Medium
The three experimental data sets for determining the differential rate of the reaction $2 NO_{(g)} + Cl_{2_{(g)}} \rightarrow 2 NOCl_{(g)}$ at a definite temperature are given below. (Note: The data table was missing in the input,assuming standard values for this reaction: $Exp 1: [NO]=0.1, [Cl_2]=0.1, Rate=0.18$; $Exp 2: [NO]=0.1, [Cl_2]=0.2, Rate=0.36$; $Exp 3: [NO]=0.2, [Cl_2]=0.1, Rate=0.72$).
$(a)$ Derive the differential rate law of the reaction.
$(b)$ Calculate the order of the reaction.
$(c)$ Calculate the value of the rate constant.
$(a)$ Derive the differential rate law of the reaction.
$(b)$ Calculate the order of the reaction.
$(c)$ Calculate the value of the rate constant.
(C) The differential rate law is given by: $-\frac{1}{2} \frac{d[NO]}{dt} = -\frac{d[Cl_2]}{dt} = k[NO]^x[Cl_2]^y$.
$(b)$ Comparing $Exp 1$ and $Exp 2$: Keeping $[NO]$ constant,doubling $[Cl_2]$ doubles the rate,so $y=1$. Comparing $Exp 1$ and $Exp 3$: Keeping $[Cl_2]$ constant,doubling $[NO]$ quadruples the rate,so $x=2$. The rate law is $Rate = k[NO]^2[Cl_2]^1$. The order of reaction $= 2 + 1 = 3$.
$(c)$ Using $Exp 1$: $0.18 = k(0.1)^2(0.1)^1$ $\Rightarrow 0.18 = k(0.001)$ $\Rightarrow k = 180 \ L^2 \ mol^{-2} \ s^{-1}$.
$(b)$ Comparing $Exp 1$ and $Exp 2$: Keeping $[NO]$ constant,doubling $[Cl_2]$ doubles the rate,so $y=1$. Comparing $Exp 1$ and $Exp 3$: Keeping $[Cl_2]$ constant,doubling $[NO]$ quadruples the rate,so $x=2$. The rate law is $Rate = k[NO]^2[Cl_2]^1$. The order of reaction $= 2 + 1 = 3$.
$(c)$ Using $Exp 1$: $0.18 = k(0.1)^2(0.1)^1$ $\Rightarrow 0.18 = k(0.001)$ $\Rightarrow k = 180 \ L^2 \ mol^{-2} \ s^{-1}$.
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In a reaction $A \rightarrow B$,if the concentration of reactant is increased by $9$ times,then the rate of reaction increases $3$ times. What is the order of reaction?
$[A]_0 / \text{mol } L^{-1}$
$t_{1/2} / \text{min}$
$0.100$
$200$
$0.025$
$100$
For a given reaction $R \rightarrow P$,$t_{1/2}$ is related to $[A]_0$ as given in the table:
Given: $\log 2 = 0.30$
Which of the following is true?
$A.$ The order of the reaction is $1/2$.
$B.$ If $[A]_0$ is $1 \text{ M}$,then $t_{1/2}$ is $200 \sqrt{10} \text{ min}$.
$C.$ The order of the reaction changes to $1$ if the concentration of reactant changes from $0.100 \text{ M}$ to $0.500 \text{ M}$.
$D.$ $t_{1/2}$ is $800 \text{ min}$ for $[A]_0 = 1.6 \text{ M}$.
Choose the correct answer from the options given below:
| $[A]_0 / \text{mol } L^{-1}$ | $t_{1/2} / \text{min}$ |
| $0.100$ | $200$ |
| $0.025$ | $100$ |
For a given reaction $R \rightarrow P$,$t_{1/2}$ is related to $[A]_0$ as given in the table:
Given: $\log 2 = 0.30$
Which of the following is true?
$A.$ The order of the reaction is $1/2$.
$B.$ If $[A]_0$ is $1 \text{ M}$,then $t_{1/2}$ is $200 \sqrt{10} \text{ min}$.
$C.$ The order of the reaction changes to $1$ if the concentration of reactant changes from $0.100 \text{ M}$ to $0.500 \text{ M}$.
$D.$ $t_{1/2}$ is $800 \text{ min}$ for $[A]_0 = 1.6 \text{ M}$.
Choose the correct answer from the options given below:
For the $2^{nd}$ order reaction $R \rightarrow P$,the following graph holds true. Identify the incorrect statement.
Medium
View SolutionThe mechanism of reaction $2 \ NO + Cl_2 \rightarrow 2 \ NOCl$ is given as :
$(i) 2 NO \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} N_2 O _2$ (fast)
$(ii) N_2O_2 + Cl_2 \xrightarrow{K_3} 2 \ NOCl$ (slow)
The rate expression of the reaction will be :
$(i) 2 NO \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} N_2 O _2$ (fast)
$(ii) N_2O_2 + Cl_2 \xrightarrow{K_3} 2 \ NOCl$ (slow)
The rate expression of the reaction will be :
Difficult
View Solution$A_{(g)} + 2B_{(g)} \to$ product is an elementary reaction. Which of the following is incorrect?
Medium
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