Solutions Tutorial 1

1st Law of thermodynamics and other basic concepts (04-03-2015)

1. w = nRTln(p1/p2) = 1 x 8.314 x 273 ln (1013.25/41) = 7.28 KJ Isothermal process, so, dT = 0 ∆U = nCvdT = 0 ∆H = nCpdT = 0 Now q = U + w = w = 7.28 kJ

2. w = nRTln(v2/v1) = 41.96 KJ

3. w = nRTln(v2/v1) = 4 x 1.98 x T x ln (10v1/v1) 40000 Cal = 4 x 1.98 x T x ln (10v1/v1) Or, T = 2193.4 K V1 = nRT1/P1 = 7.202 lit.

4. C2H4(g) + 3O2(g) = 2CO2(g) + 2H2O(l)……………(i) 2H2O(g) + O2(g) = 2H2O(l)…………………………(ii) 2C2H6(g) + 7O2(g) = 4CO2(g) + 6H2O(l)…………..(iii) (i) + (ii)/2 - (iii)/2

C2H4(g) + H2(g) = C2H6(g) ∆H = ∆H(i) + ∆H(ii)/2 - ∆H(iii)/2 = -32.8 Kcal.

5. The reaction is

Zn + 2H+ → Zn2+ + H2

Hence 1 mole H2 gas is liberated for every 1 mole Zn used.

Therefore work done at constant pressure is given by

w = -p∆V = -pVgas = - nRT = (5g / 65.4 g mol-1)* (8.3145 JK-1 mol-1) * (23+ 273) K = -188J

6. ∆n=1-1.5 = -0.5

∆H =∆U + ∆ (PV) ∆H =∆U + ∆nRT ∆U°= ∆H° - ∆nRT = (-282.97 kJ mol-1) – (-0.5 ˣ 8.314 JK-1 mol-1 ˣ 298.15 K) = - 281.73 KJmol-1

7. Final volume of the cylinder is

V(final) =

(1.000 mol)(8.314 KJ mol-1)( 273.15 K)

= 0.0151 m3

1.50 × 105 Pa

Since P (transmitted) cannot exceed Pext,

W (max.) = - Pext ΔV = - (1.50 × 105 Pa) (0.0151 m3−0.0224 m3) = 2270 J

SinceP (transmitted) will exceed the equilibrium pressure that the gas would have at any volume,

W (min.) = - (1.000 mol) (8.314 KJ mol-1) (273.15)

= 890 J

(The actual value must lie between these values, since it depends on the rate of the process and other things that we cannot evaluate using thermodynamics)

8. Balanced equation of the reaction is

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)

ln(0.0151 m3) (0.0224 m3)

0 = CO2 (g) + 2H2O (l) - CH4 (g) - 2O2 (g)

So,

−890.36 KJ mol-1 = ∆fH° (CO2) +2∆fH° (H2O) + (-1) ∆fH° (CH4) + (-1) ∆fH° (O2)

Since standard enthalphy of gaseous O2 [∆fH° (O2)] = 0, at 273.15 K. and applying the given enthalpy values of CH4 and H2O.

∆fH° (CH4) = −890.36 KJ mol-1 + (−393.522 KJ mol-1) + 2(−285.830 KJ mol-1) + (−2) (0)

= −74.82 KJ mol-1

9. Assume supercooled water is metastable and treat it as though it were at equilibrium. We calculate ∆H along the reversible path.

Step 1 is the reversible heating of the supercooled liquid to 0 °C, the equilibrium freezing temperature.

Step 2 is the reversible freezing of the system at 0 °C, and

Step 3 is the reversible cooling of the solid to−15 °C. The enthalpy change for the irreversible process is equal to the enthalpy change of this reversible process.

(Figure taken from text book)

Enthalpy change in step 1:

∆H1 = = Cp(l) ∆T

= (2.000 mol) (75.48 KJ mol-1) (15 K) = 2264 J

Enthalpy change in step 2:

∆H2 = (2.000mol) (18.02) (−333.5 KJ mol-1) = -12020 J

Enthalpy change in step 3:

∆H3 = =

= (2.000 mol) (37.15 KJ mol-1) (15 K) = 1114 J

The enthalpy change of total process is

∆H = ∆H1+∆H2+∆H3 = −10870 J = −10.87 kJ

Since process is at constant pressure, q = ∆H

10.

= ( ) =( ) = 1.062

T2 = (1.062) (298 K) = 316 K

tC,2 = 43 °C

Tutorial 2 Phase Equilibria, 2nd law,chemical equilibria

Solution 11:

Solution 12:

Solution 13: Spontaneous means ∆G should be negative

∆G = nRTln(p2/p1)

= 2.0 x 8.314 x 273 x ln (1/2)

= -3.147.06 J

Solution 14: We have 50 g copper at 393 K and 100 g copper at 303 K. The temperature of the two bodies when they have come to thermal equilibrium can be calculated as follows.

