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Concentration of a reactant ‘A’ is changed from 0.044M to 0.032M in 25 minutes, the rate of the average reaction during this interval is

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$4.8 \times 10^{- 4} m o l e / l i t / \min$

0.00048mole/lit/sec

0.0048 mole/lit/min

0.0048 mol/lit/sec

answer is C.

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Detailed Solution

Average rate $= \frac{[A_{2}] - [A_{1}]}{t_{2} - t_{1}}$
$= \frac{0.044 - 0.032}{25} = 4.8 \times 10^{- 4} M \min^{- 1}$

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