Home
Class 12
CHEMISTRY
a process A rarr D is difficult to occur...

a process A rarr D is difficult to occur directly instead it takes place in three successive steps, DeltaS(ArarrB)=40 e.u., DeltaS(BrarrC)=30 e.u., DeltaS(DrarrC)=20 e.u. where e.u. is entropy unit then the entropy change DeltaS for the process (ArarrD) is

A

+ 90 e.u.

B

+ 50 e.u.

C

-90 e.u.

D

- 50 e.u.

Text Solution

AI Generated Solution

The correct Answer is:
To calculate the total entropy change (ΔS) for the process A → D, we will add the entropy changes for each of the three successive steps: A → B, B → C, and D → C. ### Step-by-Step Solution: 1. **Identify the given entropy changes**: - ΔS(A → B) = 40 e.u. - ΔS(B → C) = 30 e.u. - ΔS(D → C) = 20 e.u. 2. **Recognize the relationship between D and C**: - Since we need to find ΔS(A → D), we can express it in terms of the steps we have: - ΔS(A → D) = ΔS(A → B) + ΔS(B → C) + ΔS(C → D) 3. **Calculate ΔS(C → D)**: - The process D → C has an entropy change of +20 e.u. - Therefore, the reverse process C → D will have an entropy change of: \[ ΔS(C → D) = -ΔS(D → C) = -20 \text{ e.u.} \] 4. **Combine all the entropy changes**: - Now we can substitute the values into the equation: \[ ΔS(A → D) = ΔS(A → B) + ΔS(B → C) + ΔS(C → D) \] \[ ΔS(A → D) = 40 \text{ e.u.} + 30 \text{ e.u.} - 20 \text{ e.u.} \] 5. **Perform the calculation**: \[ ΔS(A → D) = 40 + 30 - 20 = 50 \text{ e.u.} \] ### Final Answer: The total entropy change ΔS for the process A → D is **50 e.u.**. ---
Promotional Banner

Topper's Solved these Questions

  • MOCK TEST 10

    AAKASH INSTITUTE ENGLISH|Exercise Example|30 Videos

  • MOCK TEST 12

    AAKASH INSTITUTE ENGLISH|Exercise Exercise|14 Videos

Similar Questions

Explore conceptually related problems

The direct conversion of A to B is difficult, hence it is carried out as A rarr C rarrD rarr B Given, DeltaS_((A^rarrC)) = 50 eU,DeltaS_((C^rarrD))=30eU,DeltaS_((BrarrD))=20 eU , where eU is entropy unit. Thus the change in entropy in (A rarrB) is:

The direct conversion of A to B is difficult, hence it is carried out as A rarr C rarrD rarr B Given, DeltaS_((A^rarrC)) = 50 eU,DeltaS_((C^rarrD))=30eU,DeltaS_((BrarrD))=20 eU , where eU is entropy unit. Thus the change in entropy in (A rarrB) is:

The direct conversion of A or B is difficult, hence it is carried out by the following shown path Given, DeltaS_(A to C) = 50 eu , DeltaS_(C to D) = 30 eu, DeltaS_(B to D) =20 eu Where eu is entropy unit, then DeltaS_(A to B) is .

P - V diagram of a cyclic process ABCA is as shown in Fig. Choose the correct alternative (A) Delta Q_(ArarrB)= negative (B) Delta U_(BrarrC)= positive (C) DeltaU_(CrarrA)= negative (D) DeltaW_(CAB)= negative

A reversible heat engine carries 1 mole of an ideal monatomic gas around the cycle 1-2-3-1. Process 1-2 takes place at constant volume, process 2-3 is adiabatic, and process 3-1 takes place at constant pressure. Complete the values for the heat Delta Q , the change in internal energy Delta U , and the work done Delta , for each of the three processes and for the cycle as a whole.

Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. For the reaction at 298 K ,2A + B rarr C DeltaH =100 kcal and DeltaS=0.050 kcal K^(-1) . If DeltaH and DeltaS are assumed to be constant over the temperature range, above what temperature will the reaction become spontaneous?

Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. For the reaction at 25^(@), X_(2)O_(4)(l) rarr 2XO_(2)(g) DeltaH=2.1 Kcal and DeltaS = 20 cal K^(-1) . The reaction would be

Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. A reaction has a value of DeltaH =-40 Kcal at 400 k cal mol^(-1) . The reaction is spontaneous, below this temperature , it is not . The values fo DeltaG and DeltaS at 400 k are respectively

Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. When CaCO_(3) is heated to a high temperature , it undergoes decomposition into CaO and CO_(2) whereas it is quite stable at room temperature . The most likely explanation of it, is

As per second law of thermodynamics a process taken place spontaneously if and only if the entropy of the universe increases due to the process. Change in entropy is given by Delta S = (Q_(rev))/(T) What is the change in entropy of the universe due to the following reaction occuring at 27^(@)C ? A + 2B rarr C + 2D, Delta H = +1.8 kJ mol^(-1) Molar entropy values of A, B, C and D are 1,2,3 and 4 JK^(-1) mol^(-1) respectively.