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$\left.3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_{3} \mathrm{(~g}\right)$

for the above reaction $298\mathrm{K},\mathrm{K_{c}}$ is found to be $3.0 \times 10^{-59}$ . If the concentration of $\mathrm{O}_{2}$ at equilibrium is $0.040 \mathrm{M}$ then concentration of $\mathrm{O}_{3} \, \mathrm{in} \: \mathrm{M}\: \mathrm{is}$

Option: 1
$1.9 \times 10^{-63}$

Option: 2
$2.4 \times 10^{31}$

Option: 3
$1.2 \times 10^{21}$

Option: 4
$4.38 \times 10^{-32}$

Answers (1)

best_answer

$\mathrm{ 3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{g})} \\$

$\mathrm{K_{c}=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]^{3}}} \\$

$\mathrm{\Rightarrow 3 \times 10^{-59}=\frac{\left[\mathrm{O}_{3}\right]^{2}}{(0.04)^{3}} }\\$

$\mathrm{\Rightarrow {\left[\mathrm{O}_{3}\right]^{2}=3 \times 10^{-59} \times(0.04)^{3} }} \\$

$\mathrm{\Rightarrow {\left[\mathrm{O}_{3}\right]^{2}=1.92 \times 10^{-63} }} \\$

$\mathrm{\Rightarrow {\left[\mathrm{O}_{3}\right]=4.38 \times 10^{-32} }}$

hence correct option is 4

Posted by

shivangi.bhatnagar

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