Thermodynamics
5.1 – 5.5 Fundamentals & Enthalpy
5.1 A thermodynamic state function is a quantity…
Correct Answer: (ii) whose value is independent of path.
Explanation: A state function (like Pressure, Volume, Temperature, Enthalpy, Internal Energy) depends only on the initial and final states of the system, not on how the change was accomplished (the path).
5.2 Condition for adiabatic process.
Correct Answer: (iii) q = 0.
Explanation: An adiabatic process is defined as a process where there is no exchange of heat between the system and its surroundings. Therefore, heat ($q$) is zero.
5.3 Enthalpies of all elements in standard states.
Correct Answer: (ii) zero.
Explanation: By convention, the standard enthalpy of formation ($\Delta_f H^\ominus$) of an element in its standard reference state (the most stable form at 298 K and 1 bar) is taken as zero.
5.4 Relation between $\Delta H$ and $\Delta U$ for methane combustion.
Correct Answer: (iii) < $\Delta U^\ominus$
Reasoning:
Combustion Reaction: $CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)$.
$\Delta n_g = (\text{moles gaseous products}) – (\text{moles gaseous reactants})$
$\Delta n_g = 1 – (1 + 2) = 1 – 3 = -2$.
Formula: $\Delta H = \Delta U + \Delta n_g RT = \Delta U – 2RT$.
Since $T$ is absolute (positive), $-2RT$ is a negative value.
$\Delta H$ is obtained by subtracting a positive quantity from $\Delta U$.
However, combustion is exothermic ($\Delta U$ is negative, e.g., -X).
$\Delta H = -X – (\text{positive val})$. The result is a larger negative number, meaning $\Delta H$ is “more negative” or mathematically smaller than $\Delta U$.
5.5 Enthalpy of formation of CH₄.
Target Reaction: $C(s) + 2H_2(g) \to CH_4(g)$
Given Combustion Data ($\Delta_c H^\ominus$):
(1) $CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)$; $\Delta H_1 = -890.3$ kJ/mol
(2) $C(s) + O_2(g) \to CO_2(g)$; $\Delta H_2 = -393.5$ kJ/mol
(3) $H_2(g) + \frac{1}{2}O_2(g) \to H_2O(l)$; $\Delta H_3 = -285.8$ kJ/mol
Calculation:
To get target eq: Eq(2) + 2 $\times$ Eq(3) – Eq(1)
$\Delta_f H = \Delta H_2 + 2(\Delta H_3) – \Delta H_1$
$\Delta_f H = -393.5 + 2(-285.8) – (-890.3)$
$\Delta_f H = -393.5 – 571.6 + 890.3$
$\Delta_f H = -965.1 + 890.3 = -74.8$ kJ/mol.
Answer: (i) -74.8 kJ mol⁻¹.
5.6 – 5.10 Spontaneity, First Law & Calorimetry
5.6 Spontaneity of A + B → C + D + q (Positive Entropy).
Reaction: $A + B \to C + D + q$.
Heat ($q$) is released, so the reaction is Exothermic ($\Delta H < 0$).
Given: Entropy change is positive ($\Delta S > 0$).
Gibbs Free Energy Equation: $\Delta G = \Delta H – T\Delta S$.
Since $\Delta H$ is negative and $\Delta S$ is positive:
$\Delta G = (\text{negative}) – T(\text{positive}) = (\text{negative}) – (\text{positive})$.
The result is always negative, regardless of $T$.
Answer: (v) Possible at any temperature.
5.7 Change in Internal Energy (First Law).
Given:
Heat absorbed by system ($q$) = +701 J.
Work done by system ($w$) = -394 J (Expansion work is negative by convention).
$\Delta U = 701 + (-394) = 701 – 394 = \mathbf{307 \text{ J}}$.
5.8 Reaction of Cyanamide (Bomb Calorimeter).
Reaction: $NH_2CN(s) + \frac{3}{2}O_2(g) \to N_2(g) + CO_2(g) + H_2O(l)$
Given: $\Delta U = -742.7$ kJ/mol at 298 K.
Formula: $\Delta H = \Delta U + \Delta n_g RT$.
Calculate $\Delta n_g$:
Gaseous Products: 1 ($N_2$) + 1 ($CO_2$) = 2 moles.
Gaseous Reactants: 1.5 ($O_2$) moles.
$\Delta n_g = 2 – 1.5 = 0.5$.
Calculate $\Delta H$:
$R = 8.314 \times 10^{-3}$ kJ K⁻¹ mol⁻¹.
$\Delta H = -742.7 + (0.5 \times 8.314 \times 10^{-3} \times 298)$
$\Delta H = -742.7 + 1.239 = \mathbf{-741.5 \text{ kJ mol}^{-1}}$.
5.9 Heat for Aluminium Block.
Mass $m = 60.0$ g. Atomic Mass $Al = 27$ g/mol.
Moles $n = \frac{60}{27} = 2.22$ mol.
