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

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Recap </primary>
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- 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 $\Delta U$ and $\Delta H$
- Change in enthalpy in different processes / reactions
- Enthalpy of reaction
- Enthalpy of formation
- Enthalpy of phase transition
- Enthalpy of other processes / reactions

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<footer style = "font-size: 18px"> $\rightarrow$ THERMODYNAMICS $\rightarrow$ <span style = "color:blue"> Recap </span> $\rightarrow$ Enthalpy of reaction $\rightarrow$ Enthalpy of reaction </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Enthalpy of reaction </primary>
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- $\Delta_r H_T^{\circ}=\sum_i a_i H _{m,T}^{\circ}-\sum b_i H _{m,T}^{\circ}$ 
- Conventional temperature considered as $25^{\circ} {C}, 298.15 K$
- $\Delta_r H^{\circ}=\sum_i a_i H_m^{\circ}-\sum b_i H_m^{\circ}$
- Standard heat of formation / Standard enthalpy of formation
- $\Delta_f H_T^{\circ}, \Delta_f H^{\circ} \quad T \Rightarrow 25^{\circ} \mathrm{C}$

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<footer style = "font-size: 18px">THERMODYNAMICS $\rightarrow$ Recap $\rightarrow$ <span style = "color:blue"> Enthalpy of reaction </span> $\rightarrow$ Enthalpy of reaction $\rightarrow$ Enthalpy of reaction </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Enthalpy of formation </primary>
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- Standard heat of formation of water, $H_2O$ (l) -298K

- $\Delta_rH^o$ at 298K

- Product - $H_2O$ (1 mole) at standard state at 298K (1 bar)

- Reactant-
 - $H_2$(g)- 1 bar , 298K
 - $O_2$(g)- 1 bar, 298K

- Reaction:
 - $H_2(g)+O_2(g) \rightarrow H_2O (l)$ 1 bar, 298K
 - $\Delta_rH^o=\Delta_fH^o _{H_2O}$ 298K
 - -286 KJ $mol^-1$

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<footer style = "font-size: 18px">Enthalpy of reaction $\rightarrow$ Enthalpy of reaction $\rightarrow$ <span style = "color:blue"> Enthalpy of formation </span> $\rightarrow$ Enthalpy of formation $\rightarrow$ Enthalpy of formation </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Enthalpy of formation </primary>
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- $CH_4$
- $C(graphite)+2H_2(g) \rightarrow CH_4(s)$
- $\Delta_rH^o(298K)=-74.8 KJ mol^-1$
- $\Delta_fH^o _{CH_4}(298K)=-74.8 KJ mol^-1$
- $C_2H_5OH$
- $2C(graphite)+3H_2(g)+ \frac{1}{2}O_2(l) \rightarrow C_2H_5OH(l)$
- $\Delta_fH^o(C_2H_5OH)= -277 KJ mol^-1$
- $HBr(g)$
- At $25^oC$, 1 bar, $Br_2(l)$
- $\frac{1}{2}H_2(g)+\frac{1}{2}Br_2(l) \rightarrow HBr(g)$
- $\Delta_fH^o(HBr(g))= -36.4 KJ mol^-1$

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<footer style = "font-size: 18px">Enthalpy of reaction $\rightarrow$ Enthalpy of formation $\rightarrow$ <span style = "color:blue"> Enthalpy of formation </span> $\rightarrow$ Enthalpy of formation $\rightarrow$ Enthalpy of reaction </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Enthalpy of formation </primary>
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- $CaO(s)+CO_2(g) \rightarrow CaCO_3(s)$
- $\Delta_rH^o \ne \Delta_fH^o(CaCO_3)(s)$
- $\Delta_fH^o$ of an element in its reference state is taken as zero
- $\Delta_fH^o(graphite)$
- $C(graphite) \rightarrow C(graphite)$ 1 bar
- $\Delta_rH^o = 0$
- $\Delta_fH^{\circ}$ for elements at thier reference state = 0

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<footer style = "font-size: 18px">Enthalpy of formation $\rightarrow$ Enthalpy of formation $\rightarrow$ <span style = "color:blue"> Enthalpy of formation </span> $\rightarrow$ Enthalpy of reaction $\rightarrow$ Enthalpy of reaction </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Enthalpy of reaction </primary>
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- $\begin{gathered}
\Delta_r H^{\circ}=\sum a_i h_m^{\circ}-\sum b_i H_m^{\circ} \\
a A+b B \rightarrow c C+d D
\end{gathered}$

- $Reactants(aA+bB) \xrightarrow I Products(cC+dD) \xrightarrow {III}$ constituent elements in their reference state at T
- $Reactants(aA+bB) \xrightarrow {II}$ constituent elements in their reference state at T $\xrightarrow {III} Products(cC+dD)$

