Now, question says : rate becomes 100 times when pH change 2 to 1 . we know, Then, when pH = 2 , 2 = - log[H⁺] ⇒[H⁺] = 10⁻² M

similarly , when pH = 1 1 = -log[H⁺] ⇒ [H⁺] = 10⁻¹ M

Now, Let rate = r₁ , when pH = 2 then, r₁ = (10⁻²)ⁿ --------------------(1)

when pH changes then, rate will be 100 times of r₁so, 100r₁ = (10⁻¹)ⁿ ------------(2)

dividing equations (2) and (1) , 1/100 = (10⁻²)ⁿ/(10⁻¹)ⁿ 1/100 = (1/10)ⁿ1/10² = (1/10)ⁿ ⇒n = 2

Hence, order of reaction is 2

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b is write jidkdndifi

before solving this you have to know that

pH=−log[H+]pH=−log[H+]

ielog[H+]=−pHlog[H+]=−pH

ie[H+]=10^−pH[H+]=10^−pH

∴whenpH=1then[H+]=10^−1=0.1

whenpH=2then[H+]=10^−2=0.01

given that whenpHchanges from2to1rate becomes100times

ie when concentration changes from0.01to0.1rate becomes100times

consider an equationdx/dt=k[H+]n

Let when concentration is0.01rate isdx/dt1=k(0.01)n…….(1)

∴when rate is0.1rate is

100dx/dt1=k(0.1)n….…(2)

from(1)and(2)

100=0.1^n/0.01^n

∴100=10^n

∴n=2

∴order=2