Class 12th

Reaction Kinetics, also known as chemical kinetics, is the study of the rate of chemical reaction and the factors affecting the reaction rate, such as temperature, concentration, and catalyst.

A ->2B + C

t = 0          a      0       0

t = 10       a – x  2x     x

t ->    $\infty$ 0       2a    a

$\therefore$ $\begin{array}{l}a+2x\propto 176\\ 3a\propto 270\\ a\propto 90\\ 2x\propto 86\\ x\propto 43\\ a-x\propto 90-43\\ a-x\propto 47\end{array}$                       

$\mathrm{\Delta }{E}_1=\frac{12400\mathrm{e}\mathrm{V}\text{?}}{1085\text{?}}=11.43\mathrm{e}\mathrm{V}$

$\mathrm{\Delta }{E}_2=\frac{12400}{304}\mathrm{e}\mathrm{V}=40.80\mathrm{e}\mathrm{V}$

$\mathrm{\Delta }{E}_1+\mathrm{\Delta }{E}_2=54.4\mathrm{e}\mathrm{V}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\Rightarrow 52.23=54.4\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

If $n_1=1\mathrm{}{n}_2=5$

S₃O₉ has no

S - S linkage

Process of active methylene group  acidic strength and the side chain should be e^- withdrawing.

$\frac{1}{20}=(1.55-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f}=(1.50-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Divide (1) by (2), $\frac{f}{20}=\frac{0.55}{0.50}$

$f=22\mathrm{c}\mathrm{m}$

(I? /2)cos²θ = (18/100)I?
⇒ cos²θ = 9/25
⇒ cosθ = 3/5 ⇒ θ = 53°

    Potential difference across R?
(R? / (R? +R? ) * V is greater than zenor voltage
⇒ i? = V_z/R? = 20/1000 A = 20 mA
Current through R? , i? = (40-V_z)/R? = 20/500 = 40 mA

Current through zener diode = i? - i? = 20 mA

Neighboring group participation by a sulphide involved.

at t = 0, I = 0; E = L (dI/dt) ⇒ dI/dt = E/L = 1.5/0.1 = 15 A/s