The following initial rate data are for the reaction of ammonium ion with nitrit

Question

The following initial rate data are for the reaction of ammonium ion with nitrite ion in aqueous solution:

NH4+ + NO2- --> N2 + 2 H2O

Complete the rate law for this reaction in the box below.
Use the form k[A]m[B]n , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n

From these data, the rate constant is ______M-1s-1.

Experiment [NH4+]o, M [NO2-]o, M Initial Rate, Ms-1 1 0.706 2.74×10-2 6.02×10-6 2 0.706 5.47×10-2 1.20×10-5 3 1.41 2.74×10-2 1.20×10-5 4 1.41 5.47×10-2 2.40×10-5

Explanation / Answer

Sol . As Rate = K [NH4+]m [NO2-]n where k = rate constant , m and n are the order of reaction with respect to NH4+ and NO2- respectively .

So , using the data of experiment no. 1 and put in rate law equation and then , using using the data of experiment no. 2 and put in rate law equation , then , dividing them .

Therefore ,

(6.02 × 10-6)/(1.20 × 10-5) = (0.706/0

706)m × (2.74×10-2 / 5.47 × 10-2 )n

So , 0.5 = (0.5)n , therefore , n = 1.

Now , using the data of experiment no. 2 and put in rate law equation and using the data of experiment no. 4 and put in rate law equation , then , dividing them ,

(1.20 × 10-5)/(2.40 × 10-5) = (0.706/1.41)m × (5.47 × 10-2 /5.47 × 10-2)n

So , 0.5 = (0.5)m , Therefore , m = 1 .

Hence , Rate = k [NH4+] [NO2-]

Now , taking the data of experiment no. 1 , we have ,

6.02 × 10-6 = k × 0.706 × 2.74 × 10-2

So , K = (6.02 × 10-6 ) / ( 0.706 x 2.74 x 10-2 )

K = 3.11 × 10-4 M-1 s-1 .

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