For A→B [A] changed from 4.4 × 10–2M to 3.2 × 10–2M in 25 min. Now −Δ[A]Δt will be

−Δ[A]Δt = 4.4 × 10-2-3.2 × 10-225−Δ[A]Δt = (4.4 -3.2 ) 10-225−Δ[A]Δt = (1.2) 10-225= 0.000048 = 4.8×10−4M min-1

−Δ[A]Δt = 4.4 × 10-2-3.2 × 10-225−Δ[A]Δt = (4.4 -3.2 ) 10-225−Δ[A]Δt = (1.2) 10-225= 0.000048 = 4.8×10−4M min-1

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For A $\to$ B [A] changed from 4.4 × 10^–2M to 3.2 × 10^–2M in 25 min. Now $\frac{- Δ [A]}{Δt}$ will be

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$4.8 \times 10^{- 4} M . s^{- 1}$

$2.4 \times 10^{- 4} M^{- min^{- 1}}$

$4 .8 \times 10^{- 4} M \min^{- 1}$

$9.6 \times 10^{- 3} M . s^{- 1}$

answer is A.

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$\frac{- Δ [A]}{Δt}$ = $\frac{4.4 \times 10^{- 2} - 3.2 \times 10^{- 2}}{25}$

$\frac{- Δ [A]}{Δt}$ = $\frac{(4.4 - 3.2) 10^{- 2}}{25}$

$\frac{- Δ [A]}{Δt}$ = $\frac{(1.2) 10^{- 2}}{25}$ = 0.000048 = $4 .8 \times 10^{- 4} M \min^{- 1}$

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