chemistry-class-11-unit-07-chapter-01-chemical-equillibrium-l-1-8-e2l-ewyrx5m

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Chemical equilibrium </primary>
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- Consider a reaction
 - $A \rightarrow B$

- Pure A having six molecules

- After some time number of molecules of A are decreasing and number of molecules of B are increasing

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<footer style = "font-size: 18px">Chemical equilibrium $\rightarrow$ Equilibrium $\rightarrow$ <span style = "color:blue"> Chemical equilibrium </span> $\rightarrow$ Chemical equilibrium $\rightarrow$ Chemical equilibrium </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Chemical equilibrium </primary>
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- For reverse  reaction - $\Delta G < 0 $
 - $B \rightarrow A$

- Pure A having six molecules

- After some time number of molecules of A are decreasing and number of molecules of B are increasing

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<footer style = "font-size: 18px">Equilibrium $\rightarrow$ Chemical equilibrium $\rightarrow$ <span style = "color:blue"> Chemical equilibrium </span> $\rightarrow$ Chemical equilibrium $\rightarrow$ Chemical equilibrium </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Chemical equilibrium </primary>
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- Reaction has achieved equilibrium

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<footer style = "font-size: 18px">Chemical equilibrium $\rightarrow$ Chemical equilibrium $\rightarrow$ <span style = "color:blue"> Chemical equilibrium </span> $\rightarrow$ Chemical equilibrium $\rightarrow$ Importance of equilibrium </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Chemical equilibrium </primary>
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- $A + B \rightleftarrows C$

- Initially A and B will decrease and C will increase with time

- After some time,concentration of A, B and C will not change
- When concentration of reactant and product do not change then equillibrium state is achieved
- Similarly
 - C $\rightleftharpoons$ A+B

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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-07-chapter-01-chemical-equillibrium-l-1_8-e2l-ewyrx5m-9.png)

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<footer style = "font-size: 18px">Chemical equilibrium $\rightarrow$ Chemical equilibrium $\rightarrow$ <span style = "color:blue"> Chemical equilibrium </span> $\rightarrow$ Importance of equilibrium $\rightarrow$ Reason for equilibrium </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Reason for equilibrium </primary>
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- A(g) $\longrightarrow$ B(g)

* Reaction is spontaneous only when

- $\Delta G < 0, G_B < G_A$

* If $G_B < G_A \Rightarrow$ why reaction does not go to completion

* If $G_B<G_{A} \Rightarrow$ why reverse reaction happens if we start from pure B

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<footer style = "font-size: 18px">Chemical equilibrium $\rightarrow$ Importance of equilibrium $\rightarrow$ <span style = "color:blue"> Reason for equilibrium </span> $\rightarrow$ Reason for equilibrium $\rightarrow$ Reason for equilibrium </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Reason for equilibrium </primary>
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- A (g) $\Rightarrow$ B (g)
- A (g) $\Rightarrow$ B (g)
- $ A(g) \rightarrow B(g)$
- 1  $\quad \quad\quad$  0
- $1-\xi$ $\quad \xi$ 
- $G_ {II }=(1-\xi) \times G_A+\xi \times G_B$

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<footer style = "font-size: 18px">Importance of equilibrium $\rightarrow$ Reason for equilibrium $\rightarrow$ <span style = "color:blue"> Reason for equilibrium </span> $\rightarrow$ Reason for equilibrium $\rightarrow$ Reasons for equilibrium </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Reason for equilibrium </primary>
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- $G_{II} = (1-\xi) G_A+\xi \times G_B=G_A-\xi G_A+ \xi G_B$
- $\Delta G=(1-\xi) G_A+\xi \times G_B-1 \times G_A$
- $\Delta G=G_A-\xi G_A+\xi G_B-G_A$
- $\Delta G=\xi\left(G_B-G_A\right)=$ -ve
- $G \hspace{0.5mm}$ (II) $= Ga+ \xi (G_B-G_A)$
- Reaction should go to completion