- $\Delta_r H^{\circ}(I)=\Delta_r H^{\circ}(\text { II })+\Delta_r H^{\circ} \text { (III) }$

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<footer style = "font-size: 18px">Enthalpy of formation $\rightarrow$ Enthalpy of formation $\rightarrow$ <span style = "color:blue"> Enthalpy of reaction </span> $\rightarrow$ Enthalpy of reaction $\rightarrow$ Enthalpy of reaction </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Enthalpy of reaction </primary>
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- $-\Delta_r H^{\circ}(\mathbb{I})=a \Delta_f H_A^0+b \Delta_f H_B^0 $ 
- $\Delta_r H^{\circ}(\underline{Q})=C \Delta_f H_c^0+d \Delta_f H_D^0 $ 
- $\Delta_r H^{\circ}(I)=\Delta_r H^{\circ}(I)+\Delta_r H^0\left(\mathbb{I}^0\right)$
- $c \Delta_f H_C^0+d \Delta_f H_D^0-\left(a \Delta_f H_A^0+b \Delta_f H_l^0\right)$
- $\Rightarrow \Delta_r H^{\circ}=\sum a_i \Delta_f H_i^{\circ}-\sum b \Delta_f H_i^0 \quad (\text{at } \operatorname{temp} T)$
- $\Rightarrow \mathrm{CH}_4(9)+2 \mathrm{O}_2(9) \rightarrow \mathrm{CO}_2(9)+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \text { (298K) }$
- $\Delta_r H^0=\Delta_f H _{{CO} _2(g)}^0+2 \Delta_f H _{{H}_2 {O}(l)}^0-\Delta_f {H} _{{CH} _4(g)}^0-2 \Delta_f {H} _{{O} _2(g)}^0$
- $-393.5 + 2(-285.8) - (-74.8)$
- $-890.4 \, \mathrm{kJ \, mn}^{-1}$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-06-chapter-05-chemical-thermodynamics-l-5_8-gnshi4lv62c-540-2031.6-1.jpg)

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<footer style = "font-size: 18px">Enthalpy of formation $\rightarrow$ Enthalpy of reaction $\rightarrow$ <span style = "color:blue"> Enthalpy of reaction </span> $\rightarrow$ Enthalpy of reaction $\rightarrow$ Enthalpy of reaction </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Enthalpy of reaction </primary>
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- $CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

- $\Delta_rH^o=178.3 KJ mol^{-1}$

- $2CaO(s)+2CO_2(g) \rightarrow 2CaCO_3(s)$

- $\Delta_rH^o = -2\times 178.3 = -356.6 KJ mol^{-1}$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/chemistry-class-11-unit-06-chapter-05-chemical-thermodynamics-l-5_8-gnshi4lv62c-576-2137.2-1.jpg)

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<footer style = "font-size: 18px">Enthalpy of reaction $\rightarrow$ Enthalpy of reaction $\rightarrow$ <span style = "color:blue"> Enthalpy of reaction </span> $\rightarrow$ Change in enthalpy $\rightarrow$ Change in enthalpy </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Change in enthalpy </primary>
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- Enthalpy change during  phase transition 
- Standard enthalpy of phase transition
- $\Delta _{trs} H^{\circ}(T)$ often  called latent heat
- Phase change
- Solid $\rightarrow$  liquid $\quad$ fusion $\Delta_fus H^{\circ}(T)$
- Liquid $\rightarrow$  gas $\quad$ vapourisation  $\Delta _{ vap } H^{\circ} (T)$
- Solid $\rightarrow$  gas $\quad$ sublimation $\Delta _{sub}H^{\circ}(T)$ 
- Phase transition - generally happens at different temperature and pressure

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<footer style = "font-size: 18px">Enthalpy of reaction $\rightarrow$ Enthalpy of reaction $\rightarrow$ <span style = "color:blue"> Change in enthalpy </span> $\rightarrow$ Change in enthalpy $\rightarrow$ Change in enthalpy </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Change in enthalpy </primary>
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- $H_2 O(l) \rightarrow H_2 O(g) \quad \Delta_r H^{\circ}=40.66\quad KJ mol^{-1}$

- Pure $H_2O(l)-$ 1bar,373 k

- Pure $H_2O (g)-$ 1bar,373 K

- $\Delta_{rap}H^{\circ}_{H_2} o(l)\hspace{0.5mm}$(at 373 k)

- $=40.66 KJ mol^{-1}$

- 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 $\quad\Delta _{trs} H^{\circ}(T)$
- Standard enthalpy of sublimation

