   Chapter 13, Problem 73AE

Chapter
Section
Textbook Problem

# Consider the decomposition of the compound C5H6O3 as follows: C 5 H 6 O 3 ( g ) → C 2 H 6 ( g ) + 3 CO ( g ) When a 5.63-g sample of pure C5H6O3(g) was sealed into an otherwise empty 2.50-L flask and heated to 200°C, the pressure in the flask gradually rose to 1.63 atm and remained at that value. Calculate K for this reaction.

Interpretation Introduction

Interpretation: The decomposition reaction, mass of C5H6O3 , volume of flask and rise in pressure is given. The value of equilibrium constant is to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of pressure, it is represented as Kp .

To determine: The value of equilibrium constant.

Explanation

Explanation

Given

The rise in pressure of given reaction is 1.63atm .

Mass of C5H6O3 is 5.63g .

Volume of flask is 2.50L .

The given temperature is 200°C .

The molar mass of C5H6O3 ,

[5(12.01)+(6×1.008)+(3×15.999)]g=114.10g/mol

The conversion of degree Celsius (°C) into Kelvin (K) is done as,

T(K)=T(°C)+273

Hence,

The conversion of 200°C into Kelvin is,

T(K)=T(°C)+273T(K)=(200+273)K=473K

The stated reaction is,

C5H6O3(g)C2H6(g)+3CO(g)

The number of moles is calculated using the formula,

Numberofmoles=Givenmass×1moleofcompoundMolarmassofcompound

Substitute the values of given mass and molar mass of C5H6O3 in above equation.

Numberofmoles=Givenmass×1moleofcompoundMolarmassofcompound=5.63g×1moleofC5H6O3114.10g/mol=0.0493mol

It is the initial moles of C5H6O3 .

The initial moles of product is,

C2H6(g)=03CO(g)=0

The formula of ideal gas law is,

PV=ntotalRT

Where,

• P is total pressure.
• V is volume.
• ntotal is total moles.
• R is the universal gas constant (0.08206Latm/Kmol) .
• T is absolute temperature.

Substitute the values of P,V,R and T in above equation.

PV=ntotalRT1.63atm×2.50L=ntotal×0.08206Latm/Kmol×473Kntotal=1.63atm×2.50L0.08206Latm/Kmol×473Kntotal=0.105mol

Simplify the above equation,

ntotal=1.63atm×2.50L0.08206Latm/Kmol×473Kntotal=0.105mol

It is assumed that the change in pressure of C2H6(g) is x .

On the reaction with one molecule of C5H6O3 , the equilibrium reaction is,

C5H6O3(g)®C2H6(g)3CO(g)Initialmoles:0.049300Change(moles):-xx3xEquilibrium(moles):0

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