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For a reaction (d x)/(d t) = K[H^(+)]^(n). If pH of reaction medium changes from two to one rate becomes 100 times of value at pH = 2, The order of reaction is

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Solution :pH = 2, `r_(1)=k XX (10^(-2))^(n)` {`:. [H^(+)]=10^(-pH)`}
pH = 1, `r_(2)=k xx (10^(-1))^(n)`
Given `r_(2) = 100r_(1)`
`IMPLIES ((10^(-1))/(10^(-2)))^(n)=100`
`10^(n)=100`
`:.` n = 2


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