Consider the following reaction, 2NO2⟶2NO+O2If rate of disappearance of NO2 is 6.0×10−12mol L−1s−1then 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

answer is A.

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Detailed Solution

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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