CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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Chapter 17, Problem 17.103P

(a)

Interpretation Introduction

Interpretation:

Equilibrium pressure of N2, O2 and NO for the given reaction has to be calculated.

Concept Introduction:

Equilibrium constant: It is the ratio of products to reactants has a constant value when the reaction is in equilibrium at a certain temperature. And it is represented by the letter K.

For a reaction,

aA+bBcC+dD

The equilibrium constant in terms of  partial pressure is, K=(PC)c(PD)d(PA)a(PB)b

where,

a, b, c and d are the stoichiometric coefficients of reactant and product in the reactions

(a)

Expert Solution
Check Mark

Answer to Problem 17.103P

  • Equilibrium pressure of N2 is 0.780 atm.
  • Equilibrium pressure of O2 is 0.210 atm.
  • Equilibrium pressure of NO is 2.669×10-16atm.

Explanation of Solution

Given data is shown below:

  N2(g) + O2(g)  2NO(g)Atmospheric partial pressure of N2= 0.780 atmAtmospheric partial pressure of O2= 0.210 atmKP =  4.35×1031

Equilibrium constant for the given reaction is given in terms of partial,

  Kp = (PNO)2(PN2)(PO2)

Equilibrium pressure of N2, O2 and NO can be determined by constructing ICE table.

                      N2(g)    +         O2(g)  2NO(g)Initial0.7800.210      0     Changexx+2xEquilibrium  (0.780x)     (0.210x)  2x

Substitute these values for the equation of equilibrium constant as given,

  Kp = (PNO)2(PN2)(PO2) 4.35×1031 = (2x)2(0.780x)(0.210x)

Here, assume that x<<K. Hence,

  4.35×1031 = (2x)2(0.780)(0.210)x = 1.3345×1016

Therefore,

Equilibrium pressure of N2, O2 and NO is calculated as follows,

  PN2(Equlibrium) = (0.7801.3345×1016)= 0.780 atm N2PO2(Equlibrium) = (0.2101.3345×1016)= 0.210 atm O2PNO(Equlibrium) = 2(1.3345×1016)= 2.669×1016atm NO

Equilibrium pressure of N2 is 0.780 atm.

Equilibrium pressure of O2 is 0.210 atm.

Equilibrium pressure of NO is 2.669×10-16atm.

(b)

Interpretation Introduction

Interpretation:

Pcontainer in the container N2(g) + O2(g)  2NO(g) has to be calculated.

Concept Introduction:

According to Dalton’s law of partial pressures, the total pressure is the sum partial pressure of all the gaseous component present.

(b)

Expert Solution
Check Mark

Answer to Problem 17.103P

Kp for the given reaction is 4.9×105.

Explanation of Solution

Given data is shown below:

Total pressure is the sum of partial pressure of N2, O2 and NO

Therefore,

Pcontainer is determined as follows,

  Pcontainer = PN2(Equlibrium)+PO2(Equlibrium)+PNO(Equlibrium)= 0.780 atm N2+ 0.210 atm O2 + 2.669×1016atm NO= 0.990 atm

Pcontainer in the container is 0.990 atm.

(c)

Interpretation Introduction

Interpretation:

KC for the given reaction N2(g) + O2(g)  2NO(g) has to be calculated.

Concept Introduction:

The relation between Kp and Kc is given by the following equation.

Kp = Kc(RT)Δngas Kp = Equilibrium constant in terms of partial pressureKc = Equilibrium constant in terms of concentrationΔngas = moles of gaseous product  moles of gaseous reactant

Only moles of gaseous products and reactants are used for calculating Δngas.

(c)

Expert Solution
Check Mark

Answer to Problem 17.103P

KC for the given reaction is 4.9×105.

Explanation of Solution

Given data is shown below:

  2NO(g) + O2(g)  2NO2(g)KP =  4.35×1031T = 298 K

  • Determine Δngas:

The total number of moles of gaseous reactants is 3 and the moles of gaseous product is 2.  Δngas is determined as follows,

  Δngas  = moles of gaseous product - moles of gaseous reactant =22 =0

  • Determine KC:

KC of the reaction can be determined from given Kp as below,

  Kp = Kc(RT)Δngas4.35×1031= Kc[(0.0821 atm.L/mol.K)(298 K)]0 Kc= 4.35×1031

Therefore,

KC for the given reaction is 4.35×10-31.

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Chapter 17 Solutions

CHEMISTRY >CUSTOM<

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