Title DPP-4 SOLUTION.pdf ✅

CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 4 1. (a) Ice takes up heat to melt and thus, enthalpy change is +ve. 3 (a) When ∆H = ― ve, ∆S = + ve and ∆G = ― ve than reaction is spontaneous 4 (d) KE = (3/2)RT 5 (b) C(s) + O2(g)→CO2(g);∆H = r ...(i) CO(g) + 1 2 O2(g)→CO2(g); ∆H = s ...(ii) C(s) + 1 2 O2(g)→CO(g); ∆H = ? Subtract Eq. (ii) from Eq. (i) C(s) + O2(g)→CO2(g);∆H = r CO(g) + 1 2 O2→CO2(g); ∆H = s ― ― ― ― _______________________________________ C(s) + 1 2 O2(g)→CO(g); ∆H = r ― s 6 (d) Cu(g)→Cu +(g) + e, ∆H = 745 kJ mol―1 I(g) +e →I ―(g); ∆H = ―295 kJ mol―1 Adding Cu +(g) + I ―(g)→CuI(g); ∆H = ―446 kJ mol―1 Cu(g) + I(g) →CuI(g); ∆H = 4 kJ mol―1 7 (a) Topic :- THERMODYNAMICS Solutions
Entropy of universe is tending towards maximum. 9 (a) ∆H(reaction) = ∆Hf(diamond) ― ∆Hf(graphite) = 1.896 – 0.23 = 1.666 kJ/ mol 10 (b) p=1 atm ∆V = (50 ― 15) = 35 L ∴ W = ―p.∆V = ―1 × 35 = ―35 Latm Hence, work done by the system on the surroundings is equal to 35 L-atm. 11 (d) The product possesses maximum energy and thus, least stable. 12 (d) By eq. [(i) + 2 × (ii)] – (iii), C + 2H2 ⟶CH4; ∆H = 74.1 kJ 13 (a) For the equation, H2 + S + 2O2→H2SO4 Eqs. (i) +(ii)+(iii)+(iv) ∆H = ―287.3 + (298.2) + ( ―98.7) + ( ― 130.2) = 814.4 kJ 14 (d) (a) For isochoric process, ∆V = 0 W = p∆V = 0 ∴ ∆E = Q (b) For adiabatic process, Q = 0 ∆E = W (c) For isothermal process, ∆T = 0 and ∆E = 0 Q = ― W (d) For cyclic process, state functions like ∆E = 0
Q = ― W 16 (a) ∆G = ∆H ― T∆S = ―ve ― ve = ―ve 17 (a) F2 + 1 2 O2 ⟶F2O; ∆H = + ve. 18 (c) Two equivalent of each are used. 19 (a) Isothermally (at constant temperature) and reversible work. W = 2.303 nRTlog p2 p1 = 2.303 × 1 × 2 × 300log 10 2 = 2.303 × 600 × log 5 = 965.84 At constant temperature, ∆E = 0 ∆E = q + W, q = ―W = ―965.84 cal 20 (d) The gaseous phase have more entropy and thus, ∆S is +ve in (a) and (b). Also decrease in pressure increases disorder and thus, ∆S is +ve in (c). In (d) the disorder decreases in liquid state due to decrease in temperature. Thus, ∆S = ― ve.