${∆}_{{\mathrm{n}}_{\mathrm{g}}}={\mathrm{n}}_{\mathrm{p}}-{\mathrm{n}}_{\mathrm{r} }= 1-\frac{3}{2}$

${∆}_{{\mathrm{n}}_{\mathrm{g}}}= \frac{-1}{2}$. Hence K_P = K_c(RT)^-1/2

$\frac{{\mathrm{K}}_{\mathrm{p}}}{{\mathrm{K}}_{\mathrm{c}}}=\frac{1}{(\mathrm{RT}{)}^{1/2}}=\frac{1}{\sqrt{\mathrm{RT}}}$

BF₃ has zero dipole moment

Dissociation energy of methane = 360 kJ mol^-1

Bond energy of C-H bond ⁼$\frac{360}{4}=90 \mathrm{kJ}$

Bond energy of ethane,

1(C-C) 6 (C-H) =620 kJ/mol

(C-C) + 6 $\times$ 90 = 620

(C-C) + 540 = 62

C-C = 80 kJ mo1^-1
Bond dissociation of C-C bond= 80 kJ moJ-1.

Angular momentum = $\mathrm{m\nu r}= \frac{\mathrm{nh}}{2\mathrm{\pi }}$

Greater are the intermolecular forces of attraction, higher is the critical temperature.

In unfavorable reaction have Delta G values that are positive (also called endergonic reaction). When Delta G for reaction is zero, a reaction is said to be at equilibrium. Equilibrium does not mean equal concentrations. If the Delt G is zero, there is no net change in A and B, as the system is at equilibrium.

Following the conservation of energy principle,

Kinetic energy $\left[\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2\right]$ = h($\text{v}$-${\mathrm{v}}_{\circ }$)

                    =(6.626 $\text{×}$ 10^-34) (10¹⁴s^-1 - 5 $\text{×}$ 10^{13 s-1})

                    =(6.626 x 10^-34J s) (5 x 10¹³ s^-1)             

                    =3.313 x 10^-20J.

Larger the value of pK_a smaller will be its acidity. Out of the four groups, -COOH, -NO₂ and -CN are e^- withdrawing which makes benzoic acid more acidic whereas -OCH₃ is e^- donating which reduces the acidity (makes H less easily available). pK_a value increases if OCH₃ is present at para-position of benzoic acid.

NH₄Cl solution is acidic, its pH < 7. NaNO₃ solution is neutral, its pH = 7. CH₃COOK and Na₂CO₄solutions are basic their pH > 7. But Na₂CO₃ solution is more basic, its pH > pH of CH₃COOK solution.

             ${\mathrm{N}}_2{\mathrm{O}}_4 \rightleftharpoons 2{\mathrm{NO}}_2$

Initially     1          0

At equili.  1- $\mathrm{\alpha }$      2$\mathrm{\alpha }$

N₂O₄ is 50% dissociated, so $\mathrm{\alpha }$ =$\frac{1}{2}$

${\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{p}}_{{\mathrm{NO}}_2}^2}{{\mathrm{p}}_{{\mathrm{N}}_2{\mathrm{O}}_4}}=\frac{(2\times {\displaystyle \frac{1}{2}}{)}^2}{(1-{\displaystyle \frac{1}{2}})}=2 \mathrm{atm}$