Question

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Given the following experimental data, find the rate law and the rate constant for the reaction: NO (g) + N0_2 (g) + 0_2 (g) rightarrow N_2O_5 (g)

Answer 1

The given reaction is NO(g) + NO2(g) + O2(g) ----> N2O5(g)

Let the rate law be r = k[NO]^x[NO₂]^y[O₂]^z     ----(1)

Where

x = order of the reaction with respect to NO

y = order of the reaction with respect to NO₂

z = order of the reaction with respect to O₂

From the table of content ,

From (1) & (2) , On doubling the concentration of NO(2x0.10= 0.20M) kept the concentration of the NO₂ & O₂ constant,the rate of the reaction becomes doubles(2x2.1x10^-2 = 4.2x10^-2 ) so the value of x = 1

From (1) & (4) , On doubling the concentration of O₂(2x0.10= 0.20M) kept the concentration of the NO₂ & NO constant,the rate of the reaction becomes the same (1x2.1x10^-2 = 2.1x10^-2 ) so the value of z = 0

From (2) & (3) , On tripling the concentration of NO₂ (3x0.10= 0.30M) & doubling the concentration of O2(2x0.1=0.20M)( it has no effect on rate from above) kept the concentration of the NO constant,the rate of the reaction becomes triples(3x4.2x10^-2 = 1.26x10^-1 ) so the value of y = 1

Therefore the rate law be r = k[NO]¹[NO₂]¹[O₂]⁰

   So the rate law is r = k[NO][NO₂]     ----(2)

Plug the first values in the given tabular form we get

2.1x10^-2 Ms^-1 = k(0.10M ) x ( 0.10M)

      k = 2.1M^-1 s^-1

Therefore the rate constant is 2.1M^-1 s^-1