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Question

# Ammonium ion (NH+4) reacts with nitrite ion (NO−2) in aqueous solution according to the equationNH+4(aq)+NO−2(aq)→N2(g)+2H2O(l)The following initial rates of reaction have been measured for the given reactant concentrations.Expt No.[NH+4],(M)[NO−2],(M)Rate (M/hr)10.0100.0200.02020.0150.0200.03030.0100.0100.005Which 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][NO−2]2Let the order of reaction wrt[NH+4],[NO−2] be x and y respectively.Then, rate law, r=k[NH−4]x[NO−2]yOn 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]yDividing Eq. (i) by Eq. (ii) gives(0.0200.030)=(0.0100.015)x(23)1=(23)x or x=1Dividing Eq. (i) by (iii) gives0.0200.005=(0.0200.010)y=4=2y(2)2=(2)yor y=2On substituting the values of x and y in the rate law, we getr=k[NH+4][NO−2]2.

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