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How would you calculate the order of the reaction 2NO_((g))+O_(2(g))rarr2NO_(2(g)) by an experiment ? (or) prove that 2NO+O_2rarr2NO_2 is a third order reaction. |
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Answer» Solution :`2NO_((g))+O_(2(g))rarr2NO_(2(g))` Series of EXPERIMENTS are conducted by keeping the CONCENTRATION of one of the reactants as constant and changing the concentration of the others.  Rate `=k[NO]^m[O_2]^n` For experiment 1, the rate law is `"Rate"_1= k [NO]^m[O_2]^n""...(1)` `19.26xx10^(-2)=k[1.3]^m[1.1]^n` For experiment 2 `"Rate"_2= k [NO]^m[O_2]^n""...(2)` `38.40xx10^(-2)=k[1.3]^m[2.2]^n` For experiment 3 `"Rate"_3= k [NO]^m[O_2]^n""...(3)` `76.8xx10^(-2)=k[1.3]^m[2.2]^n` `((2))/((1))IMPLIES(38.40xx10^(-2))/(19.26xx10^(-2))=(k[1.3]^m[2.2]^n)/(k[1.3]^m[1.1]^n)` `2=((2.2)/(1.1))^n` `2=2^nimpliesn =1 ` Therefore the reaction is FIRST order respect to `O_2` `((3))/((2))implies(76.8xx10^(-2))/(19.26xx10^(-2))=(k[2.6]^m[1.1]^n)/(k[1.3]^m[1.1]^n)` `4=((2.6)/(1.3))^n` `4 =2^m implies m = 2` Therefore the reaction is second order with respect to NO The rate law is `"Rate "_1 =k[NO]^2[O_2]^1` The overall order of the reaction = 2+1=3 |
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