In the reaction, A+2B→6C+2D, if the initial rate −dAdt at t = 0 is 2.6×10−2M sec−1 , what will be the value of −dBdtat t=0 ?

A+2B→6C+2D−dAdt=−12dBdt=+16dCdt=+12dDdt∴−dBdt=2×−dAdt=2×2.6×10−2=5.2×10−2M sec−1

In the reaction, A+2B→6C+2D, if the initial rate −dAdt at t = 0 is 2.6×10−2M sec−1 , what will be the value of −dBdtat t=0 ?

A+2B→6C+2D−dAdt=−12dBdt=+16dCdt=+12dDdt∴−dBdt=2×−dAdt=2×2.6×10−2=5.2×10−2M sec−1

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In the reaction, $A + 2 B \to 6 C + 2 D$ , if the initial rate $- \frac{d [A]}{d t}$ at t = 0 is $2.6 \times 10^{- 2} M \sec^{- 1}$ , what will be the value of $- \frac{d [B]}{d t} a t t = 0 ?$

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$8.5 \times 10^{- 2} M \sec^{- 1}$

$2.5 \times 10^{- 2} M \sec^{- 1}$

$5.2 \times 10^{- 2} M \sec^{- 1}$

$7.5 \times 10^{- 2} M \sec^{- 1}$

answer is C.

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Detailed Solution

$A + 2 B \to 6 C + 2 D$

$\frac{- d [A]}{d t} = - \frac{1}{2} \frac{d [B]}{d t} = + \frac{1}{6} \frac{d [C]}{d t} = + \frac{1}{2} \frac{d [D]}{d t}$

$\begin{array}{l} ∴ \frac{- d [B]}{d t} = 2 \times \frac{- d [A]}{d t} \\ = 2 \times 2.6 \times 10^{- 2} \\ = 5.2 \times 10^{- 2} M \sec^{- 1} \end{array}$

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In the reaction, A+2B→6C+2D, if the initial rate −dAdt at t = 0 is 2.6×10−2M sec−1 , what will be the value of −dBdtat t=0 ?

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In the reaction, $A + 2 B \to 6 C + 2 D$ , if the initial rate $- \frac{d [A]}{d t}$ at t = 0 is $2.6 \times 10^{- 2} M \sec^{- 1}$ , what will be the value of $- \frac{d [B]}{d t} a t t = 0 ?$