Chapter 3: Thermodynamic and Chemical Equilibrium¶

Problem: 1, Page no: 105¶

In :
# Variables
q = 120                     # Heat from surrounding, cal
W = 70                      # Work done, cal

# Solution
delta_E = q - W

print "Change in internal Energy", delta_E, "cals."
Change in internal Energy 50 cals.

Problem: 2, Page no: 105¶

In :
print "CH4 (g) + 2O2 (g) -> CO2 (g) + 2H20 (l)"
delta_n = 1 - (1 + 2)
solution = - 2 * 2 * 298                        # cals
print "Delta H - Delta E is:", solution, "cals"
CH4 (g) + 2O2 (g) -> CO2 (g) + 2H20 (l)
Delta H - Delta E is: -1192 cals

Problem: 3, Page no: 105¶

In :
# Variables
delta_G = -16.0                 # Kelvin cal
delta_H = -10.0                 # Kelvin cal
T = 300                         # Kelvin

# Solution
delta_S = (delta_H - delta_G) * 10 ** 3 / T    # cal/deg
new_T = 330                     # Kelvin
new_delta_G = (delta_H * 10 ** 3) - new_T * delta_S

print "The free energy at 330K is:", "{:.2e}".format(new_delta_G), "K cal"
The free energy at 330K is: -1.66e+04 K cal

Problem: 4, Page no: 106¶

In :
# Variables
delta_S = -20.7                     # cal /deg /mol
delta_H = -67.37                    # K cal
T = 25                              # deg C

# Solution
T += 273                            # K
delta_G = delta_H - (T * delta_S * 10 ** -3)
print "The change in free energy at 25deg C is:", delta_G, "K cal / mol"
The change in free energy at 25deg C is: -61.2014 K cal / mol

Problem: 5, Page no: 106¶

In :
# Variables
wt = 1                  # g
delta_H = 149           # joules

# Solution
delta_H_fusion = delta_H * (10 * 12 + 8 * 1)
print "Enthalpy of fusion of naphthalene:", delta_H_fusion * 10 ** -3, "kJ/mol"
Enthalpy of fusion of naphthalene: 19.072 kJ/mol

Problem: 6, Page no: 106¶

In :
# Variables
delta_H_acetylene = 230             # kJ/mol
delta_H_benzene = 85                # kJ/mol
T = 298                             # K

# Solution
delta_H = delta_H_benzene - 3 * delta_H_acetylene
print "The enthalpy change for the reaction is:", delta_H, "kJ/mole"
The enthalpy change for the reaction is: -605 kJ/mole

Problem: 7, Page no: 106¶

In :
# Constant
delta_H_vap = 2.0723                # kJ/g
Tb = 373                            # K

# Solution
delta_H_vap *= 18                   # kJ/mol
delta_S = delta_H_vap / Tb
delta_G = delta_H_vap - Tb * delta_S
delta_S *= 1000

print "The Entropy change is:", "{:.1f}".format(delta_S), "J/mol/K"
print "The Free Energy change is:", delta_G, "kJ/mol"
The Entropy change is: 100.0 J/mol/K
The Free Energy change is: 0.0 kJ/mol

Problem: 8, Page no: 106¶

In :
import math

# Constant
R = 1.987               # cal /K /mol

# Variables
moles = 5
Vo = 4                  # litres, Initial Volume
Vf = 40                 # litres, Final Volume
T = 27                  # deg C

# Solution
print "dS = nRln(V2 / V1)"
dS = moles * R * 2.303 * math.log10(Vf / Vo)
print "The change in entropy is:", "{:.2f}".format(dS), "cal / degree"
dS = nRln(V2 / V1)
The change in entropy is: 22.88 cal / degree

Problem: 9, Page no: 107¶

In :
# Variables
wt = 10                     # g
heat_abs = 4.5              # K

# Solution
mole = 10 / 100.0             # mol
delta_H = heat_abs / mole
print "The heat of the reaction is:", delta_H, "K cal / mol"
The heat of the reaction is: 45.0 K cal / mol

