QJEEM21Ch6177: JEE Main 2021 Chemistry - Class 11 JEEM21Ch6177 Solution: A homogeneous ideal gaseous reaction is carried out in a 25…

[p = Total pressure at equilibrium = 1.9 atm]

Now, at equilibrium pV = (1 + 2x)RT

$\Rightarrow 1 + 2x = {{pV} \over {RT}} = {{1.9 \times 25} \over {0.082 \times 300}} = 1.93$

[V = 25 L, R = 0.082 L atm mol ^{$-$ 1} K ^{$-$ 1} T = 300 K]

$\Rightarrow x = {{1.93 - 1} \over 2} = 0.465$

$\Rightarrow {K_p} = {{{p_A} \times p_B^2} \over {{p_{A{B_2}}}}} \Rightarrow {{\left( {{x \over {1 + 2x}}p} \right) \times {{\left( {{{2x} \over {1 + 2x}}p} \right)}^2}} \over {\left( {{{1 - x} \over {1 + 2x}}p} \right)}}$

$= {{4{x^3} \times {p^3}} \over {{{(1 + 2x)}^3}}} \times {{(1 + 2x)} \over {(1 - x) \times p}} = {{4{x^3} \times {p^2}} \over {{{(1 + 2x)}^2} \times (1 - x)}}$

$= {{4 \times {{(0.465)}^3} \times {{(1.9)}^2}} \over {{{(1 + 2 \times 0.465)}^2} \times (1 - 0.465)}} = 0.7285$ atm

$= 72.85 \times {10^{ - 2}}$ atm $\simeq 73 \times {10^{ - 2}} = x \times {10^{ - 2}}$

$\therefore$ $x = 73$