PHYSICAL CHEMISTRY-WEBASSIGN
PHYSICAL CHEMISTRY-WEBASSIGN
11th Edition
ISBN: 9780357087411
Author: ATKINS
Publisher: CENGAGE L
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Chapter 2, Problem 2C.7AE

(i)

Interpretation Introduction

Interpretation: The values of ΔrH° and ΔrU° at 298K has to be calculated for the given reaction.

Concept introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system.  The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by U.

(i)

Expert Solution
Check Mark

Answer to Problem 2C.7AE

The value of ΔrH° for the reaction is 131.29 kJmol_1.  The value of ΔrU° for the reaction is 128.81kJmol-1_.

Explanation of Solution

The given reaction is:

    C(graphite)+H2O(g)CO(g)+H2(g)                                                    (1)

The standard enthalpy for the reaction is calculated by the formula,

    ΔrHο=mΔHfο(products)nΔHfο(reactants)                                (2)

Where,

  • ΔHfο(products) is the standard enthalpy of formation of products.
  • ΔHfο(reactants) is the standard enthalpy of formation of reactants.
  • m is the stoichiometric coefficient of products.
  • n is the stoichiometric coefficient of reactants

For standard enthalpy of the reaction, equation (2) is written as,

    ΔrHο=(1×ΔHfο(CO)+1×ΔHfο(H2)(1×ΔHfο(C(graphite))+1×ΔHfο(H2O)))

The standard enthalpy of formation of CO is 110.53 kJ mol1.

The standard enthalpy of formation of H2 is 0 kJ mol1.

The standard enthalpy of formation of H2O is 241.82 kJ mol1.

The standard enthalpy of formation of C(graphite) is 0 kJ mol1.

Substitute the values of standard enthalpy of formation of reactants and products in the above equation to calculate the standard enthalpy of the reaction.

    ΔrHο=(1×(110.53 kJ mol1)+1×(0 kJ mol1)(1×(0 kJ mol1)+1×(241.82 kJ mol1)))=110.53 kJmol1+241.82 kJmol1=131.29 kJmol_1

Hence, the value of ΔrH° for the reaction is 131.29 kJmol_1.

The change in number of moles of the reaction is calculated by the formula,

  Numberofmoles,Δng=MolesofgaseousproductMolesofgaseousreactant

Hence, the number of moles for the reaction is,

  Δng=(1+1)(1)=1

The expression to calculate the value of ΔrU° for the reaction is,

    ΔrU°=ΔrH°ΔngRT

Substitute the values of ΔrH°, R, Δng and T=298K in the above equation to calculate the value of ΔrU°.

  ΔrU°=131.29kJmol1(1×8.314×103kJK1mol1×298K)=131.29kJmol1(2477.572×103kJmol1)=128.81kJmol-1_

Hence, the value of ΔrU° for the reaction is 128.81kJmol-1_.

(ii)

Interpretation Introduction

Interpretation: The values of ΔrH° and ΔrU° has to be calculated for the given reaction at 478K.

Concept introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system.  The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by U.

(ii)

Expert Solution
Check Mark

Answer to Problem 2C.7AE

The value of ΔrH° for the reaction is 131.29 kJmol_1.  The value of ΔrU° for the reaction is 127.31 kJmol-1_.

Explanation of Solution

The given reaction is:

    C(graphite)+H2O(g)CO(g)+H2(g)                                                     (1)

The standard enthalpy for the reaction is calculated by the formula,

    ΔrHο=mΔHfο(products)nΔHfο(reactants)                                (2)

Where,

  • ΔHfο(products) is the standard enthalpy of formation of products.
  • ΔHfο(reactants) is the standard enthalpy of formation of reactants.
  • m is the stoichiometric coefficient of products.
  • n is the stoichiometric coefficient of reactants

For standard enthalpy of the reaction, equation (2) is written as,

    ΔrHο=(1×ΔHfο(CO)+1×ΔHfο(H2)(1×ΔHfο(C(graphite))+1×ΔHfο(H2O)))

The standard enthalpy of formation of CO is 110.53 kJ mol1.

The standard enthalpy of formation of H2 is 0 kJ mol1.

The standard enthalpy of formation of H2O is 241.82 kJ mol1.

The standard enthalpy of formation of C(graphite) is 0 kJ mol1.

Substitute the values of standard enthalpy of formation of reactants and products in the above equation to calculate the standard enthalpy of the reaction.

    ΔrHο=(1×(110.53 kJ mol1)+1×(0 kJ mol1)(1×(0 kJ mol1)+1×(241.82 kJ mol1)))=110.53 kJmol1+241.82 kJmol1=131.29 kJmol_1

Hence, the value of ΔrH° for the reaction is 131.29 kJmol_1.

The change in number of moles of the reaction is calculated by the formula,

  Numberofmoles,Δng=MolesofgaseousproductMolesofgaseousreactant

Hence, the number of moles for the reaction is,

  Δng=(1+1)(+1)=1

The expression to calculate the value of ΔrU° for the reaction is,

    ΔrU°=ΔrH°ΔngRT

Substitute the values of ΔrH°, R, Δng and T=478K in the above equation to calculate the value of ΔrU°.

  ΔrU°=131.29kJmol1(1×8.314×103kJK1mol1×478K)=131.29kJmol1(3.97kJmol1)=127.31 kJmol-1_

Hence, the value of ΔrU° for the reaction is 127.31kJmol-1_.

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Chapter 2 Solutions

PHYSICAL CHEMISTRY-WEBASSIGN

Ch. 2 - Prob. 2A.4DQCh. 2 - Prob. 2A.5DQCh. 2 - Prob. 2A.1AECh. 2 - Prob. 2A.1BECh. 2 - Prob. 2A.2AECh. 2 - Prob. 2A.2BECh. 2 - Prob. 2A.3AECh. 2 - Prob. 2A.3BECh. 2 - Prob. 2A.4AECh. 2 - Prob. 2A.4BECh. 2 - Prob. 2A.5AECh. 2 - Prob. 2A.5BECh. 2 - Prob. 2A.6AECh. 2 - Prob. 2A.6BECh. 2 - Prob. 2A.1PCh. 2 - Prob. 2A.2PCh. 2 - Prob. 2A.3PCh. 2 - Prob. 2A.4PCh. 2 - Prob. 2A.5PCh. 2 - Prob. 2A.6PCh. 2 - Prob. 2A.7PCh. 2 - Prob. 2A.8PCh. 2 - Prob. 2A.9PCh. 2 - Prob. 2A.10PCh. 2 - Prob. 2B.1DQCh. 2 - Prob. 2B.2DQCh. 2 - Prob. 2B.1AECh. 2 - Prob. 2B.1BECh. 2 - Prob. 2B.2AECh. 2 - Prob. 2B.2BECh. 2 - Prob. 2B.3AECh. 2 - Prob. 2B.3BECh. 2 - Prob. 2B.4AECh. 2 - Prob. 2B.4BECh. 2 - Prob. 2B.1PCh. 2 - Prob. 2B.2PCh. 2 - Prob. 2B.3PCh. 2 - Prob. 2B.4PCh. 2 - Prob. 2B.5PCh. 2 - Prob. 2C.1DQCh. 2 - Prob. 2C.2DQCh. 2 - Prob. 2C.3DQCh. 2 - Prob. 2C.4DQCh. 2 - Prob. 2C.1AECh. 2 - Prob. 2C.1BECh. 2 - Prob. 2C.2AECh. 2 - Prob. 2C.2BECh. 2 - Prob. 2C.3AECh. 2 - Prob. 2C.3BECh. 2 - Prob. 2C.4AECh. 2 - Prob. 2C.4BECh. 2 - Prob. 2C.5AECh. 2 - Prob. 2C.5BECh. 2 - Prob. 2C.6AECh. 2 - Prob. 2C.6BECh. 2 - Prob. 2C.7AECh. 2 - Prob. 2C.7BECh. 2 - Prob. 2C.8AECh. 2 - Prob. 2C.8BECh. 2 - Prob. 2C.1PCh. 2 - Prob. 2C.2PCh. 2 - Prob. 2C.3PCh. 2 - Prob. 2C.4PCh. 2 - Prob. 2C.5PCh. 2 - Prob. 2C.6PCh. 2 - Prob. 2C.7PCh. 2 - Prob. 2C.8PCh. 2 - Prob. 2C.9PCh. 2 - Prob. 2C.10PCh. 2 - Prob. 2C.11PCh. 2 - Prob. 2D.1DQCh. 2 - Prob. 2D.2DQCh. 2 - Prob. 2D.1AECh. 2 - Prob. 2D.1BECh. 2 - Prob. 2D.2AECh. 2 - Prob. 2D.2BECh. 2 - Prob. 2D.3AECh. 2 - Prob. 2D.3BECh. 2 - Prob. 2D.4AECh. 2 - Prob. 2D.4BECh. 2 - Prob. 2D.5AECh. 2 - Prob. 2D.5BECh. 2 - Prob. 2D.1PCh. 2 - Prob. 2D.2PCh. 2 - Prob. 2D.3PCh. 2 - Prob. 2D.4PCh. 2 - Prob. 2D.5PCh. 2 - Prob. 2D.6PCh. 2 - Prob. 2D.7PCh. 2 - Prob. 2D.8PCh. 2 - Prob. 2D.9PCh. 2 - Prob. 2E.1DQCh. 2 - Prob. 2E.2DQCh. 2 - Prob. 2E.1AECh. 2 - Prob. 2E.1BECh. 2 - Prob. 2E.2AECh. 2 - Prob. 2E.2BECh. 2 - Prob. 2E.3AECh. 2 - Prob. 2E.3BECh. 2 - Prob. 2E.4AECh. 2 - Prob. 2E.4BECh. 2 - Prob. 2E.5AECh. 2 - Prob. 2E.5BECh. 2 - Prob. 2E.1PCh. 2 - Prob. 2E.2PCh. 2 - Prob. 2.1IACh. 2 - Prob. 2.3IACh. 2 - Prob. 2.4IACh. 2 - Prob. 2.5IACh. 2 - Prob. 2.6IACh. 2 - Prob. 2.7IA
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