In a reaction,n_1A + n_2B to m_1C + m_2D,5 M | vedclass

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(D) The rate of decomposition of $A$ is given by: $\frac{-\Delta[A]}{\Delta t} = - \frac{(4 \ M - 5 \ M)}{5 \ sec} = \frac{1 \ M}{5 \ sec} = 0.2 \ M \

Vedclass · Accepted Answer

$0.2 \ M \ sec^{-1} ; \left( \frac{m_2}{n_1} 	imes 0.2 ight) \ M \ sec^{-1}$

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(D) The rate of decomposition of $A$ is given by: $\frac{-\Delta[A]}{\Delta t} = - \frac{(4 \ M - 5 \ M)}{5 \ sec} = \frac{1 \ M}{5 \ sec} = 0.2 \ M \ sec^{-1}$.According to the stoichiometry of the reaction,the rate of reaction is $\frac{1}{n_1} \left( \frac{-\Delta[A]}{\Delta t} ight) = \frac{1}{m_2} \left( \frac{+\Delta[D]}{\Delta t} ight)$.Therefore,the rate of formation of $D$ is: $\frac{+\Delta[D]}{\Delta t} = \frac{m_2}{n_1} \left( \frac{-\Delta[A]}{\Delta t} ight) = \left( \frac{m_2}{n_1} 	imes 0.2 ight) \ M \ sec^{-1}$.

In a reaction,$n_1A + n_2B \to m_1C + m_2D$,$5 \ M$ of reactant $A$ are allowed to react with $3 \ M$ of reactant $B$. After $5 \ sec$,the concentration of $A$ was found to be $4 \ M$. What are the rate of decomposition of $A$ and the rate of formation of $D$,respectively?

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