# For the reaction $2NO\left(g\right)+C{l}_{2}\to 2NOCl\left(g\right)$ the following data were collected. All the measurnments were taken at 263 K. a) Write the expression for rate law, b) Calculate the value of rate constant and specify its untis. c) What is the initial rate of disappearance of $C{l}_{2}$ in exp. 4?

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## The rate law equation may be expressed as: Rate=$k{\left[A\right]}^{p}{\left[B\right]}^{q}$ Comparing experiments 1 and 2 ${\left(Rate\right)}_{1}=k{\left[0.15\right]}^{p}{\left[0.15\right]}^{q}=0.60$ ${\left(Rate\right)}_{2}=k{\left[0.15\right]}^{p}{\left[0.30\right]}^{q}=1.20$ Dividing eq. (ii) by eq. (i), $\frac{Rat{e}_{2}}{Rat{e}_{1}}=\frac{k{\left[0.15\right]}^{p}{\left[0.30\right]}^{q}}{k{\left[0.15\right]}^{p}{\left[0.15\right]}^{q}}=\frac{1.20}{0.60}=2,{\left[2\right]}^{q}={\left[2\right]}^{1}$ or q=1 Comparing experiments 1 and 3 $\left(Rat{e}_{1}\right)=k{\left[0.15\right]}^{p}{\left[0.15\right]}^{q}$ $\left(Rat{e}_{3}\right)=k{\left[0.30\right]}^{p}{\left[0.15\right]}^{q}$ Dividing eq. (iii) by eq (i) $\frac{Rat{e}_{3}}{Rat{e}_{1}}=\frac{k{\left[0.30\right]}^{p}{\left[0.15\right]}^{q}}{k{\left[0.15\right]}^{p}{\left[0.15\right]}^{q}}=\frac{2.40}{0.60}=4,{\left[2\right]}^{p}={\left[2\right]}^{2}\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}p=2$ a) Expression for rate law, Rate = $k{\left[NO\right]}^{2}\left[C{l}_{2}\right]$ b) Rate constant (k) for the reaction may be calculated as: $k{\left[0.15\right]}^{p}{\left[0.15\right]}_{q}=0.60,k{0.150}^{2}{\left[0.15\right]}^{1}=k60,k=p\frac{0.60}{0.003375}=1.78×{10}^{2}$ since, the reaction is of third order : units of rate constant k =${L}^{2}mo{l}^{-2}\stackrel{-1}{min}$ c) Initial rate of disappearancing of chlorine in experiment-4 Rate $=k{\left[0.20mol{L}^{-1}\right]}^{2}\left[0.25mol{L}^{-1}\right]=\left(1.78×{10}^{2}{L}^{2}\stackrel{-1}{min}\right)×{\left(0.25mol{L}^{-1}\right)}^{3}$ $=2.78mol{L}^{-1}\stackrel{-1}{min}$

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Updated on:21/07/2023

### Knowledge Check

• Question 1 - Select One

## The rate of the reaction 2NO+Cl2→2NOCl is given by the rate equation Rate=k[NO]2[Cl2] The value of the rate constant can be increased by

AIncreasing the temperature
BIncreasing the concentration of NO
CIncreasing the concentration of the Cl2
DDoing all of these
• Question 1 - Select One

## For the equilibrium 2NO(g)+Cl2(g)⇔2NOCl(g) Kp is related to Kc by the reaction

AKp=Kc(RT)
BKp=Kc(RT)2
CKp=Kc/RT
DKp=Kc/(RT)2
• Question 1 - Select One

## For the equilibrium 2NO(g)+Cl2(g)↔2NOCl(g) Kp is related to Kc by the reaction:

AKp=Kc(RT)
BKp=Kc(RT)2
CKp=KcRT
DKp=Kc(RT)2
• Question 1 - Select One

## For a reaction 2NO(g)+Cl2(g)→2NOCl(g), when concentration of Cl2 is doubled, the rate becomes four times. What is the order of reaction with respect to Cl2 ?

A1
B2
C3
D4

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