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    Chemistry-
    General
    Easy

    Question

    The chemical reaction 2 O subscript 3 end subscript ⟶ 3 O subscript 2 end subscript proceeds as follows
    O subscript 3 end subscript rightwards harpoon over leftwards harpoon O subscript 2 end subscript plus O (fast)
    O plus O subscript 3 end subscript ⟶ 2 O subscript 2 end subscript (slow) The rate law expression for 2 O subscript 3 end subscript ⟶ 3 O subscript 2 end subscript reaction should be [K is rate constant]

    1. Rate = K[O3]2    
    2. Rate = K[O3]2[O2]–1    
    3. Rate = K[O3][O2]    
    4. None of these    

    The correct answer is: Rate = K[O3]2[O2]–1

      The slow r.d.s gives rate = K[O][O3]
      and Kc = fraction numerator left square bracket O right square bracket left square bracket O subscript 2 end subscript right square bracket over denominator left square bracket O subscript 3 end subscript right square bracket end fraction
      or [O] = fraction numerator K subscript c end subscript left square bracket O subscript 3 end subscript right square bracket over denominator left square bracket O subscript 2 end subscript right square bracket end fraction
      from this Rate = fraction numerator K times K subscript c end subscript left square bracket O subscript 3 end subscript right square bracket left square bracket O subscript 3 end subscript right square bracket over denominator left square bracket O subscript 2 end subscript right square bracket end fraction equals fraction numerator K to the power of ´ end exponent left square bracket O subscript 3 end subscript right square bracket to the power of 2 end exponent over denominator left square bracket O subscript 2 end subscript right square bracket end fraction = K to the power of ´ end exponent left square bracket O subscript 3 end subscript right square bracket to the power of 2 end exponent left square bracket O subscript 2 end subscript right square bracket to the power of negative 1 end exponent [K¢ is overall rate constant]

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