Heat gain = Heat loss

m1Cp(∆T1) = m2Cp(∆T2)

i.e. m1∆T1 = m2∆T2

100 g x (T-303 K) = 50 g x (393K-T)

T = 333 K

∆Sh = nCp,mln (T/Th)

= (50g/63gmol-1)[(0.4184 Jg-1K-1)(63 gmol-1)] x 2.303 log (333/393)

= -3.466 JK-1

∆Sc = nCp,mln(T/Tc)

= (100g/63 gmol-1)[(0.4184 Jg-1K-1)(63 gmol-1)] x 2.303log (333/303)

= 3.951 JK-1

∆Stotal = ∆Sc + ∆Sh = 0.485 JK-1

Solution 15: a) ∆S(gas)= nRln(Vf/Vi) =(21g/39.95g mol-1)* (8.314 JK-1mol-1) ln(4.6/1.2) = +5.867 JK-1

∆S(surrounding)= -∆S(gas)= -5.867 JK-1 (REVERSIBLE)

∆S(total)= 0

b) ∆S(gas) = +5.867 JK-1 (S is a state function)

∆S(surrounding) = 0 (no change in surroundings)

∆S(total)= +5.867 JK-1

c) qrev = 0 so ∆S(gas) = 0

∆S(surrounding) = 0 (no heat is transferred to the surroundings)

∆S(total)= 0

Solution 16: a) ε = 1- Tc/Th = 1 – ( 500K/1000K) = 0.5

b) Maximum work = ε ןqh 1 * 0.5 = ן kJ = 0.5 kJ

c) εmax = εrev ןwmax ן = ןqh ן – ןqc,min ן

ן wmaxן - ן qhן = ן qc,minן

= 1.0 kJ – 0.5 kJ = 0.5kJ

Solution 17:

dP/dT = ΔvapH/ T (Vg-Vl) (Clapeyron Equation)

= (40,690 J mol-1)/ (373.15 K)(30.180 × 10-3 m3 mol-1)

= 3613 Pa K-1

liq A  and  B

A B

liqA  &  BSolid  A

liqA  &  B

Solid  AB

Eutectic  1

Solid  A  and  AB

liqA  &  B

Solid  AB

liqA  &  BSolid  B

Eutectic  2

Solid  B  and  AB

Thus dT/dP = 1/3613 =2.768 × 10-4 K Pa-1

Solution 18:

Tutorial 3 Elementary kinetics & Stat. Thermodynamics

Solution 19: a) dξ/dt = 0.001 mol/ 0.01 s = 0.1 mol s-1

b) v = (1/V) dξ/dt = 0.1 mol s-1/ 0.25 L = 0.4 mol L-1 s-1

c) d[H2] / dt = - 0.4 mol L-1 s-1

d[Br2] / dt = -0.4 mol L-1 s-1

d[HBr] / dt = 0.8 mol L-1 s-1

Solution 20:

Solution 21:

Solution 22:

Here solution method is same, but the values given in problem sheet are different. Please substitute those values and get answer.

Solution 23:

Substitute half – life in sec and all other values and solve for T.

Solution 24: Number of configurations of combined system W = W1W2 W = 2*1040 S = k lnW ; S1 = k lnW1 ; S2 = k lnW2 S = k ln(2 ˣ 1040) = 92.8k = 1.282*10-21 J K-1

S1 = k ln(1020) = 0.637*10-21 J K-1

S2 = k ln(2*1020) = 0.645*10-21 J K-1

 

Solution 25:

= = 3

= 1/e and solve for T

T = 7.26 K

Solution 26: a) The ratio of populations is given by the Boltzmann factor

= exp( ) = and =

At 25 K = 0.368 and = 0.135

b) The molecular partition function is

q = ∑ = 1 + +

At 25 K q = 1.503

c) The molar internal energy is

Um = Um(0) - where β = (kT)-1

At 25 K Um = Um(0) = 88.3 J mol-1

Solution 27:

The relative population of states is given by Boltzmann distribution

= exp( ) = exp( ) and solve for T

Having 15 % of the molecules in the upper level means

Hence T = 213 K

Tutorial 4 Kinectics of complex rxns, Reaction rate theories & Surface chemistry

Solution 28:

Solution 29:

Solution 30:

Solution 31:

Solution 32:

Solution 33:

 

 

 

Solution 34:  

Solution 35: Lamgmiur isotherm is θ = Kp/(1 + Kp)