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For a reaction (d x)/(d t) = K[H^(+)]^(n...

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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The correct Answer is:
2
`:' r=K[H^(+)]^(n)`
At `pH=2, [H^(+)]=10^(-2)`
`r_(0)=K[10^(-2)]^(n) …(1)`
At `pH=1, [H^(+)]=10^(-1)`
`:. R_(1)=K[10^(-1)]^(n) …(2)`
`r_(1)/r_(0)=100=[10]^(n) implies n=2`
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