Question
The decomposition of $\mathrm{N}_{2} \mathrm{O}_{5}$ takes place according to I order as$$2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$$Calculate :(a) The rate constant, if instantaneous rate is $1.4 \times 10^{-6}$ mol litre $^{-1}$ $\mathrm{sec}^{-1}$ when concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ is $0.04 \mathrm{M}$(b) The rate of reaction when concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ is $1.20 \mathrm{M}$(c) The concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ when the rate of reaction will be $2.45 \times 10^{-5} \mathrm{~mol}$ litre $^{-1} \mathrm{sec}^{-\mathrm{i}}$.
Step 1
Step 1: The rate of a first order reaction is given by the equation: $$ \text{Rate} = k[\text{N}_2\text{O}_5] $$ where $k$ is the rate constant and $[\text{N}_2\text{O}_5]$ is the concentration of $\text{N}_2\text{O}_5$. Show more…
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The rate constant for the reaction, $2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$, is $3.0 \times 10^{-5}$ $\mathrm{sec}^{-1}$. If the rate is $2.40 \times 10^{-5}$ mol litre $^{-1} \mathrm{sec}^{-1}$, then the concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ (in mol litre $^{-1}$ ) is [2000S] (a) $1.4$ (b) $1.2$ (c) $0.04$ (d) $0.8$
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The rate constant for the reaction, $\mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$, is $3.0 \times 10^{-5} \mathrm{~s}^{-1} .$ If the rate of reaction is $2.4 \times 10^{-5} \mathrm{~mol}$ litre $^{-1} \mathrm{~s}^{-1}$, then the concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ (in mole litre $^{-1}$ ) is (a) $1.4$ (b) $1.2$ (c) $0.04$ (d) $0.8$
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The rate constant for the reaction, $2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}$ is $3.0 \times 10^{-5} \mathrm{~s}^{-1}$. If the rate is $2.40 \times 10^{-5}$ mol litre $^{-1}$ $\mathrm{s}^{-1}$, then the concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ (in mol litre $^{-1}$ ) isa. $1.4$ b. $1.2$ c. $0.04$ d. $0.8$
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