Chapter 6 Thermodynamics Solutions
Question - 11 : - Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heatreleased upon formation of 35.2 g of CO2 from carbon anddioxygen gas.
Answer - 11 : -
Formation of CO2 from carbon anddioxygen gas can be represented as:
(1 mole = 44 g)
Heat released on formation of 44 g CO2 = –393.5 kJ mol–1
Heat released on formation of 35.2 g CO2
=–314.8 kJ mol–1
Question - 12 : - Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110 kJ mol–1, – 393 kJ mol–1, 81 kJ mol–1 and 9.7 kJ mol–1 respectively. Findthe value of ΔrH for the reaction:
N2O4(g) + 3CO(g)
N2O(g) + 3CO2(g)
Answer - 12 : -
ΔrH for a reaction isdefined as the difference between ΔfH value of productsand ΔfH value of reactants.
For the given reaction,
N2O4(g) + 3CO(g) N2O(g) + 3CO2(g)
Substituting the values of ΔfH for N2O, CO2, N2O4, and CO from thequestion, we get:
Hence,the value of ΔrH for the reaction is
.
Question - 13 : - Given
; ΔrHθ = –92.4 kJ mol–1
What is the standard enthalpy of formationof NH3 gas?
Answer - 13 : -
Standard enthalpy offormation of a compound is the change in enthalpy that takes place during theformation of 1 mole of a substance in its standard form from its constituentelements in their standard state.
Re-writing the given equation for 1 mole ofNH3(g),
Standard enthalpy of formation of NH3(g)
= ½ ΔrHθ
= ½ (–92.4 kJ mol–1)
=–46.2 kJ mol–1
Question - 14 : - Calculate the standard enthalpy of formationof CH3OH(l) from the followingdata:
CH3OH(l) + 
O2(g) CO2(g) + 2H2O(l) ; ΔrHθ = –726 kJ mol–1C(g) + O2(g) CO2(g) ; ΔcHθ = –393 kJ mol–1
H2(g) +
O2(g) H2O(l) ; ΔfHθ = –286 kJ mol–1.
Answer - 14 : -
The reaction that takes place during theformation of CH3OH(l) can be written as:
C(s) + 2H2O(g) + O2(g) CH3OH(l) (1)The reaction (1) can be obtained from the given reactions by following thealgebraic 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
ΔfHθ [CH3OH(l)] = –239 kJ mol–1
Question - 15 : - Calculate the enthalpychange for the process
CCl4(g) → C(g) + 4Cl(g)
and calculate bond enthalpy of C–Cl in CCl4(g).
ΔvapHθ (CCl4) = 30.5 kJ mol–1.
ΔfHθ (CCl4) = –135.5 kJ mol–1.
ΔaHθ (C) = 715.0 kJ mol–1, where ΔaHθ is enthalpy ofatomisation
ΔaHθ (Cl2) = 242 kJ mol–1
Answer - 15 : -
The chemical equationsimplying to the given values of enthalpies are:
ΔvapHθ = 30.5 kJ mol–1
ΔaHθ = 715.0 kJ mol–1
ΔaHθ = 242 kJ mol–1
ΔfH = –135.5 kJ mol–1
Enthalpychange for the given process 
can be calculated usingthe 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
Question - 16 : - For an isolated system, ΔU = 0,what will be ΔS?
Answer - 16 : -
ΔS will be positive i.e.,greater than zero
SinceΔU = 0, ΔS will be positive and the reaction will bespontaneous.
Question - 17 : - For the reaction at 298 K,
2A + B → C
ΔH = 400 kJ mol–1 and ΔS =0.2 kJ K–1 mol–1
At what temperature will the reaction becomespontaneous considering ΔH and ΔS to be constant overthe temperature range?
Answer - 17 : -
From the expression,
ΔG = ΔH – TΔS
Assuming the reaction at equilibrium, ΔT forthe reaction would be:
(ΔG = 0 at equilibrium)
T = 2000 K
Forthe reaction to be spontaneous, ΔG must be negative. Hence, for thegiven reaction to be spontaneous, T should be greater than2000 K.
Question - 18 : - For the reaction,
2Cl(g) → Cl2(g), what are the signsof ΔH and ΔS ?
Answer - 18 : -
ΔH and ΔS arenegative
The given reaction represents the formationof 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. Sincespontaneity is decreased, ΔS is negative for the given reaction.
Question - 19 : - For the reaction
2A(g) + B(g) → 2D(g)
ΔUθ = –10.5 kJ and ΔSθ= –44.1 JK–1.
Calculate ΔGθ for the reaction,and predict whether the reaction may occur spontaneously.
Answer - 19 : -
For the given reaction,
2 A(g) + B(g) → 2D(g)
Δng = 2 – (3)
= –1 mole
Substituting the value of ΔUθ in the expression ofΔH:
ΔHθ = ΔUθ + ΔngRT
= (–10.5 kJ) – (–1) (8.314 × 10–3 kJ K–1 mol–1) (298 K)
= –10.5 kJ – 2.48 kJ
ΔHθ = –12.98 kJ
Substituting the values of ΔHθ and ΔSθ in the expression ofΔGθ:
ΔGθ = ΔHθ – TΔSθ
= –12.98 kJ – (298 K) (–44.1 J K–1)
= –12.98 kJ + 13.14 kJ
ΔGθ = + 0.16 kJ
SinceΔGθ for the reaction is positive, thereaction will not occur spontaneously.
Question - 20 : - The equilibrium constant for a reaction is10. What will be the value of ΔGθ? R = 8.314 JK–1 mol–1, T = 300K.
Answer - 20 : -
From the expression,
ΔGθ = –2.303 RT logKeq
ΔGθ for the reaction,
= (2.303) (8.314 JK–1 mol–1) (300 K) log10
= –5744.14 Jmol–1
=–5.744 kJ mol–1