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## Entropy of Mixing, Chemistry tutorial

:Entropy of mixingAssume that that n

_{1}mol of an ideal gas initially present at pressure p and n_{2}mol of the other ideal gas as well at the similar initial pressure p are mixed at constant temperature in such a way that the total pressure is as well p. This is possible by employing a vessel of appropriate volume. Assume the partial pressure of the first gas in the mixture is p_{1}and the partial pressure of the other gas is p_{2}. Then, the change in entropy for the first gas from equation ΔS = 2.303 nR log P_{1}/P_{2}is,ΔS

_{1}= n_{1}R ln (p/p_{1})Likewise, for the other gas,

ΔS

_{2}= n_{2}R ln (p/p_{2})The net entropy of mixing, ΔS

_{mix}, is thereforeΔS

_{mix}= ΔS_{1}+ ΔS_{2}= n_{1}R ln p/p_{1}+ n_{2}R ln p/p_{2}Whenever the mole fractions of the gases in the mixture are x

_{1}and x_{2}correspondingly, then according to the Dalton's law of partial pressures,ΔS

_{mix}= n_{1}R ln p/x_{1}p + n_{2}R ln p/x_{2}pΔS

_{mix}= n_{1}R ln p/x_{1}p + n_{2}R ln 1/x_{2}ΔS

_{mix}= 2.303 n_{1}R log (l/x_{1}) + 2.303 n_{2}R log (l/x_{2})Therefore,

ΔS

_{mix}= 2.303 R[n_{1 }log(1/x_{1}) + n_{2}log (1/x_{2})]ΔS

_{mix}= - 2.303 R [n_{1}log x_{1}+ n_{2}log x_{2}]If we are familiar with n

_{1}and n_{2}, ΔS_{mix}can be computed. The mole fractions x_{1 }and x_{2}are less than one, as x_{1}+ x_{2}= 1; as an outcome of this, log 1/x_{1}and log l/x_{2}are positive. Therefore, ΔS_{mix}is positive.:Entropy changes in Phase transitionsThe change of matter from one phase or stage (that is, solid, liquid, gas, allotropic form) into the other is termed as phase transition. Such changes occur at definite temperatures known as transition temperatures (that is, melting points, boiling points and so on) at a given pressure. Such transitions are accompanied via absorption or evolution of heat (known as latent heat). As absorption or evolution of heat at constant temperature leads to the entropy change, the entropy of transition is represented as,

ΔS

_{trans }= qrev/T = ΔH_{trans}/TThe above equation is valid only if the transition occurs in a reversible manner, that is, if the two phases are in equilibrium.

Therefore, whenever one mole of a solid melts to the liquid phase, the entropy of fusion is represented by:

ΔS

_{fus }= ΔH_{fus}/T_{f}Here, T

_{f}is the melting point and ΔH_{fus }is the molar enthalpy of fusion. Likewise,ΔS

_{vap }= ΔH_{vap}/T_{B}Here, T

_{B }is the boiling point and ΔH_{vap }is the molar enthalpy of vaporizationIt is evident that the entropy of freezing and condensation (that is, vapor into liquid) will be equivalent to - ΔS

_{fus }and - ΔS_{vap}, correspondingly.In a similar way, we can define the entropy change accompanying the transition of a substance from an allotropic form to the other. If such a transition occurs at a temperature T

_{trans}and ΔH_{trans}is the molar enthalpy of transition, then the entropy change accompany the transition is,ΔS

_{trans}= ΔH_{trans}/T_{trans}Therefore, in phase transitions, ΔS values can be computed from the corresponding ΔH values.

:Entropy changes in Chemical reactionsLet us now compute the entropy change accompanying a general chemical reaction of the kind,

aA + bB+....→+ cC + dD + ...

We state the entropy change for a reaction (Δ

_{r}S) as the difference between the net entropy of the products and the total entropy of the reactants. Therefore, if S_{A}, S_{B}.... is the entropies of one mole of reactants, A, B and so on, and S_{C}, S_{D}...., of the products, C, D and so on, then;Δ

_{r}S = (cS_{C}+ dS_{D}+ ....) - (aS_{A}+ bS_{B}+ .....)Here c, d, a, b... and so on are the stoichiometric coefficients. The variation of entropy change for a reaction by temperature can be readily deduced from the equation above by differentiating with respect to temperature at constant pressure. Therefore,

[∂(Δ

_{r}S)/∂T]_{p}= [c(∂S_{C}/∂T)_{p}+ d(∂S_{D}/∂T)_{p}+ .....] - [a(∂S_{A}/∂T)_{p }+ b (∂S_{B}/∂T)_{p }+ ....]According to the equation:

dH = TdS + Vdp

Or C

_{P}‾dT = TdS + VdpAt constant pressure (dp = 0),

C

_{P}‾dT = TdS_{p}Or dS

_{p}/dT = C_{P}‾/TOr (∂S/∂T) = C

_{P}‾/TC

_{P}‾ is the molar heat capacity of the substance at constant pressure.By utilizing the above result, we obtain:

[∂(Δ

_{r}S)/∂T]_{p}= [{cC_{P}‾(C) + dC_{P}‾(D) + ....} - {aC_{P}‾(A) + bC_{P}‾(B) + ....}]/T[∂(Δ

_{r}S)/∂T]_{p}= ΔC_{p}/THere, ΔC

_{P}is the difference between the heat capacities of the products and reactants at constant pressure.Or d(Δ

_{r}S) = ΔC_{p}(dT/T)Suppose that Δ

_{r}S_{1}and Δ_{r}S_{2}are the entropy change at temperatures T_{1}and T_{2}and ΔC_{p}is independent of the temperature. Then the above equation on integration gives,_{ΔrS1}∫^{ΔrS2}d(ΔrS) = ΔC_{p}_{T1}∫^{T2}(dT/T)Or Δ

_{r}S_{2}- Δ_{r}S_{1}= C_{p}ln T_{2}/T_{1}The above equation is helpful in finding out Δ

_{r}S value of a reaction at any specific temperature, if it is recognized at any other temperature all along by C_{p}values.The entropy values of substances can be found out by using the third law of thermodynamics. The entropy values of some of the substances in their standard states at 298.15 K are represented in the table. These are termed as standard entropy (S°) values. Similar to the computation of AJS, we can compute Δ

_{r}S° from the standard entropy values of the reactants and the products.Table: Standard Entropy (S°) values at 298.15 K

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