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Consider the following reaction, 2NO22NO+O2
If rate of disappearance of NO2 is 6.0×1012mol L1s1then the rate of appearance of O2 is 


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3×10−12mol L−1s−1
6×10−12mol L−1s−1
1×10−12mol L−1s−1
1.5×10−12mol L−1s−1

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detailed solution

Correct option is A

For the reaction, 2NO2⟶2NO+O2Rate = −12dNO2dt=d[NO]2dt=dO2dt∵ −dNO2dt=6×10−12molL−1s−1∴    dO2dt=−12dNO2dt=12×6×10−12                    =3×10−12mol L−1s−1

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