The rate law for a reaction is found to be : Rate =k [NO_(2)^(-) ],[ I^(-)][H^(+)]^(2) How would the rate of reaction change when(i) Concentration of H^(+) is doubled (ii) Concentration of I^(-) is halved(iii) Concentration of each of NO_(2)^(-),I^(-) and H^(+) are tripled?

The rate law for a reaction is found to be : Rate =k [NO_(2)^(-) ],[ I

1.	The rate law for a reaction is found to be : Rate =k [NO_(2)^(-) ],[ I^(-)][H^(+)]^(2) How would the rate of reaction change when(i) Concentration of H^(+) is doubled (ii) Concentration of I^(-) is halved(iii) Concentration of each of NO_(2)^(-),I^(-) and H^(+) are tripled?
Answer» Solution :Suppose intially the CONCENTRATIONS are : `[NO_(2)^(-)]=a molL^(-1),[I^(-)]=B molL^(-1)" and "[H^(+)]=c molL^(-1):." Rate "="k a b "c^(2)` (i) New `[H^(+)]=2c:." New Rate = k a b"(2" c")^(2)=4kabc^(2)=4` times (II) New `I^(-)=(b)/(2)"New Rate "=ka(b)/(2)c^(2)=(1)/(2)kabc^(2)`, i.e., rate of REACTION is halved. (III) New `[NO_(2)^(-)]=3" a",[I^(-)]=3" b",[H^(+)]=3" cNew Rate"=k(3" a")(3" b")(3" c")^(2)=81" k "abc^(2)=81` times.