Problem: 10, Page no: 107¶

In :
import math

# Constant
R = 8.314                   # J / K

# Variables
V_O2 = 2.8                  # litres
V_H2 = 19.6                 # litres

# Solution
na = V_O2 / 22.4            # mol
nb = V_H2 / 22.4            # mol
Xa = na / (na + nb)
Xb = nb / (na + nb)
delta_S = (- R) * (na * math.log(Xa) + nb * math.log(Xb))

print "The increase in entropy on mixing is:", "{:.3f}".format(delta_S), "J / K"
The increase in entropy on mixing is: 3.132 J / K

Problem: 11, Page no: 107¶

In :
# Solution
print "For 1 mole of ideal gas,"
print "\tPV = RT or T = (PV) / R\n"
print "Differentiating with respect to V at constant P,"
print "\t[dT/dV]p = P/R\n"
print "Differentiating again with respect oto P at constant V"
print "\t[d2T/(dV*dP)] = 1/R\n"
print "Now differectiating with respect to P at constant V,"
print "\t[dT/dP]v = V/R\n"
print "Differentiating again with respect to V at constant P,"
print "\t[d2T/(dV*dP)] = 1/R\n"
print "From equations we get:"
print "\t[d2T/(dV*dP)] = [d2T/(dV*dP)]\n"
print "Hence, dT is a perfect differential."
For 1 mole of ideal gas,
PV = RT or T = (PV) / R

Differentiating with respect to V at constant P,
[dT/dV]p = P/R

Differentiating again with respect oto P at constant V
[d2T/(dV*dP)] = 1/R

Now differectiating with respect to P at constant V,
[dT/dP]v = V/R

Differentiating again with respect to V at constant P,
[d2T/(dV*dP)] = 1/R

From equations we get:
[d2T/(dV*dP)] = [d2T/(dV*dP)]

Hence, dT is a perfect differential.

Problem: 12, Page no: 108¶

In :
# Variables
delta_G_25 = - 85.77                # k J, Free Energy at 25 C
delta_G_35 = - 83.68                # k J, Free Energy at 35 C
Ti = 273 + 25                       # K
Tf = 273 + 35                       # K

# Solution
print "Equating the entropy change at both the temperatures."
print "(delta_H + delta_G_25) / Ti = (delta_H + delta_G_35) / Tf"
delta_H = - 148
print "The change in enthalpy for the process at 30C is", delta_H, "kJ"
Equating the entropy change at both the temperatures.
(delta_H + delta_G_25) / Ti = (delta_H + delta_G_35) / Tf
The change in enthalpy for the process at 30C is -148 kJ

Problem: 13, Page no: 108¶

In :
# Constants
Lv = 101                    # cal /g, Latent headt of vap.
mwt = 78                    # molecular weight of benzene

# Variable
moles = 2
Tb = 80.2                   # C, boiling point of benzene

# Solution
Tb += 273                   # K
delta_H = Lv * mwt
delta_S = delta_H / Tb
delta_G = delta_H - Tb * delta_S
print "delta_S =", "{:.2f}".format(delta_S), "cal / K"
print "delta_G = delta_A =", delta_G
delta_S = 22.30 cal / K
delta_G = delta_A = 0.0

Problem: 14, Page no: 109¶

In :
# Variables
V1 = 6                      # dm^3
V2 = 2                      # dm^3
T1 = 27                     # C
moles = 5

# Solution
print "T1*V1 ^ (gamma - 1) = T2 * V2 ^ (gamma - 1)"
T1 += 273                   # K
T2 = T1 * (V1 / V2) ** (8.314 / 20.91)
print "The final temperature is", "{:.1f}".format(T2), "K"
q = 0                       # Adiabatic process
delta_E = - moles * 20.91 * (T2 - T1)
delta_E /= 1000
print "q = ", q
print "Change is Energy is", "{:.2f}".format(delta_E), "kJ / mol"
W = - delta_E
print "W = ", "{:.2f}".format(delta_E), "kJ / mol"
T1*V1 ^ (gamma - 1) = T2 * V2 ^ (gamma - 1)
The final temperature is 464.3 K
q =  0
Change is Energy is -17.18 kJ / mol
W =  -17.18 kJ / mol

