chemistry-class-11-unit-06-chapter-05-chemical-thermodynamics-l-5-8-gnshi4lv62c

Chemical Thermodynamics L-5

Recap

Essential concepts and definitions
Heat, work & energy internal energy
Calculation of work, heat in different process for an ideal gas
Enthalpy and heat capacity
Determination of $Δ U$ and $Δ H$
Change in enthalpy in different processes / reactions
Enthalpy of reaction
Enthalpy of formation
Enthalpy of phase transition
Enthalpy of other processes / reactions

Chemical Thermodynamics L-5

Enthalpy of reaction

$Δ_{r} H_{T}^{\circ} = \sum_{i} a_{i} H_{m, T}^{\circ} - \sum b_{i} H_{m, T}^{\circ}$
Conventional temperature considered as $25^{\circ} C, 298.15 K$
$Δ_{r} H^{\circ} = \sum_{i} a_{i} H_{m}^{\circ} - \sum b_{i} H_{m}^{\circ}$
Standard heat of formation / Standard enthalpy of formation
$Δ_{f} H_{T}^{\circ}, Δ_{f} H^{\circ} T \Rightarrow 25^{\circ} C$

Chemical Thermodynamics L-5

Enthalpy of reaction

standard molar enthalpy
$Δ_{r} H_{T}^{\circ} \Rightarrow H_{m, T}^{\circ}$ $\Rightarrow$ $H_{m}^{\circ}$
Standard heat of formation
$Δ_{f} H_{T}^{\circ}$ of a pure substance at a specified temp. T is $Δ r H^{\circ}$ for the process/reaction in which one mole of substance in its standard state at T is formed from the corresponding separated elements at T, each being in its reference state/form/phase

Chemical Thermodynamics L-5

Enthalpy of reaction

Reference state/phase/form
Most stable state of the element at 1 bar pressure at the specified temperature T
- $T = 25^{\circ} C$
- Reference states at $25^{\circ} C$
- Dihydrogen = $H_{2} (g)$
- Dioxygen = $O_{2} (g)$
- Carbon = C (graphite)
- Sulphur = S (rhombic)

Chemical Thermodynamics L-5

Enthalpy of formation

Standard heat of formation of water, $H_{2} O$ (l) -298K
$Δ_{r} H^{o}$ at 298K
Product - $H_{2} O$ (1 mole) at standard state at 298K (1 bar)
Reactant-
- $H_{2}$ (g)- 1 bar , 298K
- $O_{2}$ (g)- 1 bar, 298K
Reaction:
- $H_{2} (g) + O_{2} (g) \to H_{2} O (l)$ 1 bar, 298K
- $Δ_{r} H^{o} = Δ_{f} H_{H_{2} O}^{o}$ 298K
- -286 KJ $m o l^{-} 1$

Chemical Thermodynamics L-5

Enthalpy of formation

$C H_{4}$
$C (g r a p h i t e) + 2 H_{2} (g) \to C H_{4} (s)$
$Δ_{r} H^{o} (298 K) = - 74.8 K J m o l^{-} 1$
$Δ_{f} H_{C H_{4}}^{o} (298 K) = - 74.8 K J m o l^{-} 1$
$C_{2} H_{5} O H$
$2 C (g r a p h i t e) + 3 H_{2} (g) + \frac{1}{2} O_{2} (l) \to C_{2} H_{5} O H (l)$
$Δ_{f} H^{o} (C_{2} H_{5} O H) = - 277 K J m o l^{-} 1$
$H B r (g)$
At $25^{o} C$ , 1 bar, $B r_{2} (l)$
$\frac{1}{2} H_{2} (g) + \frac{1}{2} B r_{2} (l) \to H B r (g)$
$Δ_{f} H^{o} (H B r (g)) = - 36.4 K J m o l^{-} 1$

Chemical Thermodynamics L-5

Enthalpy of formation

$C a O (s) + C O_{2} (g) \to C a C O_{3} (s)$
$Δ_{r} H^{o} \neq Δ_{f} H^{o} (C a C O_{3}) (s)$
$Δ_{f} H^{o}$ of an element in its reference state is taken as zero
$Δ_{f} H^{o} (g r a p h i t e)$
$C (g r a p h i t e) \to C (g r a p h i t e)$ 1 bar
$Δ_{r} H^{o} = 0$
$Δ_{f} H^{\circ}$ for elements at thier reference state = 0

