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First Order Reaction

Concentration...

Question

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

0.0048

mole/lit/min

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0.00048

mole/lit.sec

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4.8 \times 10^{- 4}

mole/lit/min

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0.0048

mole/lit/sec

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Solution

The correct option is B $0.00048$ mole/lit.sec
Since:
$A v e r a g e r a t e = \frac{I n i t i a l c o n c - f i n a l c o n c}{t i m e}$
Given:
Initial Conc= $0.044 M$

Final Conc= $0.032 M$

$T i m e = 25 m i n u t e s$

So, Average rate $= \frac{0.044 - 0.032}{25}$

$= 0.00048 m o l e / l i t . s e c$

Option B is correct.

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