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Consider the following reaction:$ A + 3 B \rightarrow 2 C + 2 D$. In one experiment, the rate of change of B was found to be $2.50 \times 10^{–4}\;M/s$. What is the rate of change of C in this experiment?

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Answer: $1.67 \times 10^{-4}\;M/s$
For the given reaction, rate of the reaction $ = - \large\frac{1}{3}$$ \large\frac{\Delta B}{\Delta t}$$ =\large\frac{1}{2}$$\large\frac{\Delta C}{\Delta t}$
$\Rightarrow \large\frac{\Delta C}{\Delta t}$$ = \large\frac{2}{3}$$ \large\frac{\Delta B}{\Delta t}$$ = \large\frac{2}{3}$$ \times\; 2.50 \times 10^{–4}\;M/s$$ = 1.67 \times 10^{–4}\;M/s$..
answered Jul 24, 2014 by balaji.thirumalai

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