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Question: \(A + 2B \rightarrow C + D\). If \(- \frac{d\lbrack A\rbrack}{dt} = 5 \times 10^{- 4}moll^{- 1}s^{- ...

A+2B→C+DA + 2B \rightarrow C + D. If −d[A]dt=5×10−4moll−1s−1- \frac{d\lbrack A\rbrack}{dt} = 5 \times 10^{- 4}moll^{- 1}s^{- 1}1, then

−d[B]dt- \frac{d\lbrack B\rbrack}{dt} is.

A

2.5×10−4moll−1s−12.5 \times 10^{- 4}moll^{- 1}s^{- 1}

B

5.0×10−4moll−1s−15.0 \times 10^{- 4}moll^{- 1}s^{- 1}

C

2.5×10−3moll−1s−12.5 \times 10 - 3moll^{- 1}s^{- 1}

D

1.0×10−3moll−1s−11.0 \times 10^{- 3}moll^{- 1}s^{- 1}

Answer

2.5×10−4moll−1s−12.5 \times 10^{- 4}moll^{- 1}s^{- 1}

Explanation

Solution

A+2B→C+DA + 2 B \rightarrow C + D −d[A]dt=5×10−4\frac{- d\lbrack A\rbrack}{dt} = 5 \times 10^{- 4}

−12d[B]dt=5×10−42=2.5×10−4mol−1sec⁡−1- \frac{1}{2}\frac{d\lbrack B\rbrack}{dt} = \frac{5 \times 10^{- 4}}{2} = 2.5 \times 10^{- 4}mol^{- 1}\sec^{- 1}{}