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The rate law of a chemical reaction given below: `2NO+O_(2)rarr 2NO_(2)` is given as rate `=K[NO]^(2)[O_(2)]`. How will the rate of reaction change if the volume of reaction vessel is reduced to `1//4th` of its original valur? |
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Answer» When volume is reduced to 1/3rd , then concentration will increase three times. `r_(1)=k[A][B]^(2)" ".....(i)` `r_(2)=k[3A][3B]^(2)" ".....(ii)` From eqs. (i) and (ii), `r_(1)/(r_(2))=(1)/(27)` `r_(2)=27r_(1)` `:.` Rate will increases 27 times. |
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