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Rate of Reaction

In a closed c...

Question

In a closed container at a given temperature, concentration of a product P increases from $0.1 m o l / L$ to $0.5 m o l / L$ in $10 m i n$
Then the average rate of reaction in first $10 m i n$ will be:

0.1 m o l L^{- 1} m i n^{- 1}

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0.5 m o l^{- 1} L m i n

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25 m o l^{- 1} L m i n

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4 \times 10^{- 2} m o l L^{- 1} m i n^{- 1}

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Solution

The correct option is D $4 \times 10^{- 2} m o l L^{- 1} m i n^{- 1}$
Average rate of reaction:
$r_{a v} = \frac{[P]_{f i n a l} - [P]_{i n i t i a l}}{t_{f i n a l} - t_{i n i t i a l}} = - \frac{Δ P}{Δ t}$
Here $[P]_{f i n a l}, [P]_{i n i t i a l} are the concentrations of product at t_{f i n a l}, t_{i n i t i a l} respectively$

$r_{a v} = \frac{0.5 - 0.1}{10} m o l L^{- 1} m i n^{- 1}$
$r_{a v} = 4 \times 10^{- 2} m o l L^{- 1} m i n^{- 1}$

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Rate of Reaction

Standard VII Chemistry

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In a closed container at a given temperature, concentration of a product P increases from 0.1 mol/L to 0.5 mol/L in 10 min Then the average rate of reaction in first 10 min will be:

The correct option is D 4×10−2 mol L−1min−1Average rate of reaction: rav=[P]final−[P]initialtfinal−tinitial=−ΔPΔt Here [P]final,[P]initial are the concentrations of product at tfinal,tinitial respectively rav=0.5−0.110 mol L−1min−1 rav=4×10−2 mol L−1min−1

In a closed container at a given temperature, concentration of a product P increases from $0.1 m o l / L$ to $0.5 m o l / L$ in $10 m i n$
Then the average rate of reaction in first $10 m i n$ will be: