Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 5, Problem 5.30QP

(a)

Interpretation Introduction

Interpretation: The PV work done by the gases against an external pressure of 1.00atm and the change in internal energy (ΔE) for the system are to be calculated.

Concept introduction: Internal energy of a system is defined as the sum of all the kinetic energy and the potential energy of all the compounds of the system. From the first law of thermodynamics, the change in internal energy is the sum of heat absorbed by the system and the work done on the system.

To determine: The PV work done by the gases against an external pressure of 1.00atm .

(a)

Expert Solution
Check Mark

Answer to Problem 5.30QP

Solution

The PV work during the reaction is 48.23J_ .

Explanation of Solution

Explanation

Given

Mass of KNO3 is 1.00g .

External pressure is 1.00atm .

Density of nitrogen is 1.165g/L .

Density of CO2 is 1.830g/L .

Temperature of the reaction is 20οC .

The conversion of Celsius to Kelvin is done as,

0οC=273K

Therefore, the conversion of 20οC to Kelvin is done as,

20οC=20+273K=293K

The given chemical reaction is,

2KNO3(s)+18S8(s)+3C(s)K2S(s)+N2(g)+3CO2(g)

The number of moles potassium nitrate is calculated by using the formula,

NumberofmolesofKNO3=MassMolarmass

Molar mass of KNO3 is 101.1g/mol .

Substitute the values of mass and molar mass of KNO3 in the above formula to calculate its number of moles.

NumberofmolesofKNO3=1.00g101.1g/mol=0.00989mol

As per the balanced chemical reaction, two moles of KNO3 produces one mole of nitrogen and three moles of carbon dioxide that is four moles of gases. Therefore, number of moles of gases formed is,

Numberofmolesofgases=0.00989molKNO3×4molgases2molKNO3=0.00989×42mol=0.039562=0.01978mol

The volume occupied by the gas is calculated by using ideal gas equation,

PV=nRTV=nRTP

Where,

  • P is the pressure of the container.
  • V is the volume of the container.
  • n is the number of moles.
  • R is the universal gas constant (0.0821LatmK1mol1) .
  • T is the temperature.

Substitute the values of P , n , R and T in the ideal gas equation to calculate the volume.

V=nRTP=0.01978mol×0.0821L.atm/mol.K×293K1.00atm=1.623×103×293L.atm1.00atm=0.476L.atm1.00atm

Simplify the above equation,

V=0.476L

The volume of a gas is increased by 0.476L that is ΔV=0.476L .

The conversion of L to joule is done as,

1L=101.33J

Therefore, the conversion of 0.476L to joule is done as,

0.476L=0.476×101.33J=48.23J

So the PV work done during the reaction is calculated by using the expression,

w=PΔV

Where,

  • w is the work done.
  • P is the pressure applied on the system.
  • ΔV is the change in volume of the system.

Substitute the values of P and ΔV in the above expression to calculate the work done.

w=1.00atm×48.23J/atm=48.23J_

Hence, the PV work during the reaction is 48.23J_ .

(b)

Interpretation Introduction

To determine: The change in internal energy (ΔE) for the system.

(b)

Expert Solution
Check Mark

Answer to Problem 5.30QP

Solution

The change in internal energy, ΔE is 21.64kJ_ .

Explanation of Solution

Explanation

Given

The heat released is 21.6kJ .

The work done is 48.23J .

The conversion of J to kJ is done as,

1J=103kJ

Therefore, the conversion of 48.23J to kJ is done as,

48.23J=48.23×103kJ=0.04823kJ

The change in internal energy is calculated by using the first law of thermodynamics which is represented by the equation,

ΔE=q+w

Where,

  • ΔE is the change in internal energy.
  • q is the heat absorbed by the system.
  • w is the work done on the system.

Here, heat is released so q is negative and work is done by the system on the surroundings so w is also negative.

Substitute the values of q and w in the above equation to calculate the change in internal energy.

ΔE=21.6kJ+(0.04823kJ)=21.6kJ0.04823kJ=21.64kJ_

Hence, the change in internal energy, ΔE is 21.64kJ_ .

Conclusion

  1. a. The PV work during the reaction is 48.23J_ .
  2. b. The change in internal energy, ΔE is 21.64kJ_

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Chapter 5 Solutions

Chemistry

Ch. 5.5 - Prob. 11PECh. 5.6 - Prob. 12PECh. 5.6 - Prob. 13PECh. 5.7 - Prob. 14PECh. 5.7 - Prob. 15PECh. 5.7 - Prob. 16PECh. 5.8 - Prob. 17PECh. 5 - Prob. 5.1VPCh. 5 - Prob. 5.2VPCh. 5 - Prob. 5.3VPCh. 5 - Prob. 5.4VPCh. 5 - Prob. 5.5VPCh. 5 - Prob. 5.6VPCh. 5 - Prob. 5.7VPCh. 5 - Prob. 5.8VPCh. 5 - Prob. 5.9QPCh. 5 - Prob. 5.10QPCh. 5 - Prob. 5.11QPCh. 5 - Prob. 5.12QPCh. 5 - Prob. 5.13QPCh. 5 - Prob. 5.14QPCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Prob. 5.20QPCh. 5 - Prob. 5.21QPCh. 5 - Prob. 5.22QPCh. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - Prob. 5.26QPCh. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - Prob. 5.32QPCh. 5 - Prob. 5.33QPCh. 5 - Prob. 5.34QPCh. 5 - Prob. 5.35QPCh. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - Prob. 5.38QPCh. 5 - Prob. 5.39QPCh. 5 - Prob. 5.40QPCh. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - Prob. 5.43QPCh. 5 - Prob. 5.44QPCh. 5 - Prob. 5.45QPCh. 5 - Prob. 5.46QPCh. 5 - Prob. 5.47QPCh. 5 - Prob. 5.48QPCh. 5 - Prob. 5.49QPCh. 5 - Prob. 5.50QPCh. 5 - Prob. 5.51QPCh. 5 - Prob. 5.52QPCh. 5 - Prob. 5.53QPCh. 5 - Prob. 5.54QPCh. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - Prob. 5.57QPCh. 5 - Prob. 5.58QPCh. 5 - Prob. 5.59QPCh. 5 - Prob. 5.60QPCh. 5 - Prob. 5.61QPCh. 5 - Prob. 5.62QPCh. 5 - Prob. 5.65QPCh. 5 - Prob. 5.66QPCh. 5 - Prob. 5.67QPCh. 5 - Prob. 5.68QPCh. 5 - Prob. 5.69QPCh. 5 - Prob. 5.70QPCh. 5 - Prob. 5.71QPCh. 5 - Prob. 5.72QPCh. 5 - Prob. 5.73QPCh. 5 - Prob. 5.74QPCh. 5 - Prob. 5.75QPCh. 5 - Prob. 5.76QPCh. 5 - Prob. 5.77QPCh. 5 - Prob. 5.78QPCh. 5 - Prob. 5.79QPCh. 5 - Prob. 5.80QPCh. 5 - Prob. 5.81QPCh. 5 - Prob. 5.82QPCh. 5 - Prob. 5.83QPCh. 5 - Prob. 5.84QPCh. 5 - Prob. 5.85QPCh. 5 - Prob. 5.86QPCh. 5 - Prob. 5.87QPCh. 5 - Prob. 5.88QPCh. 5 - Prob. 5.89QPCh. 5 - Prob. 5.90QPCh. 5 - Prob. 5.91QPCh. 5 - Prob. 5.92QPCh. 5 - Prob. 5.93QPCh. 5 - Prob. 5.94QPCh. 5 - Prob. 5.95QPCh. 5 - Prob. 5.96QPCh. 5 - Prob. 5.97QPCh. 5 - Prob. 5.98QPCh. 5 - Prob. 5.99QPCh. 5 - Prob. 5.100QPCh. 5 - Prob. 5.101QPCh. 5 - Prob. 5.102QPCh. 5 - Prob. 5.103APCh. 5 - Prob. 5.104APCh. 5 - Prob. 5.105APCh. 5 - Prob. 5.106APCh. 5 - Prob. 5.107APCh. 5 - Prob. 5.108APCh. 5 - Prob. 5.109APCh. 5 - Prob. 5.110APCh. 5 - Prob. 5.111APCh. 5 - Prob. 5.112APCh. 5 - Prob. 5.113APCh. 5 - Prob. 5.114APCh. 5 - Prob. 5.115APCh. 5 - Prob. 5.116APCh. 5 - Prob. 5.117APCh. 5 - Prob. 5.118APCh. 5 - Prob. 5.119APCh. 5 - Prob. 5.120APCh. 5 - Prob. 5.121APCh. 5 - Prob. 5.122APCh. 5 - Prob. 5.123APCh. 5 - Prob. 5.124APCh. 5 - Prob. 5.125APCh. 5 - Prob. 5.126APCh. 5 - Prob. 5.127APCh. 5 - Prob. 5.128APCh. 5 - Prob. 5.129APCh. 5 - Prob. 5.130APCh. 5 - Prob. 5.131APCh. 5 - Prob. 5.132APCh. 5 - Prob. 5.133APCh. 5 - Prob. 5.134APCh. 5 - Prob. 5.135APCh. 5 - Prob. 5.136APCh. 5 - Prob. 5.137APCh. 5 - Prob. 5.138APCh. 5 - Prob. 5.139APCh. 5 - Prob. 5.140APCh. 5 - Prob. 5.141APCh. 5 - Prob. 5.142APCh. 5 - Prob. 5.143AP
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