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  • For a second order reaction<img alt="2 \mathrm{NOBr} \rightarrow 2 \mathrm{NO}+\mathrm{Br}_2" src="https://entrancecorner.oncodecogs.com/gif.latex?2%20%5Cmathrm%7BNOBr%7D%20%5Crightarrow%202%20%5Cm

For a second order reaction2 \mathrm{NOBr} \rightarrow 2 \mathrm{NO}+\mathrm{Br}_2, if initial concentration of reactant is 4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} with a k value 0.810 \mathrm{~mol}^{-1} \mathrm{~L}\: \mathrm{sec}^{-1}. What is the time required to use 1.5 \times 10^{-3} \mathrm{~mol}\: L^{-1} reactant.

Option: 1

92.6 minutes


Option: 2

110.6 seconds


Option: 3

80 seconds


Option: 4

92.6 seconds


Answers (1)

best_answer

General formula for nth order reaction:

Let a general reaction: \mathrm{A \rightarrow x B}

-\frac{1}{x} \frac{d[A]}{d t}=K[A]^n

\text { Integrating above equation: } \int_{A_0}^A \frac{d[A]}{[A]^n}=-x K \int_0^t d t \text {, here } K=\text { rate constant }

\begin{aligned} & \text { For } n \neq 1 \Rightarrow\left[\frac{(A)^{-n+1}}{-n+1}\right]_{A_0}^A=-x Kt \\ & \Rightarrow \frac{A^{-n+1}-A_0^{-n+1}}{{ }^{-n+1}}=-x Kt \\ & \Rightarrow(A)^{1-n}=\left(A_0\right)^{1-n}+x(n-1) Kt \end{aligned}

For second order reaction:\frac{1}{A}=\frac{1}{A_0}+x\mathrm{K} t

Here according to question:\kappa=0.810 \mathrm{~mol}^{-1} \mathrm{Lsec}^{-1} \text { and } A_0=4 \times 10^{-3} \operatorname{mol~L} L^{-1}

And, Amount of reactant left unreacted:A=4 \times 10^{-3}-1.5 \times 10^{-3}=2.5 \times 10^{-3} \mathrm{~mol} L^{-1}

Here in the question 2 moles of reactants are used, i.e., x=2

Thus for second order reaction:\mathrm{\frac{1}{A}=\frac{1}{A_0}+x Kt=\frac{1}{A}=\frac{1}{A_0}+2 Kt}

\begin{aligned} & \Rightarrow \frac{1}{2.5 \times 10^{-3} \mathrm{molL}^{-1}}=\frac{1}{4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}}+(2 \times 0.810 \times t) \mathrm{mol}^{-1} L \\\\ & \Rightarrow t=\frac{400-250}{2 \times 0810}=92.59 \approx 92.6 \mathrm{~seconds} \end{aligned}

Posted by

Devendra Khairwa

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