Unit - 7 Thermodynamics

Unit - Thermodynamics Evaluation Yourself . Solution : Given ∆H f CO = - . kJ mol - ∆H f CO = - . kJ mol - ∆H f (H O) = - kJ mol - CO (g) + H (g) → CO(g) + H O(g) ∆H r = ?

∆H r = Σ (∆H f ) products – Σ (∆H f ) reactants ∆H r = [∆H f (CO) + ∆H f (H O)] – [∆H f (CO )+∆H f (H )] ∆H r = [– . + (– )] – [– . + ( )] ∆H r = [– . ] + .

∆H r = . ∆H r = + . kJ mol – . Solution : Given : number of moles of water n g g mol mol − = molar heat capacity of water C P = .

J K – mol – T = C = K T = C = K D H = ? D H = nC P (T – T ) D H = mol × . J mol – K – × ( – ) K D H = 56475 J D H = . kJ .

C H l O g ( ) ( ) CO (g) H O (l) → D U at C = – . kJ Solution : Given T = C = K ; D U = – . kJ mol – D H = ? D H = D U + D n g RT D H = D U + (n p – n r ) RT H .

– . D H = – . kJ mol – . Solution : Given : Mg(S) + Br ( ) → MgBr (S) D H f = – KJ mol – Sublimation : Mg(S) → Mg(g) D H = + KJ mol – Ionisation : Mg(g) → Mg + (g) + 2e – D H = KJ mol – Vapourisation : Br ( ) → Br (g) D H = + KJ mol – Dissociation : Br (g) → 2Br(g) D H = + KJ mol – Electron affinity : Br(g) + e – → Br – (g) D H = – .

KJ mol – Solution : Mg MgBr s H f (s) Br (l) ( )  →  ∆ D H D H D H Br (g) u Br g Br g H ( ) ( ) ∆  →  Mg g Mg g H ( ) ( ) ∆  →  D H f = D H + D H + D H + D H + D H + u – = + + + + ( × – . ) + u – = + u u = – – u = – kJ mol – . Solution: Given T h = C = + = K T C = C = + = K % efficiency η = ? h C h % .

Solution: Given: S (urea) = . J mol – K – S (H O) = J mol – K – S (CO ) = . J mol – K – S (NH ) = . J mol – K – NH – CO – NH + H O → 2NH + CO D S r = Σ (S ) products – Σ (S ) reactants D S r = [ S (NH ) + S (CO )] – [S (urea) + S (H O)] D S r = [ × .

] D S r = . J mol – K – . Solution: Given: T b = K D H vap = 39840 J mol – D S v = ? ∆ = ∆ ∆ ∆ S H S S J K mol v vap b v v 39840 .

Solution: Given: D H = – kJ mol – = –10000 J mol – D S = – J K – mol – T = K D G = ? D G = D H – T D S D G = – kJ mol – – K × (–20x10 - ) kJ K – mol – D G = (– + ) kJ mol – D G = – kJ mol – At K D G = – kJ mol – – K × (–20x10 - ) kJ K – mol – D G = (– + ) kJ mol – D G = + kJ mol – The value of ∆G is negative at 300K and the reaction is spontaneous, but at 600K the value ∆G becomes positive and the reaction is non spontaneous. EVALUATION I Choose the best answer: . (b) D H .

(d) decrease in free energy . (b) q = . (d) = . (a) w = – D U .

(d) mass volume . (a) – J . (b) negative . (b) – .

kcal . (a) graphite is more stable than diamond . (d) – kJ . (d) frictional energy .

(c) kJ mol – . (a) D H < and D S > . (c) adiabatic expansion . (d) (–, –, +) .

(b) ⁰C . (d) CaCO (S) → CaO(S) + CO (g) . (a) 300K Keys to multiple choice questions: . w = – P D V w = – ( × Nm – ) ( × – m – × – m ) w = – ( – – – ) Nm w = – ( – ) – ) J w = – ( × – ) J w = – × J w = – J .

CO g O g CO g ( ) ( ) ( ) → D H C (CO) = D H f (CO ) – D H f (CO) + D H f (O )] D H C (CO) = – KCal – [– . KCal + ] D H C (CO) = – KCal + . KCal D H C (CO) = – . KCal .

2Al + Cr O → 2Cr + Al O D H r = [ D H f (Cr) + D H f (Al O )] – [ D H f (Al) + D H f (Cr O )] D H r = [ + (– kJ)] – [ + (– )] D H r = – kJ + kJ D H r = – kJ . D U = q + w D U = – kJ + kJ D U = + 3kJ . Fe + 2HCl → FeCl + H mole of Iron liberates mole of Hydrogen gas . g Iron = mole Iron ∴ n = T = C = K w = – P D V w P nRT P = −       w = – nRT w = – × .

× J w = – . J w = – . kJ . T i = C = K T f = C = K D H = nC p (T f – T i ) ∆ H R ( ) D H = – R .

C + O → CO D H = – a KJ ....................... (i) 2CO + O 2CO D H = –b kJ ....................... (ii) C O CO H → ∆ ? (i) × 2C + 2O → 2CO D H = – 2a kJ .......................

