9:58Consider the equilibrium system described bythe chemical reaction below. At equilibrium, asample of gas from the system is collected intoa 1.25 L flask at 25.0 °C. The flask is found tocontain 2.35 g of NO2, 9.60 g of N₂O, and 4.50 gof O₂. What is the value of Kc for this reaction?4 NO₂(g) 2 N₂O(g) + 3 O₂(g)Question 8 of 10Kc =Based on the given data, set up the expressionfor Kc. Each reaction participant must berepresented by one tile. Do not combine terms.[0.0409]²[0.218][0.113] ³[0.0409]¹0.480Once the expression is constructed, solve forKc.3.57 x 10³1[0.175]²[0.141][0.0409][0.175]40.0642Submit[0.113]²[0.0409]³[0.175][0.113]428.7Tap here or pull up for additional resourcesRESET[0.0511][0.175]³[0.113]15.813.6

Given reaction is- 4NO2(g)⇔2N2O(g) +3O2(g) At equilibrium conditions- Sample volume in the flask is,…

Answered: Consider the equilibrium system…

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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Related questions

Question

9:58
Consider the equilibrium system described by
the chemical reaction below. At equilibrium, a
sample of gas from the system is collected into
a 1.25 L flask at 25.0 °C. The flask is found to
contain 2.35 g of NO2, 9.60 g of N₂O, and 4.50 g
of O₂. What is the value of Kc for this reaction?
4 NO₂(g) 2 N₂O(g) + 3 O₂(g)
Question 8 of 10
Kc =
Based on the given data, set up the expression
for Kc. Each reaction participant must be
represented by one tile. Do not combine terms.
[0.0409]²
[0.218]
[0.113] ³
[0.0409]¹
0.480
Once the expression is constructed, solve for
Kc.
3.57 x 10³
1
[0.175]²
[0.141]
[0.0409]
[0.175]4
0.0642
Submit
[0.113]²
[0.0409]³
[0.175]
[0.113]4
28.7
Tap here or pull up for additional resources
RESET
[0.0511]
[0.175]³
[0.113]
15.8
13.6

Expert Solution

Step 1

Given reaction is-

$4 N O_{2} (g) \Leftrightarrow 2 N_{2} O (g) + 3 O_{2} (g)$

At equilibrium conditions-

Sample volume in the flask is,

V=1.25 L

Mass of NO₂=2.35 g

Mass of N₂O=9.60 g

Mass of O₂=4.50 g

Molar mass of NO₂=46g/mol

Molar mass of N₂O=44 g/mol

Molar mass of O₂=32 g/mol

Hence, moles of gases in the sample are-

$\begin{array}{rcl} M o l e s o f N O_{2} & = & \frac{2.35 g}{46 g / m o l} = 0.051 m o l \\ M o l e s o f N_{2} O & = & \frac{9.6 g}{44 g / m o l} = 0.218 m o l \\ M o l e s o f O_{2} & = & \frac{4.5 g}{32 g / m o l} = 0.1406 m o l \end{array}$

Also, the concentration of the gases in the sample(at equiibrium)-

$\begin{array}{rcl} [N O_{2}] & = & \frac{M o l e s}{v o l u m e o f s a m p l e} = \frac{0.051 m o l}{1.25 L} = 0.0408 M \\ [N_{2} O] & = & \frac{M o l e s}{v o l u m e o f s a m p l e} = \frac{0.218 m o l}{1.25 L} = 0.1744 M \\ [O_{2}] & = & \frac{M o l e s}{v o l u m e o f s a m p l e} = \frac{0.1406 m o l}{1.25 L} = 0.1125 M \end{array}$