$\Delta T = 55 – 35 = 20^\circ \text{C} = 20 \text{ K}$.
Molar Heat Capacity $C_m = 24$ J mol⁻¹ K⁻¹.
$q = 2.22 \times 24 \times 20 = 1065.6 \text{ J}$.
In kJ: $\mathbf{1.07 \text{ kJ}}$.
5.10 Enthalpy change: Water (10°C) to Ice (-10°C).
This process occurs in 3 steps:
- Cooling liquid water (10°C to 0°C):
$\Delta H_1 = n C_p(l) \Delta T = 1 \times 75.3 \times (0 – 10) = -753 \text{ J}$. - Freezing at 0°C:
$\Delta H_2 = – \Delta_{fus}H = -6.03 \text{ kJ} = -6030 \text{ J}$ (Exothermic). - Cooling ice (0°C to -10°C):
$\Delta H_3 = n C_p(s) \Delta T = 1 \times 36.8 \times (-10 – 0) = -368 \text{ J}$.
$\Delta H = -753 – 6030 – 368 = -7151 \text{ J}$.
$\Delta H = \mathbf{-7.151 \text{ kJ}}$.
5.11 – 5.15 Hess’s Law & Bond Enthalpy
5.11 Heat released in CO₂ formation.
Combustion eq: $C + O_2 \to CO_2$, $\Delta H = -393.5$ kJ/mol.
Molar mass of $CO_2 = 44$ g/mol.
Given mass = 35.2 g.
Moles of $CO_2 = \frac{35.2}{44} = 0.8$ mol.
Heat released = $0.8 \text{ mol} \times 393.5 \text{ kJ/mol} = \mathbf{314.8 \text{ kJ}}$.
5.12 Calculate $\Delta_r H$ for reaction.
Reaction: $N_2O_4(g) + 3CO(g) \to N_2O(g) + 3CO_2(g)$
Given $\Delta_f H^\ominus$:
$CO = -110$, $CO_2 = -393$, $N_2O = 81$, $N_2O_4 = 9.7$ (kJ/mol).
$= [\Delta_f H(N_2O) + 3\Delta_f H(CO_2)] – [\Delta_f H(N_2O_4) + 3\Delta_f H(CO)]$
$= [81 + 3(-393)] – [9.7 + 3(-110)]$
$= [81 – 1179] – [9.7 – 330]$
$= -1098 – (-320.3)$
$= -1098 + 320.3 = \mathbf{-777.7 \text{ kJ}}$.
5.13 Standard Enthalpy of Formation of NH₃.
Reaction given: $N_2(g) + 3H_2(g) \to 2NH_3(g)$; $\Delta_r H^\ominus = -92.4$ kJ/mol.
This enthalpy corresponds to the formation of 2 moles of Ammonia.
Standard enthalpy of formation ($\Delta_f H^\ominus$) is for 1 mole.
$\Delta_f H^\ominus (NH_3) = \frac{-92.4}{2} = \mathbf{-46.2 \text{ kJ mol}^{-1}}$.
5.14 Standard Enthalpy of Formation of CH₃OH(l).
Goal Equation: $C(s) + 2H_2(g) + \frac{1}{2}O_2(g) \to CH_3OH(l)$
Given Data:
(1) $CH_3OH + \frac{3}{2}O_2 \to CO_2 + 2H_2O$; $\Delta H = -726$
(2) $C + O_2 \to CO_2$; $\Delta H = -393$
(3) $H_2 + \frac{1}{2}O_2 \to H_2O$; $\Delta H = -286$
Operations:
We need C (reactant) $\to$ Take Eq(2).
We need 2H₂ (reactant) $\to$ Take 2 $\times$ Eq(3).
We need CH₃OH (product) $\to$ Reverse Eq(1).
$= -393 + 2(-286) – (-726)$
$= -393 – 572 + 726$
$= -965 + 726 = \mathbf{-239 \text{ kJ mol}^{-1}}$.
5.15 Bond Enthalpy of C–Cl.
Goal Process: $CCl_4(g) \to C(g) + 4Cl(g)$ (Atomisation of molecule).
Steps to reach gaseous atoms from elements:
We construct a thermodynamic cycle.
Total Enthalpy = Energy to form atoms – Energy of formation of compound – Vaporization.
Actually, let’s look at the reaction:
$C(g) + 4Cl(g) \to CCl_4(g)$ would be $-\text{Bond Energy}$.
Let’s calculate $\Delta H$ for $CCl_4(g) \to C(g) + 4Cl(g)$.
$\Delta H_{reaction} = \Delta_a H(C) + 4\Delta_a H(Cl) – \Delta_f H(CCl_4(g))$
Note: $\Delta_f H(CCl_4)$ given is likely for liquid (-135.5). $\Delta_{vap}H = 30.5$.
So, $\Delta_f H(CCl_4, g) = -135.5 + 30.5 = -105.0$ kJ/mol.
