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Q.

In a reaction $2 A \to$ products, the concentration of A decreases from $0.5 mol L^{- 1} to 0.4 {molL}^{- 1}$ in 10 minutes. Average rate of reaction is

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a

$0.01 {molL}^{- 1} S^{- 1}$

b

$1.67 \times 10^{- 4} {moL}^{- 1} S^{- 1}$

c

$0.01 mol L^{- 1} {hr}^{- 1}$

d

$8.33 \times 10^{- 5} {moL}^{- 1} S^{- 1}$

answer is D.

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Detailed Solution

$Rate = \frac{1}{2} [\frac{- d [A]}{dt}] = \frac{1}{2} (\frac{0.5 - 0.4}{600}) = 8.33 \times 10^{- 5} mol L^{- 1} S^{- 1}$

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