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For the reaction `2H_(2)+O_(2)rarr 2H_(2)O`, the rate law expression is , `r = k[H_(2)]^(n)`. When the concentration of `H_(2)` is doubled, the rate of reaction found to be quadrupled. The value of n is |
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Answer» Correct Answer - C `r=k[H_(2)]^(n)` On doubling the concentration of `H_(2),` `4r=k[2H_(2)]^(n)` On dividing Eq. (ii) by (i) , we get, `4=2^(n)` `(2)^(2)=(2)^(n)rArrn=2` |
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