EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684651
Author: Katz
Publisher: VST
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Chapter 22, Problem 80PQ
To determine

The heat absorbed or released in the reaction.

Expert Solution & Answer
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Answer to Problem 80PQ

The heat released in the reaction is 3.1×104J.

Explanation of Solution

Write the expression to calculate the entropy of reactant A.

  SA=0K300KcAdTT                                                                                                           (I)

Here, SA is the entropy of reactant A, cA is the molar specific heat capacity of A at constant pressure and T is the temperature.

Write the expression to calculate the entropy of reactant B.

  SB=0K300KcBdTT                                                                                                          (II)

Here, SB is the entropy of reactant A and cB is the molar specific heat capacity of B at constant pressure.

Write the expression to calculate the entropy of reactant D.

  SD=0K300KcDdTT                                                                                                        (III)

Here, SD is the entropy of reactant A, cD is the molar specific heat capacity of A at constant pressure.

Write the expression to calculate the entropy change for the reaction.

  Δs=SD(SA+SB)                                                                                               (IV)

Here, Δs is the change in entropy of the reaction.

Write the expression to calculate the heat.

  Q=TΔs                                                                                                                (V)

Here, Q is the amount of heat.

Conclusion:

Substitute 2T for cA in the above equation (I) to calculate SA.

  SA=0K300K(2T)dTT=20K300KdTT=2[2T]0K300K=69.3J/K

Substitute 4T for cB in the above equation (II) to calculate SB.

  SA=0K300K(4T)dTT=40K300KdTT=4[2T]0K300K=138.6J/K

Substitute 3T for cD in the above equation (III) to calculate SD.

  SA=0K300K(3T)dTT=30K300KdTT=3[2T]0K300K=103.9J/K

Substitute 69.3J/K for SA, 138.6J/K for SB and 103.9J/K for SD in the above equation (IV) to calculate Δs.

  Δs=103.9J/K(69.3J/K+138.6J/K)=104J/K

Substitute 104J/K for Δs and 300K for T in the equation (V) to calculate Q.

  Q=300K(104J/K)=3.1×104J

The value of Q is the negative thus, the heat will release the system after the reaction.

Therefore, the heat released in the reaction is 3.1×104J.

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Chapter 22 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

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