Ammonium ion (NH4+) reacts with nitrite ion (NO2-) to yield nitrogen gas and liq

Question

Ammonium ion (NH4+) reacts with nitrite ion (NO2-) to yield nitrogen gas and liquid water. The following initial rates of reaction have been measured for the given reactant concentrations. Expt. # [NH4+] [NO2-] Initial rate (M/hr) 1 0.010 0.010 0.020 2 0.020 0.010 0.040 3 0.010 0.020 0.080 Which of the following is the rate law (rate equation) for this reaction? a. rate = k [NH4+]2 [NO2-? b. rate = k [NH4+ ] [NO2-] c. rate = k [NH4+] [NO-2]2 d. rate = k [NH4+]-1[N02-] e. rate = k [NH4+] [NO2-]-1 Please explain how to solve!!

Explanation / Answer

The Best way to solve it is: Consider rate (R) = [NH4+] ^a [NO2-]^b 0.02 = (0.01)^a (0.01)^b ..........................(1) in next case 0.04 = ( 0.02)^a (0.01)^b ..........................(2) So , Divide eqn.2 by eqn.1 We Get, 2 = (0.02/0.01)^a = (2)^a Hence, a=1 Also, we know 0.08 = (0.01)^a (0.02)^b ........................(3) Divide Eqn.3 by eqn.1 We get, 0.08/0.02 = (0.02/0.01)^b = 2^b = 4 So, b = 2 Hence, Rate of reaction (R) = [NH4+] [NO2-]^2 Thus, the reaction is of first order with respect to Ammonium ion and of second order with respect to nitrite ion.

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