Ammonium ion reacts slowly with nitrite ion: NH^+_4 (aq) + NO_2 (aq) rightarrow

Question

Ammonium ion reacts slowly with nitrite ion: NH^+_4 (aq) + NO_2 (aq) rightarrow N_2 (g) + 2H_2O (l) Rate data for this reaction, measured at a certain temperature, are as follows: (a) Determine the rate law for the reaction. Enter your result in the equation below, by putting the appropriate exponent next to each species. (Enter 0 if the rate does not depend on the concentration of that species.) Rate = k[NH^+_4] [NO_2] (b) Now use your rate law, and the rate and concentrations for any of the experiments, to calculate the rate constant for the reaction. Please also enter the right exponents for its units. (For example, enter 0 and 1 if the right units for k are s^1: enter 2 and 1 for M^2 s.) k = M s

Explanation / Answer

Ans rate, r = k[NH4]^x [NO2]^y

r1/r2 = (0.120/0.480)^x = 54/2.16 = 25

0.25^x = 25

x cannot be determined

r2/r3 = (0.280/0.140)^y = 2

y = 1

rate = k [NH4]^0 [NO2]^1

B) k= rate/ [ NO2]= 216000/0.280 = 7.71 * 10^5 s-1

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