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 $:-$

Similar Questions

What is the rate of reaction?

In the reaction $2N_2O_5(g) \rightarrow 4 NO_2(g) + O_2(g)$, $N_2O_5$ disappears at a rate of $0.06 \text{ mol dm}^{-3} \text{s}^{-1}$. Calculate the rate of formation of $O_2(g)$.

Find the rate of the following reaction: $2 \ N_2O_{5(g)} \rightarrow 4 \ NO_{2(g)} + O_{2(g)}$ if the concentration of $NO_2$ increases to $5.2 \times 10^{-3} \ M$ in $100 \ s$.

For the reaction $2 NH_{3(g)} \rightarrow N_{2(g)} + 3 H_{2(g)}$,the rate of disappearance of $NH_3$ is $1.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$. What is the rate of formation of $N_2$ and $H_2$?

Thermal decomposition of $N_2O_5$ occurs as per the equation below:
$2N_2O_5 \longrightarrow 4NO_2 + O_2$
The correct statement is:

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