The second order reaction is given by −dAdt=k[A]2\frac{-dA}{dt}=k[A]^2dt−dA=k[A]2
−dA[A]2=kdt\frac{-dA}{[A]^2}=kdt[A]2−dA=kdt
integrating we get
1[A]−1[A]o=kt\frac{1}{[A]}-\frac{1}{[A]_o}=kt[A]1−[A]o1=kt
Hence graph of inverse of concentration at time t vs time t will give the straight line
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