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<footer style = "font-size: 18px">Reason for equilibrium $\rightarrow$ Reason for equilibrium $\rightarrow$ <span style = "color:blue"> Reason for equilibrium </span> $\rightarrow$ Reasons for equilibrium $\rightarrow$ Reason for equilibrium </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Reasons for equilibrium </primary>
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- $\Delta G=\xi(G_B-G_A)+$ $\Delta G_{mix}$
- $\Delta G=\Delta H-T \Delta S$
- $\Delta G_{mix}=-T$ $\Delta S_{mix}$
- $\Delta G_{mix}={+ve}$
- $\Delta G=\xi\left(G_B-G_A\right)+\Delta G_{\text {mix }}$
- $-ve \quad \quad -ve$
- $-ve \quad  \quad +ve$

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<footer style = "font-size: 18px">Reason for equilibrium $\rightarrow$ Reason for equilibrium $\rightarrow$ <span style = "color:blue"> Reasons for equilibrium </span> $\rightarrow$ Reason for equilibrium $\rightarrow$ Types of equilibrium </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Physical equilibrium </primary>
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- $H_2O (1,0$ $^\circ C$,1 atm)$\rightarrow$ $H_2O (S,0$ $^\circ C$,1 atm)

- Freezing of $H_2O$

- $0$ $^\circ C$, ice ,$H_2O$

- Solid-Liquid equilibrium

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<footer style = "font-size: 18px">Reason for equilibrium $\rightarrow$ Types of equilibrium $\rightarrow$ <span style = "color:blue"> Physical equilibrium </span> $\rightarrow$ Physical equilibrium $\rightarrow$ Solubility of solute in solution </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Physical equilibrium </primary>
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- $H_2O (1,100$ $^\circ C$,1 atm) $\rightarrow H_2O (V,100$ $^\circ C$, 1 atm)

- Liquid-Vapour equilibrium

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<footer style = "font-size: 18px">Types of equilibrium $\rightarrow$ Physical equilibrium $\rightarrow$ <span style = "color:blue"> Physical equilibrium </span> $\rightarrow$ Solubility of solute in solution $\rightarrow$ Chemical equilibrium </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Solubility of solute in solution </primary>
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- Solubility of $AgCl$ in water

- $Ag^{+}$

- $Cl^{-}$

- $AgCl$

- $Ag^{+}$ + $Cl^{-}$ $\rightleftharpoons$ $AgCl$

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<footer style = "font-size: 18px">Physical equilibrium $\rightarrow$ Physical equilibrium $\rightarrow$ <span style = "color:blue"> Solubility of solute in solution </span> $\rightarrow$ Chemical equilibrium $\rightarrow$ Homogeneous reaction </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Chemical equilibrium </primary>
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- $N_2+3 H_2\rightleftharpoons2NH_3$
- $K_c=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{N}_2\right]\left[\mathrm{H}_2\right]^3}$
- Q = Reaction quotient
- $K_c =\frac{\left[\mathrm{NH}_3\right]_t^2}{\left[\mathrm{N}_2\right]_t\left[\mathrm{H}_2\right]_t^3}$

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### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Chemical equilibrium </primary>
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- $K_P, K_c \Rightarrow$ Relationship
- $PCl_5(g)$  $\rightarrow PCl_3 (g)+ Cl_2(g)$
- $K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$
- $K_p=\frac{P_{PCl_3}P_{Cl_2}}{P_{PCl_5}}$
- $\frac{ [PCl_3] (RT) [Cl_2] (RT)}{[PCl_5] (RT)}$
- $K_C(RT)^{'}$
- $P V=n R T$
- $P=\frac{n}{V} R T$ $ ~ $ $=C R T$

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### <Success style="font-size:25px"> Chemical Equilibrium L-1 </Success>
### <primary style="font-size:24px"> Chemical equilibrium </primary>
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- $aA+bB\rightleftharpoons cC+dD$
- $K_P=\frac{\left(P_C\right)^c(D)^d}{\left(P_A\right)^a\left(P_B\right)^b}$
- $\frac{([C] R T)^c([D] R T)^d}{([A] R T)^a([B] R T)^b}$
- $\Delta_n=n_p-n_r$
- $\frac{[C]^c[D]^a}{[A]^a[B]^b} \times R T^{(c+d-a-b)}$
- $K_c \times R T^{\Delta n}$

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