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<footer style = "font-size: 18px">Enthalpy of reaction $\rightarrow$ Change in enthalpy $\rightarrow$ <span style = "color:blue"> Change in enthalpy </span> $\rightarrow$ Change in enthalpy $\rightarrow$ Hess's law </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Change in enthalpy </primary>
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- $\Delta H$ - only depends on initial and final states
- **Sublimation**: solid $\rightarrow$ gas
- solid  $\rightarrow$ liquid  $\rightarrow$  gas
- $H_2 O(s) \rightarrow H_2 O(g) \quad \Delta _{sub } H^{\circ}$
- $\Delta _{sub } H^{\circ}=\Delta _{fus} H^{\circ}+\Delta _{vap } H^{\circ}$

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<footer style = "font-size: 18px">Change in enthalpy $\rightarrow$ Change in enthalpy $\rightarrow$ <span style = "color:blue"> Change in enthalpy </span> $\rightarrow$ Hess's law $\rightarrow$ Hess's law </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Hess's law </primary>
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- Hess's law of constant heat summation

- $\Delta_f{H}^{\circ}$ (298k)  of ethane gas

- $2 C (graphite) + 3H_2 (s) \rightarrow C_2H_6$ (g) at 298k
 - $\Delta_rH^{\circ}$

- (I) $C_2 H_6(g)+\frac{7}{2} O_2(g) \rightarrow 2 CO_2(g)+3 H_2 O(l)$
 - $\Delta_rH^{\circ}$ (298 k)=-1560 
- (II) $C  (graphite)+O_2(g) \rightarrow CO_2(g) $
 - $\Delta_rH^{\circ}=-393.5$
- (III) $H_2(g)+\frac{1}{2} O_2(g) \rightarrow H_2 O(l)$ 
 - $\Delta_r H^{\circ}=-286 KJ mol^{-1}$

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<footer style = "font-size: 18px">Change in enthalpy $\rightarrow$ Change in enthalpy $\rightarrow$ <span style = "color:blue"> Hess's law </span> $\rightarrow$ Hess's law $\rightarrow$ Hess's law </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Hess's law </primary>
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- $(I) \times (-1)$
 - $ 2 Co_ 2(g)+3 H_2O \rightarrow C_2 H_6 (g)+\frac{7}{2} O_2 (g)$
- $ \Delta r H^{\circ}=1560$
 - $(II) \times 2$
- $2 C(graphite )+2 o_2(g) \rightarrow 2 C o_ 2(g)$ 
 - $\Delta r H^{\circ}=2 \times(-343.5)$
- $(III) \times 3$
 - $3H_2(s) + \frac{3}{2} O_2 (s) \rightarrow 3 H_2o (l)$
 - $\Delta_rH^{\circ} = 3 \times (-286)$
- $2 c (graphite) +3 H_2(g) \rightarrow C_2 H_6(g)-85 KJ mol^{-1}$
- $\Delta_f H^{\circ}C_2H_6(g) \quad (298K)=-85 KJ mol^{-1}$

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<footer style = "font-size: 18px">Change in enthalpy $\rightarrow$ Hess's law $\rightarrow$ <span style = "color:blue"> Hess's law </span> $\rightarrow$ Hess's law $\rightarrow$ Hess's law </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Hess's law </primary>
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- $A \xrightarrow{\Delta r H^{\circ}} B$
- $A \xrightarrow{\Delta_rH^{\circ} _{1}} C$
- $C \xrightarrow{\Delta_rH^{\circ} _{2}} D$
- $D \xrightarrow{\Delta_rH^{\circ} _{3}} B$
- $\Delta_r H^0=\Delta r H_1^0+\Delta_r H_2^0+\Delta_r H_3^0$
- $ C (graphite)+ O_2 \rightarrow C$

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<footer style = "font-size: 18px">Hess's law $\rightarrow$ Hess's law $\rightarrow$ <span style = "color:blue"> Hess's law </span> $\rightarrow$ Hess's law $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Chemical Thermodynamics L-5 </Success>
### <primary style="font-size:24px"> Hess's law </primary>
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- Given that the standard enthalpy of combustion of graphite is -393.51 kJ ${mol}^{-1}$ and that of 
diamond is -395.41 kJ ${mol}^{-1}$, calculate the enthalpy of the graphite-to-diamond transition.

- $C (graphite)+ O_2 \rightarrow CO_2 (g)\Delta _rH^{\circ} =-393 . 5kJ {mol}^{-1}$

- $C  (diamond) +O_2 \rightarrow CO_2 (g)\Delta _rH^{\circ} =-395 . 4 {mol}^{-1}$

- $C (graphite) \rightarrow C  (diamond)$

- $\Delta_rH^{\circ} =-393 . 5 + 395.4$

- $= 1.90 \quad KJ mol^{-1}$

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<footer style = "font-size: 18px">Hess's law $\rightarrow$ Hess's law $\rightarrow$ <span style = "color:blue"> Hess's law </span> $\rightarrow$ Thank you $\rightarrow$  </footer>