(iii) Reverse of equation (ii) will be 2CO → 2CO + O D H = + b kJ ....................... (iv) (iii) + (iv) 2C + O → 2CO D H = b – 2a kJ ....................... (v) (v) ÷ C O CO H b a kJ → ∆ ( ) . Given : D H C (CH ) = – kJ mol – D H C (C H ) = – kJ mol – Let the mixture contain lit of and lit of propane.

CH + 2O → CO + 2H O x 2x C H + 5O → 3CO + 4H O ( . – x) ( . – x) Volume of oxygen consumed = 2x + ( . – x) = lit 2x + .

– 5x = l . – 3x = 3x = . l x = . l Given mixture contains .

lit of methane and . lit of propane, hence H H H CH lit x lit H C H lit C C C ( ) ( ) ( ) x lit H kJ mol lit C lit lit lit H C kJ mol kJ mol – . KJ mol – C . 4E C–H = kJ mol – E C–H = kJ mol – E C–C + E C–H = kJ mol – E C–C + × = kJ mol – E C–C + = kJ mol – E C–C = kJ mol – .

During compression, energy of the system increases, in isothermal ­condition, to main- tain temperature constant, heat is liberated from the system. Hence q is negative. During compression entropy decreases. During compression work is don e on the system, hence w is positive .

∆ = ∆ = ∆ ∆ S H H S J mol J mol K K C b b . In CaCO (S) → CaO(S) + CO (g), entropy change is positive. in option d, A solid reactant gives a gaseous product. Hence the entropy change is expected to be maximum for this process.

. D G = D H – T D S At 300K D G = 30000 J mol – – K × J K – mol – D G = above K ; ∆G will be negative and reaction becomes spontaneous. I Keys to the short answer questions: . SOLUTION : Given : n = moles V i = ml = .

× × . w = − J w = − . kJ . SOLUTION : Given : T i = K T f = .

K k = . kJ K − m = .5g M m = heat evolved = k ΔT = k (T f − T i ) = . kJ K − ( . − )K = .125kJ ΔH c = × kJ mol − ΔH c = kJ mol − .

SOLUTION : Given : T sys = C = ( + ) = K T surr = C = ( + ) = K q = J ΔS sys = q T sys = − = − . JK − ΔS surr = q T sys = + = + . JK − ΔS univ = ΔS sys + ΔS surr ΔS univ = − . JK − + .

JK − ΔS univ = . JK − . SOLUTION : Given : n = 1mole P = . atm V= Lit T = ?

q= J ΔS = q ΔS = q PV nR       ΔS = nRq PV ∆= × S lit atm K J atm lit ∆= × S lit atm K J atm lit ΔS = . JK − . SOLUTION: Given : ΔH f (NaCl) = . kJ = 30400 J mol − ΔS f (NaCl) = .

JK − mol − T f = ? ∆ = ∆ = ∆ ∆ S H H S f f f f f f J mol J K mol f 30400 T f = . K . SOLUTION : Given C H + 5O → 3CO + 4H O ∆ H C = − .

kJ mol − −−−−−( ) C + O → CO ∆ H f = − . kJ mol − −−−−−( ) H + O → H O ∆ H f = − . kJ mol − −−−−− ( ) 3C + 4H → C H ∆ H C = ? ( ) × ⇒ 3C + 3O → 3CO ∆ H f = − .

kJ −−−−−( ) ( )× ⇒ H + 2O → 4H O ∆ H f = − . kJ −−−−−( ) ( ) + ( ) − ( ) ⇒ 3C + 3O + 4H + 2O + 3CO + 4H O → 3CO + 4H O + C H + 5O ∆ H f = − . − . − (− .

) kJ 3C + H → C H ∆ H f = − . kJ Standard heat of formation of propane is ∆ H f (C H ) = − . kJ . S.

No Liquid Boiling points ( C) ΔH ( kJ mol − ) . Ethanol . + . .

Toluene . + . SOLUTION : For ethanol : Given : T b = . C = ( .

+ ) = . K ΔH V (ethanol) = + . kJ mol − ∆ = ∆ S H b ∆ = + S kJ mol K S J mol K 42400 ΔS V = + . J K − mol − For Toluene : Given : T b = .

C = ( . + ) = . K ΔH V (toluene) = + . kJ mol − ∆ = ∆ S H b ∆ = + S kJ mol K ∆ = + S J mol K 35200 ΔS V = + .

J K − mol − . Solution : Given : ΔH = . kJ mol − = 30560 J mol − ΔS = . × - kJK − mol − T = ?

at which ΔG= ΔG = ΔH − TΔS = ΔH − TΔS H S = ∆ ∆ kJmol kJK mol . × - T = K (i) At 4589K ; ΔG = the reaction is in equilibrium. (ii) at temperature below K , ΔH > T Δ S ΔG = ΔH − T Δ S > , the reaction in the forward direction, is non spontaneous. In other words the reaction occurs in the backward direction.

. Solution : Given T = 400K ; ΔH = . kJ mol − =77200 J mol − ; ΔS = JK − mol − ΔG = − . RT log K eq log K G RT eq = – ∆ log K H T S RT eq = − ∆ −∆