partbq3 -ra (mol/dm³ s): 0.012.T (K):300ХАA + 2B= 4.000.064327moldm³→ 3C,XB=moldm³6.00-0.9635411.66380 (B) The value of kinetic rate constant k using T = 300 K as the base case. From there, write out theArrhenius equation in terms of two temperatures, with this k constant substituted in along with T1 =300.

Answered: (B) The value of kinetic rate constant…

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

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partbq3

-ra (mol/dm³ s): 0.012
.
T (K):
300
ХА
A + 2B
= 4.00
0.064
327
mol
dm³
→ 3C
,XB
=
mol
dm³
6.00-
0.96
354
11.66
380

(B) The value of kinetic rate constant k using T = 300 K as the base case. From there, write out the
Arrhenius equation in terms of two temperatures, with this k constant substituted in along with T1 =
300.

Expert Solution

Step 1: Value of kinetic rate constant

The Arrhenius equation describes the temperature dependence of reaction rates.

It is given by

$k equals A e to the power of negative fraction numerator E subscript a over denominator R T end fraction end exponent$

where, k is the rate constant.

A is the Arrhenius factor.

E_a is the activation energy.

R is the universal gas constant (8.314 J/mol-K).

T is the absolute temperature in Kelvin.

Reaction taking place is

$A plus 2 B rightwards arrow 3 C$

This reaction is elementary reaction. Therefore, rate of reaction is

$negative r subscript A equals k C subscript A C subscript B superscript 2$

Given, concentrations of A and B are

$C subscript A equals 4 space fraction numerator m o l over denominator d m cubed end fraction C subscript B space equals space 6 space fraction numerator m o l over denominator d m cubed end fraction$

Putting value in rate equation we get

$negative r subscript A equals k cross times 4 cross times open parentheses 6 close parentheses squared space equals space 144 k k space equals space fraction numerator negative r subscript A over denominator 144 end fraction$