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Question: For a given reaction \(3A + B \rightarrow C + D\) the rate of reaction can be represented by....

For a given reaction 3A+B→C+D3A + B \rightarrow C + D the rate of reaction can be represented by.

A

−13d[A]dt=−d[B]dt=+d[C]dt=+d[D]dt- \frac{1}{3}\frac{d\lbrack A\rbrack}{dt} = \frac{- d\lbrack B\rbrack}{dt} = \frac{+ d\lbrack C\rbrack}{dt} = \frac{+ d\lbrack D\rbrack}{dt}

B

−13d[A]dt=d[C]dt=K[A]m[B]n- \frac{1}{3}\frac{d\lbrack A\rbrack}{dt} = \frac{d\lbrack C\rbrack}{dt} = K\lbrack A\rbrack^{m}\lbrack B\rbrack^{n}

C

+13d[A]dt=−d[C]dt=K[A]n[B]m+ \frac{1}{3}\frac{d\lbrack A\rbrack}{dt} = \frac{- d\lbrack C\rbrack}{dt} = K\lbrack A\rbrack^{n}\lbrack B\rbrack^{m}

D

None of these

Answer

−13d[A]dt=−d[B]dt=+d[C]dt=+d[D]dt- \frac{1}{3}\frac{d\lbrack A\rbrack}{dt} = \frac{- d\lbrack B\rbrack}{dt} = \frac{+ d\lbrack C\rbrack}{dt} = \frac{+ d\lbrack D\rbrack}{dt}

Explanation

Solution

−13d[A]dt=−d[B]dt=+d[C]dt=+d(D)dt- \frac{1}{3}\frac{d\lbrack A\rbrack}{dt} = - \frac{d\lbrack B\rbrack}{dt} = \frac{+ d\lbrack C\rbrack}{dt} = \frac{+ d(D)}{dt}.