This chapter explains the terms - system and surroundings. It also helps in discriminating between close, open and isolated systems. It also gives explanation about internal energy, work and heat. It also states first law of thermodynamics and express it mathematically. Calculation of energy changes as work and heat contributions in chemical systems are also carried out. It also explains state functions - U, H along with their correlation and experimental measures. Calculation of enthalpy changes for various types of reactions can also be carried out. It also states Hess's law of constant heat summation along with its application. It also helps in differentiating between extensive and intensive properties. Definition of spontaneous and non -spontaneous processes are also given. Explanation of entropy as a thermodynamic state function and its application for spontaneity is also given. It also explains Gibbs energy change and establishes relationship between Gibbs energy change and spontaneity.
In thermodynamics, function of state, state quantity, or state variable is a property of a system that depends only on the current state of the system, not on the way in which the system acquired that state. A state function describes the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of a path.
Functions like p, V, T etc. depend only on the state of a system and not on the path.
Hence, alternative (ii) is correct.
An adiabatic process is one that occurs without transfer of heat or matter between a system and its surroundings & it helps in explaining first law of thermodynamics
Hence, under adiabatic conditions, q = 0.
Therefore, alternative (iii) is correct.
(ii) zero
(iii) < ΔU0
According to the question,
Thus, the desired equation is the one that represents the formation of CH4(g) i.e.,
[-393.5 + 2(-285.8) - (-890.3)] kJ Mol-1
= -74.8 kJ Mol-1
So,Enthalpy of formation of CH4(g) is -74.8 kJ Mol-1
That means answer is (i).
For a reaction to be spontaneous, ΔG should be negative.
ΔG = ΔH- TΔS
According to the question, for the given reaction,
ΔS = positive
ΔH= negative (since heat is evolved)
⇒ ΔG= negative
Therefore, the reaction is spontaneous at any temperature. Hence, answer is (iv).
According to the first law of thermodynamics,
ΔU= q + W (i)
Where, ΔU = change in internal energy for a process
q = heat
W = work
Given, q = + 701 J (Since heat is absorbed)
W = -394 J (Since work is done by the system)
Substituting the values in expression (i), we get
ΔU= 701 J + (-394 J) ΔU = 307 J
Hence, the change in internal energy for the given process is 307 J.
Enthalpy change for a reaction (ΔH) is given by the expression,
ΔH = ΔU + ΔngRT
Where, ΔU = change in internal energy
Δng = change in number of moles
For the given reaction,
Δng = 
= (2 - 1.5) moles
Δng = 0.5 moles
And,
ΔU = -742.7 kJ mol-1
T = 298 K
R = 8.314 x 10-3 kJ mol-1 K-1
Substituting the values in the expression of ΔH:
ΔH = (-742.7 kJ mol-1) + (0.5 mol) (298 K) (8.314 x 10-3 kJ mol-1 K-1)
= -742.7 + 1.2
ΔH = -741.5 kJ mol-1
From the expression of heat (q),
q = m. c. ΔT
Where,
c = molar heat capacity
m = mass of substance
ΔT = change in temperature
Substituting the values in the expression of q:
q = (60/27 mol) (24 Jmol–1 K–1) (20K)
q = 1066.7 J
q = 1.07 kJ
Total enthalpy change involved in the transformation is the sum of the following changes:
(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.
(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.
(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at -10°C.
Total ΔH = Cp [H2OCI] ΔT + ΔHfreezing + Cp[H2O(s)] ΔH
= (75.3 J mol-1 K-1) (0 - 10)K + (-6.03 × 103 J mol-1) + (36.8 J mol-1 K-1) (-10 - 0)K
= -753 J mol-1 - 6030 J mol-1 - 368 J mol-1
= -7151 J mol-1
= -7.151 kJ mol-1
Hence, the enthalpy change involved in the transformation is -7.151 kJ mol-1.
Formation of CO2 from carbon and dioxygen gas can be represented as:
C(s) + O2(g) → CO2(g) ΔfH = -393.5 kJ mol-1
(1 mole = 44 g)
Heat released on formation of 44g CO2 = -393.5 kJ mol-1
= [ -393.5kJ mol-1 / 44g ] x 35.2g
= -314.8 kJ mol-1
ΔrH for a reaction is defined as the difference between ΔfH value of products and ΔfH value of reactants.
