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Question

Ammonium ion (NH+4) reacts with nitrite ion (NO2) in aqueous solution according to the equation
NH+4(aq)+NO2(aq)N2(g)+2H2O(l)
The following initial rates of reaction have been measured for the given reactant concentrations.

Expt No.[NH+4],(M)[NO2],(M)Rate (M/hr)
10.0100.0200.020
20.0150.0200.030
30.0100.0100.005
Which of the following is the rate law for this reaction?

A
Rate=k[NH+4][NO2]4
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B
Rate=k[NH+4][NO2]
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C
Rate=k[NH+4][NO2]2
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D
Rate=k[NH+4]2[NO2]
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E
Rate=k[NH+4]12[NO2]14
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Solution

The correct option is C Rate=k[NH+4][NO2]2
Let the order of reaction wrt[NH+4],[NO2] be x and y respectively.
Then, rate law, r=k[NH4]x[NO2]y

On substituting the values of different experiments,
(i) 0.020=k[0.010]x[0.020]y
(ii) 0.030=k[0.015]x[0.020]y
(iii) 0.005=k[0.010]x[0.010]y

Dividing Eq. (i) by Eq. (ii) gives
(0.0200.030)=(0.0100.015)x
(23)1=(23)x or x=1

Dividing Eq. (i) by (iii) gives
0.0200.005=(0.0200.010)y
=4=2y
(2)2=(2)y
or y=2
On substituting the values of x and y in the rate law, we get
r=k[NH+4][NO2]2.

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