The chemical reaction $2{{\rm{O}}_3} \to 3{{\rm{O}}_2}$ proceeds as follows ${{\rm{O}}_3}\;\leftrightharpoons\;{{\rm{O}}_2} + {\rm{O}}(fast)$ ${\rm{O}} + {{\rm{O}}_3} \to 2{{\rm{O}}_2}(slow)$ The rate law expression should be

$r = k{\left[ {{{\rm{O}}_3}} \right]^2}{\left[ {{{\rm{O}}_2}} \right]^{ - 1}}$

$r = k{\left[ {{{\rm{O}}_3}} \right]^2}$

$r = k\left[ {{{\rm{O}}_3}} \right]\left[ {{{\rm{O}}_2}} \right]$

NEET UG CHEMISTRY MCQ: CHEMICAL KINETICS

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Analyzing +4 | -1

The chemical reaction

2{\mathrm{O}_3} \to 3{\mathrm{O}_2}

proceeds as follows

{\mathrm{O}_3}\;\leftrightharpoons\;{\mathrm{O}_2} + \mathrm{O}(fast)

\mathrm{O} + {\mathrm{O}_3} \to 2{\mathrm{O}_2}(slow)

The rate law expression should be

View answer and explanation

Correct option: B

r = k{\left[ {{\mathrm{O}_3}} \right]^2}{\left[ {{\mathrm{O}_2}} \right]^{ - 1}}

Option B (

r = k{\left[ {{\mathrm{O}_3}} \right]^2}{\left[ {{\mathrm{O}_2}} \right]^{ - 1}}

) is correct because the rate law is determined by the slow step and the equilibrium of the fast step.