Also, $\Delta_a H(Cl)$ is half of $\Delta_a H(Cl_2)$? No, $\Delta_a H(Cl_2)$ means $Cl_2 \to 2Cl$. So energy to get 4Cl is $2 \times 242$.
$= 715.0 + 2(242) – (-105.0)$ (using gas phase formation)
Alternatively using standard cycle:
Energy required to break bonds = [Formation of atoms] – [Formation of CCl4(l)] + [Vaporization]
$\Delta H = 715.0 + 2(242) – (-135.5) – (-30.5)$? No.
Let’s sum the paths:
$CCl_4(g) \to CCl_4(l)$ ($\Delta H = -30.5$)
$CCl_4(l) \to C(s) + 2Cl_2(g)$ ($\Delta H = +135.5$)
$C(s) \to C(g)$ ($\Delta H = +715.0$)
$2Cl_2(g) \to 4Cl(g)$ ($\Delta H = 2 \times 242 = +484$)
Total Energy $= -30.5 + 135.5 + 715.0 + 484 = \mathbf{1304 \text{ kJ mol}^{-1}}$.
There are 4 C-Cl bonds.
Bond Enthalpy $= 1304 / 4 = \mathbf{326 \text{ kJ mol}^{-1}}$.
5.16 – 5.22 Entropy & Gibbs Energy
5.16 For isolated system, $\Delta U = 0$, what is $\Delta S$?
For an isolated system, energy is constant ($\Delta U = 0$).
According to the Second Law of Thermodynamics, a spontaneous process in an isolated system is accompanied by an increase in entropy.
Therefore, $\Delta S > 0$.
5.17 Temperature for spontaneity.
Reaction: $2A + B \to C$.
$\Delta H = 400$ kJ/mol (Positive/Endothermic).
$\Delta S = 0.2$ kJ K⁻¹ mol⁻¹ (Positive).
Condition for spontaneity: $\Delta G = \Delta H – T\Delta S < 0$.
$\Delta H < T\Delta S \implies T > \frac{\Delta H}{\Delta S}$.
The reaction becomes spontaneous above 2000 K.
5.18 Signs of $\Delta H$ and $\Delta S$ for $2Cl(g) \to Cl_2(g)$.
Reaction: Formation of a bond between two atoms.
- $\Delta H$: Bond formation releases energy. So, $\Delta H$ is Negative (< 0).
- $\Delta S$: Two separate gaseous atoms combine to form one molecule. Disorder/Randomness decreases. So, $\Delta S$ is Negative (< 0).
5.19 Calculate $\Delta G^\ominus$ and predict spontaneity.
Reaction: $2A(g) + B(g) \to 2D(g)$.
$\Delta U^\ominus = -10.5$ kJ. $\Delta S^\ominus = -44.1$ J/K = $-0.0441$ kJ/K.
$T = 298$ K.
First calculate $\Delta H^\ominus$:
$\Delta n_g = 2 – (2+1) = -1$.
$\Delta H^\ominus = \Delta U^\ominus + \Delta n_g RT = -10.5 + (-1 \times 8.314 \times 10^{-3} \times 298)$.
$\Delta H^\ominus = -10.5 – 2.48 = -12.98$ kJ.
$= -12.98 – [298 \times (-0.0441)]$
$= -12.98 – (-13.14)$
$= -12.98 + 13.14 = \mathbf{+0.16 \text{ kJ}}$.
Since $\Delta G^\ominus$ is positive, the reaction is non-spontaneous under standard conditions.
5.20 Equilibrium constant K=10. Value of $\Delta G^\ominus$?
Formula: $\Delta G^\ominus = -2.303 RT \log K$.
$R = 8.314$ J/K mol. $T = 300$ K. $K = 10$.
$= -2.303 \times 8.314 \times 300 \times 1$
$= -5744.14 \text{ J}$
$= \mathbf{-5.74 \text{ kJ mol}^{-1}}$.
5.21 Thermodynamic Stability of NO.
Given:
(1) Formation of NO: $\Delta_f H = +90$ kJ/mol (Endothermic).
(2) Formation of NO₂ from NO: $\Delta_r H = -74$ kJ/mol (Exothermic).
Comment:
Since the enthalpy of formation of NO is positive (energy is absorbed), it means NO has higher energy than its constituent elements ($N_2$ and $O_2$). Therefore, NO is thermodynamically unstable with respect to decomposition into $N_2$ and $O_2$.
5.22 Entropy change in surroundings for H₂O formation.
Reaction: $H_2(g) + \frac{1}{2}O_2(g) \to H_2O(l)$; $\Delta_f H^\ominus = -286$ kJ/mol.
This reaction is exothermic. The heat released by the system is absorbed by the surroundings.
$q_{surr} = -q_{sys} = -\Delta H_{sys} = -(-286) = +286 \text{ kJ} = 286000 \text{ J}$.
Standard conditions imply $T = 298$ K.