Medium
The three experimental data for determining the differential rate of reaction $Cl_{2(g)} + 2NO_{(g)} \rightarrow 2NOCl_{(g)}$ at $310 \ K$ temperature are provided. $(a)$ Derive the differential rate of reaction. $(b)$ Calculate the order of reaction. $(c)$ Calculate the value of the rate constant.
(N/A) The differential rate expression is given by: $-\frac{d[Cl_2]}{dt} = -\frac{1}{2}\frac{d[NO]}{dt} = k[NO]^2[Cl_2]^1$.
$(b)$ The order of reaction is the sum of the powers of the concentration terms in the rate law: $2 + 1 = 3$.
$(c)$ Based on the experimental data provided for the reaction,the rate constant $k$ is calculated as $175 \ L^2 \ mol^{-2} \ s^{-1}$.
$(b)$ The order of reaction is the sum of the powers of the concentration terms in the rate law: $2 + 1 = 3$.
$(c)$ Based on the experimental data provided for the reaction,the rate constant $k$ is calculated as $175 \ L^2 \ mol^{-2} \ s^{-1}$.
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Similar Questions
The mechanism for the reaction is given below:
$2P + Q \to S + T$
$P + Q \to R + S$ (slow)
$P + R \to T$ (fast)
The rate law expression for the reaction is:
$2P + Q \to S + T$
$P + Q \to R + S$ (slow)
$P + R \to T$ (fast)
The rate law expression for the reaction is:
Medium
View SolutionReaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$
takes place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$
The order of the reaction is
takes place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$
The order of the reaction is
Difficult
View SolutionSelect the rate law for the reaction $A + B \longrightarrow C$ based on the following data:
$Exp$ $[A]$ $[B]$ $Rate$ $1$ $0.012$ $0.035$ $0.10$ $2$ $0.024$ $0.070$ $0.80$ $3$ $0.024$ $0.035$ $0.10$ $4$ $0.012$ $0.070$ $0.80$
| $Exp$ | $[A]$ | $[B]$ | $Rate$ |
| $1$ | $0.012$ | $0.035$ | $0.10$ |
| $2$ | $0.024$ | $0.070$ | $0.80$ |
| $3$ | $0.024$ | $0.035$ | $0.10$ |
| $4$ | $0.012$ | $0.070$ | $0.80$ |
Difficult
View SolutionThe following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur?
Experiment $[S]$ $[Nu]$ Rate $1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$ $2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$ $3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$
| Experiment | $[S]$ | $[Nu]$ | Rate |
|---|---|---|---|
| $1$ | $0.1$ | $0.1$ | $2.2 \times 10^{-3}$ |
| $2$ | $0.2$ | $0.1$ | $4.4 \times 10^{-3}$ |
| $3$ | $0.1$ | $0.2$ | $4.4 \times 10^{-3}$ |
What is the order of reaction for $A + B \to C$?
$Observation$ $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ $Rate \ (mol \ L^{-1} \ sec^{-1})$ $1$ $0.1$ $0.1$ $2 \times 10^{-3}$ $2$ $0.2$ $0.1$ $4 \times 10^{-3}$ $3$ $0.1$ $0.2$ $8 \times 10^{-3}$
| $Observation$ | $[A] \ (mol \ L^{-1})$ | $[B] \ (mol \ L^{-1})$ | $Rate \ (mol \ L^{-1} \ sec^{-1})$ |
| $1$ | $0.1$ | $0.1$ | $2 \times 10^{-3}$ |
| $2$ | $0.2$ | $0.1$ | $4 \times 10^{-3}$ |
| $3$ | $0.1$ | $0.2$ | $8 \times 10^{-3}$ |
Medium
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