Chemical Thermodynamics L-5

Enthalpy of reaction

$\begin{matrix} Δ_{r} H^{\circ} = \sum a_{i} h_{m}^{\circ} - \sum b_{i} H_{m}^{\circ} a A + b B \to c C + d D \end{matrix}$
$R e a c t a n t s (a A + b B) \overset{I}{\to} P r o d u c t s (c C + d D) \overset{I I I}{\to}$ constituent elements in their reference state at T
$R e a c t a n t s (a A + b B) \overset{I I}{\to}$ constituent elements in their reference state at T $\overset{I I I}{\to} P r o d u c t s (c C + d D)$
$Δ_{r} H^{\circ} (I) = Δ_{r} H^{\circ} (II) + Δ_{r} H^{\circ} (III)$

Chemical Thermodynamics L-5

Enthalpy of reaction

$- Δ_{r} H^{\circ} (I) = a Δ_{f} H_{A}^{0} + b Δ_{f} H_{B}^{0}$
$Δ_{r} H^{\circ} (\underset{―}{Q}) = C Δ_{f} H_{c}^{0} + d Δ_{f} H_{D}^{0}$
$Δ_{r} H^{\circ} (I) = Δ_{r} H^{\circ} (I) + Δ_{r} H^{0} (I^{0})$
$c Δ_{f} H_{C}^{0} + d Δ_{f} H_{D}^{0} - (a Δ_{f} H_{A}^{0} + b Δ_{f} H_{l}^{0})$
$\Rightarrow Δ_{r} H^{\circ} = \sum a_{i} Δ_{f} H_{i}^{\circ} - \sum b Δ_{f} H_{i}^{0} (at temp T)$
$\Rightarrow {CH}_{4} (9) + 2 O_{2} (9) \to {CO}_{2} (9) + 2 H_{2} O (l) (298K)$
$Δ_{r} H^{0} = Δ_{f} H_{{C O}_{2} (g)}^{0} + 2 Δ_{f} H_{H_{2} O (l)}^{0} - Δ_{f} H_{{C H}_{4} (g)}^{0} - 2 Δ_{f} H_{O_{2} (g)}^{0}$
$- 393.5 + 2 (- 285.8) - (- 74.8)$
$- 890.4, {kJ, mn}^{- 1}$

Chemical Thermodynamics L-5

Enthalpy of reaction

Balanced equation along with $Δ_{r} H^{o}$ value
Thermochemical equation (thermochemistry)
Physical states and allotropic state must be mentioned
$C H_{4} (s) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$
$Δ_{r} H^{o} (298 K) = - 890.4 K J m o l^{- 1}$
stoichiometric coefficients = number of moles
$Δ_{r} H^{o} \to$ extensive quantity
Reverse chemical reaction will have opposite sign but equal magnitude in $Δ_{r} H^{o}$

Chemical Thermodynamics L-5

Enthalpy of reaction

$C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$
$Δ_{r} H^{o} = 178.3 K J m o l^{- 1}$
$2 C a O (s) + 2 C O_{2} (g) \to 2 C a C O_{3} (s)$
$Δ_{r} H^{o} = - 2 \times 178.3 = - 356.6 K J m o l^{- 1}$

Chemical Thermodynamics L-5

Change in enthalpy

Enthalpy change during phase transition
Standard enthalpy of phase transition
$Δ_{t r s} H^{\circ} (T)$ often called latent heat
Phase change
Solid $\to$ liquid fusion $Δ_{f} u s H^{\circ} (T)$
Liquid $\to$ gas vapourisation $Δ_{v a p} H^{\circ} (T)$
Solid $\to$ gas sublimation $Δ_{s u b} H^{\circ} (T)$
Phase transition - generally happens at different temperature and pressure

Chemical Thermodynamics L-5

Change in enthalpy

$H_{2} O (l) \to H_{2} O (g) Δ_{r} H^{\circ} = 40.66 K J m o l^{- 1}$
Pure $H_{2} O (l) -$ 1bar,373 k
Pure $H_{2} O (g) -$ 1bar,373 K
$Δ_{r a p} H_{H_{2}}^{\circ} o (l)$ (at 373 k)
$= 40.66 K J m o l^{- 1}$
Standand enthalpy of vapourisation is the amount of heat required to vaporize one mole of a liquid at a constant temperature and under standard pressure (1 bar)
Standard enthalpy of fusion $Δ_{t r s} H^{\circ} (T)$
Standard enthalpy of sublimation

