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Q.

N2O5→2NO2+12O2  In a definite time interval the rate of decomposition of N2O5  is 1.8×10-3mol L-1min-1,  then the rate of formation of O2 will be

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a

1.8×10-3mol L-1min-1

b

9×10-4mol L-1min-1

c

3.6×10-3mol L-1min-1

d

3.6×10-4mol L-1min-1

answer is B.

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Detailed Solution

dN2O5dt=1.8×10-3mol L-1min-1 dO2dt=?-dN2O5dt= +2×dO2dt 1.8×10-3 = +2×dO2dt dO2dt=1.8×10-32 dO2dt= 9×10-4mol L-1 min-1
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