What is the value of the specific rate constant, k, for this reaction: CO (g) + NO2 (g) → NO (g) + CO2 (g)? Expt.[CO] (M)[NO2] (M)Initial Rate (M/s)10.1000.1000.002120.1000.2000.008230.2000.2000.0083 0.21 8.2 0.0083 21

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Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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Question

What is the value of the specific rate constant, k, for this reaction:

CO_(g) + NO_{2 (g)} → NO_(g) + CO_{2 (g)}?

Expt.	[CO] (M)	[NO₂] (M)	Initial Rate (M/s)
1	0.100	0.100	0.0021
2	0.100	0.200	0.0082
3	0.200	0.200	0.0083

		0.21
		8.2
		0.0083
		21

Expert Solution

Step 1

Given reaction:

CO_(g) + NO_{2 (g)} → NO_(g) + CO_{2 (g)}

We have to determine the value of the specific rate constant, k.

Also, the data for the concentration of reactants, turning different rates of reaction has been provided from 3 experimental trials.

Let us write the rate law first for the given equation:

$r = k {[C O]}^{x} {[N O_{2}]}^{y} . . . (1) where, r = rate of reaction, k = specific rate constant, [C O] = concentration of CO [N O_{2}] = concentration of {NO}_{2} x and y are degree to which the reaction rate is sensitive to concentration of CO and {NO}_{2} respectively . x + y = n (Overall order of reaction)$

The order of the reaction and values of x,y, and k could be found with the help of provided experimental data.

Step 2

y determination:

As from the table, it is apparent that, after the first trial being performed at certain concentrations of reactant, in the second trial concentration of CO has been kept constant, and change is rate is observed based on the change in concentration of NO₂. So from power law:

$r \propto {[N O_{2}]}^{y} \Rightarrow \frac{r_{2}}{r_{1}} = {(\frac{{[N O_{2}]}_{2}}{{[N O_{2}]}_{1}})}^{y} Here the substricts denotes the trial numbers . Substituting the values from given table : \Rightarrow \frac{0.0082}{0.0021} = {(\frac{0.200}{0.100})}^{y} \Rightarrow l n (3.904) = y l n (2) \Rightarrow y = 1.96 ≃ 2 Please note that the ratios are being used here in same unit . Also, exponents doesn' t have unit .$

x determination:

Similarly, after the second trial being performed, in the third trial concentration of NO₂ has been kept constant, and change is rate is observed based on the change in concentration of CO. So from power law:

$r \propto {[C O]}^{y} \Rightarrow \frac{r_{3}}{r_{2}} = {(\frac{{[C O]}_{3}}{{[C O]}_{2}})}^{x} Here the substricts denotes the trial numbers . Substituting the values from given table : \Rightarrow \frac{0.0083}{0.0082} = {(\frac{0.200}{0.100})}^{x} \Rightarrow l n (1.012) = y l n (2) \Rightarrow x = 0.01 ≃ 0$

$r = k {[C O]}^{0} {[N O_{2}]}^{2} \Rightarrow r = k {[N O_{2}]}^{2} - rate of reaction is almost independent of change in concentration of CO Substituing values from 1 st trial : \Rightarrow 0.0021 M / s = k {[0.100]}^{2} M^{2} \Rightarrow k = 0.21 M^{- 1} s^{- 1} - you can verify putting another set of value, you will get similar results .$

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