ΔrH =
For the given reaction,
N2O4(g)+ 3CO(g) → N2O(g) + 3CO2(g)
ΔrH = [{ΔfH (N2O) + 3ΔfH(CO2)} - {ΔfH(N2O4) + 3ΔfH(CO)}]
Substituting the values of ΔfH for N2O, CO2, N2O4,and CO from the question, we get:
ΔrH = [{81 KJ mol-1 + 3(-393) KJ mol-1} - {9.7KJ mol-1 + 3(-110)KJ mol-1}]
ΔrH = -777.7 KJ mol-1
Hence, the value of ΔrH for the reaction is -777.7 KJ mol-1.
Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.
Re-writing the given equation for 1 mole of NH3(g),
½ N2(g) + 3/2 H2(g) → NH3(g)
= ½ ΔrHθ
= ½ (-92.4 kJ mol-1)
= - 46.2 kJ mol-1
The reaction that takes place during the formation of CH3OH(l) can be written as:
C(s) + 2H2O(g) + ½O2(g) → CH3OH(l) (1)
The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 × equation (iii) - equation (i)
ΔfHθ [CH3OH(l)] = ΔcHθ + 2ΔfHθ [H2O(l)] - ΔrHθ
= (-393 kJ mol-1) + 2 (-286 kJ mol-1) - (-726 kJ mol-1)
= (-393 - 572 + 726) kJ mol-1
= -239 kJ mol-1
The chemical equations implying to the given values of enthalpies are:
(i) CCl4(l) → CCL4(g) ΔvapHθ = 30.5 kJ mol-1
(ii) C(s) → C(g) ΔaHθ = 715.0 kJ mol-1
(iii) Cl2(g) → 2Cl(g) ΔaHθ = 242 kJ mol-1
(iv) C(g) + 4Cl(g) → CCl4(g) ΔfH = -135.5 kJ mol-1
Enthalpy change for the given process C(g) + 4Cl(g) → CCl4(g) can be calculated using the following algebraic calculations as:
Equation (ii) + 2 × Equation (iii) - Equation (i) - Equation (iv)
ΔH = ΔaHθ(C) + 2ΔaHθ (Cl2) - ΔvapHθ - ΔfH
= (715.0 kJ mol-1) + 2(242 kJ mol-1) - (30.5 kJ mol-1) - (-135.5 kJ mol-1)
∴ΔH = 1304 kJ mol-1
Bond enthalpy of C-Cl bond in CCl4(g) = 326 kJ mol-1
ΔS will be positive i.e., greater than zero
Since ΔU= 0, ΔS will be positive and the reaction will be spontaneous.
From the expression,
ΔG = ΔH - TΔS
Assuming the reaction at equilibrium, ΔTfor the reaction would be:
T = (ΔH - ΔG) / ΔS
T = ΔH / ΔS (ΔG = 0 at equilibrium)
= 400 kJ mol-1 / 0.2 kJ K-1mol-1
T= 2000 K
For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.
ΔH and ΔS are negative The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, ΔH is negative.
Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔS is negative for the given reaction.
For the given reaction,
2 A(g) + B(g) → 2D(g)
Δng = 2 - (3) = -1 mole
Substituting the value of ΔU0¸ in the expression of ΔH:
ΔHº = ΔUº + ΔngRT
= (-10.5 kJ) - (-1) (8.314 x 10-3 kJ K-1 mol-1) (298 K)
= -10.5 kJ - 2.48 kJ
ΔHº = -8.02 kJ
Substituting the values of ΔHº and ΔSº¸ in the expression of ΔH:
ΔGº = ΔHº - TΔSº
= -8.-02 kJ - (298 K) (-44.1 J K-1)
= -8.02 kJ + 13.14 kJ
ΔGº = + 5.12 kJ
Since ΔGº for the reaction is positive, the reaction will not occur spontaneously.
From the expression, ΔG0 = -2.303 RTlogKeq
ΔG0 for the reaction,
= (2.303) (8.314 JK-1mol-1) (300 K) log10
= -5744.14 Jmol-1
= -5.744 kJ mol-1
The positive value of ΔrH indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (N2 and O2). Hence, NO(g) is unstable. The negative value of ΔrH indicates that heat is evolved during the formation of NO2(g) from NO(g) and O2(g). The product, NO2(g) is stabilized with minimum energy. Hence, unstable NO(g) changes to stable NO2(g).
It is given that 286 kJ mol-1of heat is evolved on the formation of 1 mol of H2O(l). Thus, an equal amount of heat will be absorbed by the surroundings.
qsurr = +286 kJ mol-1
Entropy change (ΔSsurr) for the surroundings = qsurr / 7
= 286 kJ mol-1 / 298k