In a reaction,n_1 A + n_2 B rightarrow m_1 C | vedclass

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

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 $\frac{-\Delta[A]}{\Delta t}$.$\frac{-\Delta[A]}{\Delta t} = -\frac{(4 \ M - 5 \ M)}{5 \ s} = \frac{1 \ M}{5 \ s} = 0.2 \ M \ sec^{-1}$.According to the stoichiometry of the reaction,$\frac{1}{n_1} \left( \frac{-\Delta[A]}{\Delta t} ight) = \frac{1}{m_2} \left( \frac{+\Delta[D]}{\Delta t} ight)$.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,$n_1 A + n_2 B \rightarrow m_1 C + m_2 D$,$5 \ M$ solution of reactant $A$ is allowed to react with $3 \ M$ solution of reactant $B$. After $5 \ s$,the concentration of $A$ was found to be $4 \ M$. The rate of decomposition of $A$ and the rate of formation of $D$ are respectively $:-$

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