo-first-order reaction looks to be an order, it belongs to another order. It's a second-order reaction because it involves two reactants.

Let's have a look at a reaction.

CH₃Br + OH→CH₃OH + Br^- 

So, the rate law for the reaction is

Rate = k [OH] [CH₃Br]

Rate = k [OH^- ] [CH₃Br] = k (constant) [CH₃Br] = K' [CH₃Br]

Only the concentration of CH₃Br will change during the reaction, and the rate will be determined by the reaction's modifications.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Correct option C    

In the presence of a catalyst, the value of ΔG cannot be changed for any reaction.

ΔG=−RtlnQ

Where Q is the Reaction Quotient, which is determined by the product and reactant concentrations. As a result, ΔG has no connection to the catalyst. Only when the reaction is spontaneous, which must be negative, is it checked. As a result, ΔG cannot be changed.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option C    

In the presence of a catalyst, the value of ΔG cannot be changed for any reaction.

ΔG=−RtlnQ

Where Q is the Reaction Quotient, which is determined by the product and reactant concentrations. As a result, ΔG has no connection to the catalyst. Only when the reaction is spontaneous, which must be negative, is it checked. As a result, ΔG cannot be changed.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option B

Let order of A and B be x and y.

r = k [A]^x [B]^y

0.1 = k (0.3)^x (0.3)^y_______1

0.4 = 0.1 = k (0.3)^x (0.6)^y_______2

0.2 = 0.1 = k (0.6)^x (0.3)^y______3

Divide 2 by 1

$\frac{0.4}{0.1}$  = $\frac{{\left(0.6\right)}^{\mathrm{y}}}{{\left(0.3\right)}^{\mathrm{y}}}$

Divide 3 by 1

$\frac{0.2}{0.1}$  = $\frac{{\left(0.6\right)}^{\mathrm{x}}}{{\left(0.3\right)}^{\mathrm{x}}}$

Hence Rate law is

r = k [A]¹ [B]²

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option D

Note. It is not possible to perform 100% of the reaction because is half-life

Basically, it is called half-line period of the reaction. So, the time taken for 100% completion of the reaction is infinite.

This is a Fill in the blanks Type Question as classified in NCERT Exemplar

Ans: Correct option C

In the case of gases, collision theory is used to predict the speeds of chemical processes. According to this idea, the reacting chemicals must collide with enough energy and in the right direction. The molecules travel quicker and the frequency increases as the temperature rises.

This proves that all three other possibilities, A, B, and D, are correct. Option C does not correspond to the correct answer.