What is the rate of formation of $O_2$ for the reaction stated below?
$2 N_2O_{5(g)} \longrightarrow 4 NO_{2(g)} + O_{2(g)}$
Given: $\left| \frac{d[N_2O_5]}{dt} \right| = 0.02 \ mol \ dm^{-3} \ s^{-1}$
$2 N_2O_{5(g)} \longrightarrow 4 NO_{2(g)} + O_{2(g)}$
Given: $\left| \frac{d[N_2O_5]}{dt} \right| = 0.02 \ mol \ dm^{-3} \ s^{-1}$
- A$0.01 \ mol \ dm^{-3} \ s^{-1}$
- B$0.02 \ mol \ dm^{-3} \ s^{-1}$
- C$0.03 \ mol \ dm^{-3} \ s^{-1}$
- D$0.04 \ mol \ dm^{-3} \ s^{-1}$
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Similar Questions
From the concentrations of $C_{4}H_{9}Cl$ (butyl chloride) at different times given below,calculate the average rate of the reaction:
$C_{4}H_{9}Cl + H_{2}O \rightarrow C_{4}H_{9}OH + HCl$
during different intervals of time.
$t/s$ $0$ $50$ $100$ $150$ $200$ $300$ $400$ $700$ $800$ $[C_{4}H_{9}Cl]/mol\ L^{-1}$ $0.100$ $0.0905$ $0.0820$ $0.0741$ $0.0671$ $0.0549$ $0.0439$ $0.0210$ $0.017$
$C_{4}H_{9}Cl + H_{2}O \rightarrow C_{4}H_{9}OH + HCl$
during different intervals of time.
| $t/s$ | $0$ | $50$ | $100$ | $150$ | $200$ | $300$ | $400$ | $700$ | $800$ |
| $[C_{4}H_{9}Cl]/mol\ L^{-1}$ | $0.100$ | $0.0905$ | $0.0820$ | $0.0741$ | $0.0671$ | $0.0549$ | $0.0439$ | $0.0210$ | $0.017$ |
Difficult
View SolutionThe rate of disappearance of $SO_2$ in the reaction $2SO_2 + O_2 \to 2SO_3$ is $1.28 \times 10^{-3} \ g/sec$. What is the rate of formation of $SO_3$?
Medium
View SolutionIf the rate of reaction is given as $\frac{1}{3} \frac{d[X]}{dt} = -\frac{1}{2} \frac{d[Y]}{dt} = -\frac{d[Z]}{dt}$,the reaction can be represented as
In the reaction,$N_{2} + 3H_{2} \longrightarrow 2NH_{3}$,the rate of disappearance of $H_{2}$ is $0.02 \ M/s$. The rate of appearance of $NH_{3}$ is (in $M/s$)
In the following reaction: $xA \to yB$,$\log \left[ -\frac{d[A]}{dt} \right] = \log \left[ \frac{d[B]}{dt} \right] + \log 2$,where the negative sign indicates the rate of disappearance of the reactant. Thus,$x : y$ is:
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