Problem: 15, Page no: 109¶

In :
import math

# Constant
R = 8.314                       # J / K mol

# Variables
mole = 1
V1 = 5                          # dm^3
V2 = 10                         # dm^3
T = 300                         # K

# Solution
print "For isothermal and reversible process,"
delta_E = delta_H = 0
delta_A = delta_G = - 2.303 * mole * R * T * math.log10(V2 / V1)
q = W = - delta_G
print "delta_E = delta_H =", delta_H
print "delta_G = delta_A =", "{:.3f}".format(delta_G), "J / mol\n"
print "For isothermal and reversible expansion"
print "q = W = -delta_G =", "{:.3f}".format(W), "J / mol"
For isothermal and reversible process,
delta_E = delta_H = 0
delta_G = delta_A = -1729.159 J / mol

For isothermal and reversible expansion
q = W = -delta_G = 1729.159 J / mol

Problem: 16, Page no: 109¶

In :
import math

# Constant
R = 8.314                       # J / K mol

# Variables
n = 5                           # moles
T = 27                          # C
V1 = 50.0                         # L, Initial Volume
V2 = 1000                       # L, Final Volume

# Solution
T += 273
delta_G = 2.303 * n * R * T * math.log10(V1 / V2)
delta_G /= 1000
print "The free energy change is", "{:.3f}".format(delta_G), "k J"
The free energy change is -37.367 k J

Problem: 17, Page no: 110¶

In :
# Variables
delta_H_neu = - 51.46                   # k J/mol, neutralization
delta_H_ion = - 57.1                    # k J/mol, ionization

# Solution
delta_H = - delta_H_ion + delta_H_neu
print "The head of ionization for NH4OH is", delta_H, "kJ / mol"
The head of ionization for NH4OH is 5.64 kJ / mol

Problem: 18, Page no: 110¶

In :
print "For 1 mole of an ideal gas,"
print "PV = RT or V = (RT)/P"
print "(dV/ dP) = -(RT/P^2)\t (at constant temperature)"
print "(d^2V/ (dP*dT)) = -(R/ P^2)"
print "(dV/ dT) = (R/ P)\t (at constant pressure)"
print "(d^2V/ (dT*dP)) = -(R/ P^2)\n"
print "(d^2V/ (dT*dP)) = (d^2V/ (dP*dT))\t[From above equations]"
print "Hence, dV is an exact differential."
For 1 mole of an ideal gas,
PV = RT or V = (RT)/P
(dV/ dP) = -(RT/P^2)	 (at constant temperature)
(d^2V/ (dP*dT)) = -(R/ P^2)
(dV/ dT) = (R/ P)	 (at constant pressure)
(d^2V/ (dT*dP)) = -(R/ P^2)

(d^2V/ (dT*dP)) = (d^2V/ (dP*dT))	[From above equations]
Hence, dV is an exact differential.

Problem: 19, Page no: 110¶

In :
print "Let P1, V1, T1 and P2, V2, T2 b the initial and final",
print "state, respectively of the system"
print "W rev = nRT[P1/ P2 - 1]"
print "W irr = nRT[1 - P2/ P1]"
print "W rev - W irr = [nRT/(P1P2)]*(P1 - P2)^2"
print "Because RHS of the above equation is always positive,"
print "W rev > W irr"
Let P1, V1, T1 and P2, V2, T2 b the initial and final state, respectively of the system
W rev = nRT[P1/ P2 - 1]
W irr = nRT[1 - P2/ P1]
W rev - W irr = [nRT/(P1P2)]*(P1 - P2)^2
Because RHS of the above equation is always positive,
W rev > W irr

Problem: 20, Page no: 111¶

In :
# Solution
Eq_HI = 1.56 / 2
Eq_H2 = 0.22 / 2
Eq_I2 = 0.22 / 2
Kc = Eq_H2 * Eq_I2 / (Eq_HI ** 2)
print "The equilibrium constant for the dissociation reaction",
print "{:.4f}".format(Kc)
The equilibrium constant for the dissociation reaction 0.0199

Problem: 21, Page no: 111¶

In :
# Variables
Kc = 0.5                    # / mole^2 litre^2
T = 400                     # K
R = 0.082                   # litre atm degree^-1 mole^-1

