The decomposition of hydrogen peroxide in the presence of iodide ion has been found to be first order in `H_2O_2`:
`2H_2O_2 (aq) overset(1^(-)(aq))(rarr) 2H_2O(l) + O_2 (g)`.
The rate constant has been found to be `1.01 xx10^(-2) " min"^(-1)` :
(a) Calculate the rate of reaction when `[H_2O_2] = 0.4" mol lit"^(-1)`.
(b) What concentration of `[H_2O_2]` would give a rate of `1.12 xx 10^(-2)" mol lit"^(-1)" min"^(-1)` ?

Explanation

Rate law.expression for the reaction is :
rate=`k[H_2O_2]`
rate=`1.0xx10^(-2)min^(-1)[H_2O_2]`
(a) Here `[H_2O_2]=0.4" mol L"^(-1)min^(-1)`,Rate=?
`therefore` rate=`(1.01xx10^(-2)min^(-1))xx(0.4" mol L"^(-1))`
`=4.04xx10^(-3)" mol L"^(-1)min^(-1)`
(b) Here `[H_2O_2]=0.15" mol L"^(-1)`,Rate=?
`therefore` rate=`(1.01xx10^(-2)min^(-1))xx(0.15" mol L"^(-1))`
`=1.5xx10^(-3)" mol L"^(-1)min^(-1)`
(c) Here are =`1.2xx10^(-2)mol" L"^(-1)min^(-1)`,
`[H_2O_2]=?`
`therefore[H_2O_2]=("rate")/(k)`
`1(1.12xx10^(-2)mol" L"^(-1)min^(-1))/(1.01xx10^(-2)min^(-1))`
`1.11" mol L"^(-1)`