Chemical Thermodynamics L-5

Change in enthalpy

$Δ H$ - only depends on initial and final states
Sublimation: solid $\to$ gas
solid $\to$ liquid $\to$ gas
$H_{2} O (s) \to H_{2} O (g) Δ_{s u b} H^{\circ}$
$Δ_{s u b} H^{\circ} = Δ_{f u s} H^{\circ} + Δ_{v a p} H^{\circ}$

Chemical Thermodynamics L-5

Hess's law

Hess's law of constant heat summation
$Δ_{f} H^{\circ}$ (298k) of ethane gas
$2 C (g r a p h i t e) + 3 H_{2} (s) \to C_{2} H_{6}$ (g) at 298k
- $Δ_{r} H^{\circ}$
(I) $C_{2} H_{6} (g) + \frac{7}{2} O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)$
- $Δ_{r} H^{\circ}$ (298 k)=-1560
(II) $C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g)$
- $Δ_{r} H^{\circ} = - 393.5$
(III) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
- $Δ_{r} H^{\circ} = - 286 K J m o l^{- 1}$

Chemical Thermodynamics L-5

Hess's law

$(I) \times (- 1)$
- $2 C o_{2} (g) + 3 H_{2} O \to C_{2} H_{6} (g) + \frac{7}{2} O_{2} (g)$
$Δ r H^{\circ} = 1560$
- $(I I) \times 2$
$2 C (g r a p h i t e) + 2 o_{2} (g) \to 2 C o_{2} (g)$
- $Δ r H^{\circ} = 2 \times (- 343.5)$
$(I I I) \times 3$
- $3 H_{2} (s) + \frac{3}{2} O_{2} (s) \to 3 H_{2} o (l)$
- $Δ_{r} H^{\circ} = 3 \times (- 286)$
$2 c (g r a p h i t e) + 3 H_{2} (g) \to C_{2} H_{6} (g) - 85 K J m o l^{- 1}$
$Δ_{f} H^{\circ} C_{2} H_{6} (g) (298 K) = - 85 K J m o l^{- 1}$

Chemical Thermodynamics L-5

Hess's law

$A \overset{Δ r H^{\circ}}{\to} B$
$A \overset{Δ_{r} H_{1}^{\circ}}{\to} C$
$C \overset{Δ_{r} H_{2}^{\circ}}{\to} D$
$D \overset{Δ_{r} H_{3}^{\circ}}{\to} B$
$Δ_{r} H^{0} = Δ r H_{1}^{0} + Δ_{r} H_{2}^{0} + Δ_{r} H_{3}^{0}$
$C (g r a p h i t e) + O_{2} \to C$

Chemical Thermodynamics L-5

Hess's law

Given that the standard enthalpy of combustion of graphite is -393.51 kJ ${m o l}^{- 1}$ and that of diamond is -395.41 kJ ${m o l}^{- 1}$ , calculate the enthalpy of the graphite-to-diamond transition.
$C (g r a p h i t e) + O_{2} \to C O_{2} (g) Δ_{r} H^{\circ} = - 393.5 k J {m o l}^{- 1}$
$C (d i a m o n d) + O_{2} \to C O_{2} (g) Δ_{r} H^{\circ} = - 395.4 {m o l}^{- 1}$
$C (g r a p h i t e) \to C (d i a m o n d)$
$Δ_{r} H^{\circ} = - 393.5 + 395.4$
$= 1.90 K J m o l^{- 1}$

Chemical Thermodynamics L-5

THERMODYNAMICS

Chemical Thermodynamics L-5

Recap

Chemical Thermodynamics L-5

Enthalpy of reaction

Chemical Thermodynamics L-5

Enthalpy of reaction

Chemical Thermodynamics L-5

Enthalpy of reaction

Chemical Thermodynamics L-5

Enthalpy of formation

Chemical Thermodynamics L-5

Enthalpy of formation

Chemical Thermodynamics L-5

Enthalpy of formation

Chemical Thermodynamics L-5

Enthalpy of reaction

Chemical Thermodynamics L-5

Enthalpy of reaction

Chemical Thermodynamics L-5

Enthalpy of reaction

Chemical Thermodynamics L-5

Enthalpy of reaction

Chemical Thermodynamics L-5

Change in enthalpy

Chemical Thermodynamics L-5

Change in enthalpy

Chemical Thermodynamics L-5

Change in enthalpy

Chemical Thermodynamics L-5

Hess's law

Chemical Thermodynamics L-5

Hess's law

Chemical Thermodynamics L-5

Hess's law

Chemical Thermodynamics L-5

Hess's law

Chemical Thermodynamics L-5

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