NEET-XII-Chemistry

c2022 year:2022

with Solutions - page 3
  • #40
    `` 3 O _{2}( g ) \rightleftharpoons 2 O _{3}( g ) `` for the above reaction at `` 298\, K , K _{ c } `` is found to be `` 3.0 \times 10^{-59} `` . If the concentration of `` O _{2} `` at equilibrium is `` 0.040 \,M `` then concentration of `` O _{3} `` in `` M `` is
    (A) `` 1.2 \times 10^{21} ``

    (B) `` 4.38 \times 10^{-32} ``
    (C) `` 1.9 \times 10^{-63} ``
    (D) `` 2.4 \times 10^{31} ``
    digAnsr:   B
    Ans : `` 3 O _{2}( g ) \rightleftharpoons 2 O _{3}( g ) ``
    `` K _{ c }=\frac{\left[ O _{3}\right]^{2}}{\left[ O _{2}\right]^{3}} ``
    `` 3 \times 10^{-59}=\frac{\left[ O _{3}\right]^{2}}{\left(4 \times 10^{-2}\right)^{3}} ``
    `` {\left[ O _{3}\right]^{2}=3 \times 10^{-59} \times 64 \times 10^{-6} } ``
    `` =19.2 \times 10^{-64} ``
    `` =4.38 \times 10^{-32} M ``