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Following is the graph between (a – x)–1 and time t for a second order reaction,
Hence, rate at the start of the reaction is (in mol L–1 min–1 ).
    Correct answer is '0.125'. Can you explain this answer?
    Most Upvoted Answer
    Following is the graph between (a – x)–1 and time t for a ...
    A= .5 =OA
    slope(tan theta)=.5
    so,
    rate= K[A]^2
    =.5×(.5)^2
    =.5×.25
    =.125 mol L-1 min-1 .
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    Following is the graph between (a – x)–1 and time t for a second order reaction,Hence, rate at the start of the reaction is (in mol L–1 min–1 ).Correct answer is '0.125'. Can you explain this answer?
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