Chemistry NCERT Exemplar Solutions Class 12th Chapter Fifteen

$C{H}_4+2O_2\to C{O}_2+2H_{2}O$

2 moles of water produced by 1 mole of methane

Or 36 gm of water produced by 1 mole of CH₄

$\therefore$  81 gm of water produced = $\frac{1*81}{36}$  = 2.25 mole of CH₄

Mole of CH₄ required = 225 * 10^-2

the nearest integer = 225

${\left[Co{\left(H_{2}O\right)}_6\right]}^{2+}+N{H}_{3\left(excess\right)}\to \underset{Diamagneticnature}{{\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}+6H_{2}O}$

$C{o}^{3+}\to 3d^{6}4s^0$

$\Rightarrow {t_{2g}}^{6}e{g}^0$

$\Rightarrow {t_{2g}}^{6}e{g}^0$ (form of paired electron)

              No. of electron in t_2g = 6

$K_{cat}=A{e}^{-}\frac{{\left(Ea\right)}_{cat}}{RT}and{K}_{uncat}=A{e}^{-}\frac{{\left(Ea\right)}_{uncat}}{RT}$

 $\frac{K_{cat}}{K_{uncat}}=e^{\frac{{\left(Ea\right)}_{uncat}-{\left(Ea\right)}_{cat}}{RT}}=e^{\frac{10*1000}{8.314*300}}=e^{4.009}=e^x$

$\therefore x=4$

cell reaction is ;

              $Cu\left(s\right)+S{n}_{\left(aq\right)}^{2+}\to C{u}_{\left(aq\right)}^{++}+S{n}_{\left(s\right)}$  

              $Q=\frac{\left[C{u}^{++}\right]}{\left[S{n}^{++}\right]}=\frac{0.01}{0.001}andn=2$  

              $E_{cell}=E_{cell}^o-\frac{0.059}{2}log\frac{0.01}{0.001}$  

              $=-0.48-\frac{0.059}{2}log10=-0.509$  

              $=983.33*10^{-1}KJ/mole$  

              Nearest integer = 983

$Hl?\frac{1}{2}{H}_2+\frac{1}{2}{l}_2$

t = eq    1 - a       $\frac{\alpha }{2}$         $\frac{\alpha }{2}\left[\alpha =0.4\right]$  

              $Kp=\frac{{\left(P_{H_2}\right)}^{\frac{1}{2}}*{\left(P_{l_2}\right)}^{\frac{1}{2}}}{P_{Hl}}=\frac{{\left(0.2\right)}^{\frac{1}{2}}*{\left(0.2\right)}^{\frac{1}{2}}}{0.6}$  

              = $-8.31*300*2.3*log\left(\frac{1}{3}\right)=2735J/mol$  

$\pi =i*c*R*T=\frac{2*0.83*300}{60*10^3}=0.00415bar$

= 415 Pa

Vapourisation of $H_{2}O,H_2{O}_{\left(\mathcal{l}\right)}\to H_2{O}_{\left(g\right)}$  

Mole of H₂O = $\frac{36}{18}=2mol$  

$\Delta E=\Delta H-\Delta nRT$  

$=41.1-\frac{1*8.31*373}{1000}=38kJ/mole$  

Water has only two lone pair and XeF₄ has two lone pair electron in opposite plane of the central atom.

After removal A = 4 - $\left(\frac{1}{2}*2\right)=3$  

              After removal B = 4 – 1 = 4 (only two atoms are removed)

              Final formula of the compound = A₃B₄

$C{H}_4+2O_2\to C{O}_2+2H_{2}O$

Mole $\frac{100}{16}\frac{208}{32}$

$\begin{array}{l}\downarrow \downarrow \\ 6.256.5(Here{O}_{2}isthelimitingreagent)\end{array}$  

Hence mole of CO₂ formed = $\frac{6.5}{2}$  

And wt of CO₂ in gm = $\frac{6.5}{2}*44=143g$