Class 11th

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

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

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

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

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

Mole of H₂O = $\frac{36}{18}=2mol$  

$\Delta E=\Delta H-\Delta nRT$  

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

ΔG = -TΔS_universe

N? , I? , NO? are linear.

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

Ease of liquefaction depends on a/b value.

(a) Li → used in bearings for motor engine.
(b) Na → used as coolant in fast breader nuclear reactor.
(c) K → used as absorbent of CO?
(d) Cs → used in photoelectric call.

Only in n = 2 → n = 1 (one single type of photon is emitted)

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

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 = $\frac{6.5}{2}*44=143g$  

Meq of K? Cr? O? is greater than all the other given species.