# Solution
Kp = Kc * (R * T) ** (-2)

print "The given equilibrium is"
print "\t\tN2(g) + 3H2(g) <--> 2NH3(g)"
print "Kp is", "{:.3e}".format(Kp)
The given equilibrium is
N2(g) + 3H2(g) <--> 2NH3(g)
Kp is 4.648e-04

Problem: 22, Page no: 111¶

In :
# Variables
solubility = 7.5 * 10 ** - 5        # mol L^-1

# Solution
Ksp = 4 * (solubility ** 3)
print "Solubility product of the salt is", "{:.4e}".format(Ksp), "mol^3 / L^-3"
Solubility product of the salt is 1.6875e-12 mol^3 / L^-3

Problem: 23, Page no: 111¶

In :
# Variables
Ti = 25                         # C
S = 0.00179            # g / L

# Solution
S /= 170                        # mol / L
Ksp = S ** 2
print "Solubility product at 25 C is", "{:.4e}".format(Ksp), "mol^2 / L^-2"
Solubility product at 25 C is 1.1087e-10 mol^2 / L^-2

Problem: 24, Page no: 111¶

In :
# Variables
Ksp = 8 * 10 ** - 5               # Solubility product PbBr2
disso = 80 / 100                  # % dissociation

# Solution
S = (Ksp / 4) ** (1 / 3.0)          # Solubility is 100%
S_80 = S * (80 / 100.0)
S_per_g = S_80 * 367 - 1.621
print "Solubility in gm per litre is", "{:.3f}".format(S_per_g), "gm / litre"
Solubility in gm per litre is 6.349 gm / litre

Problem: 25, Page no: 112¶

In :
# Variables
n_acid = 1                          # mole
n_alcohol = 1                       # mole
T = 25                              # C
x = 0.667                           # mole

# Solution
print "Kc = [CH3COOC2H][H2O] / ([CH3COOH][C2H2OH])"
Kc = 4
print "[CH3COOH] = (2 - x) / V"
print "[C2H5OH] = (1 - x) / V"
print "[CH3COOC2H5] = [H20] = x / V"
print "3x^2 - 12x + 8 = 0"
print "x =", 2.366, "or", 0.634
print "0.634 mole of ester would be formed, because the other value,",
print "x = 2.366, is not permissible."
Kc = [CH3COOC2H][H2O] / ([CH3COOH][C2H2OH])
[CH3COOH] = (2 - x) / V
[C2H5OH] = (1 - x) / V
[CH3COOC2H5] = [H20] = x / V
3x^2 - 12x + 8 = 0
x = 2.366 or 0.634
0.634 mole of ester would be formed, because the other value, x = 2.366, is not permissible.

Problem: 26, Page no: 113¶

In :
import math

# Variables
n_acid = 0.2                          # mole
n_salt = 0.10                         # mole
Ka = 1.8 * 10 ** -5

# Solution
pH = - math.log10(Ka) + math.log10(n_salt / n_acid)
print "The pH of acidic buffer is", "{:.3f}".format(pH)
The pH of acidic buffer is 4.444

Problem: 27, Page no: 113¶

In :
import math

# Variables
n_salt = 0.02               # mole
n_base = 0.2                # mole
pKb = 4.7

# Solution
pOH = pKb + math.log10(n_salt / n_base)
pH = 14 - pOH
print "pH of a buffer solution is", pH
pH of a buffer solution is 10.3

Problem: 28, Page no: 113¶

In :
# Variables
n_H2 = 8.07                     # moles
n_I2 = 9.08                     # moles
T = 448                         # C
n_eqHI2 = 13.38                 # moles

# Solution
x = n_eqHI2 / 2 + 6.69
Kc = n_eqHI2 ** 2 / (n_H2 * n_I2)

print "H2   +   I2 <--> 2HI"
print "1         0       0"
print "1 - 2xx   x       x"
print "x/(1 - 2x) = (1/Kc)^0.5"
print "Dissociation constant of HI is 106.75 x 10^-3"
H2   +   I2 <--> 2HI
1         0       0
1 - 2xx   x       x
x/(1 - 2x) = (1/Kc)^0.5
Dissociation constant of HI is 106.75 x 10^-3