chemistry-class-11-unit-07-chapter-02-chemical-equillibrium-l-2-8-qr4ueuwmy9u

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Type of equilibrium constants </primary>
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- $a A+b B \rightleftharpoons c C+d D$
- Reactants $\leftrightarrow$ Product 
- $K_c = \frac{[C]_ {eq}^{c} [D]_ {eq}^{d}}{ [A]_ {eq}^{a} [B]_ {eq}^{b}}$
- $K_p = \frac{(P_C)_ {eq}^{e} (P_D)_ {eq}^{d}}{ (P_A)_ {eq}^{a} (P_B)_ {eq}^{b}}$

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

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **Question** Calculate the equilibrium constant for the reaction below if they are present at equilibrium
- 5 mole of $A_2, 3$ mole of $B_2$ and 2 mole of $AB_2$ at 8.21 atm and 300K
    - $A_2(g)+2 B_2(g)=2 A B_2(g)+$ heat

- **Solution**
- $A_2(g)+2 B_2(g)=2 A B_2(g)+$ heat
- $K_C=\frac{[A B_2]^2}{[A_2]^{\prime}[B_2]^2}$
- $ n_{A_2} = 5$ 
- $ n_{B_2} = 3$
- $ n_{A B_2} = 2$

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<footer style = "font-size: 18px">Equilibrium problems $\rightarrow$ Equilibrium problems $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Problem with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **Question** $SrCl_2. 2 H_2 O(s)+4 H_2 O(g)=SrCl_2. 6 H_2 O(s)$</p>
    - $K_p=1\times10^{-12}atm^{-4}$ </p>
- What is the equilibrium vapour pressure (in torr) of water in a closed vessel that contains $SrCl_2. 2 H_2 O(s)$ ?

- **Solution**
- $K_p=\frac{1}{\left(P_{H_2{O}}\right)^4} = 1 \times 10^{-12}$
- $(P_{H_2O})^4=1 \times 10^{12}$
- $P_{{H}_2 {O}}= 1 \times 10^3 atm $

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<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Problem $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- **Question** Thermal decomposition of gaseous X$_{2}$ to gaseous X at 298 K takes place according to equation
- X$_{2}(g) \leftrightarrow $ 2X$(g)$
- At the start of the reaction, there is one mole of X$_{2}$ and and no X As the reaction proceeds, the number of moles of X formed is given by β Thus, βequilibrium is the number of moles of X formed at equilibrium.
- The reaction is carried out at a constant total pressure of 2 bar what is the equilibrium constant Kp for the reaction at 298 K, in terms of βequilibrium.

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<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **Solution**

- $x_2(g) \rightleftharpoons 2 x(g)$
- $ 1 \quad \quad \quad \quad0$ $\quad$- initial concentration
- $ x \quad \quad \quad \quad β_{eq}$ $\quad$- equilibrium concentration
- $K_p \rightarrow \beta_{e q}$
- $K_p=\frac{(P_x)_ {eq}^2}{(P_{x_2})_ {eq}}$

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<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Problem with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **Solution**
- $X_2 (g) \rightleftharpoons 2X (g)$
- $ 1 \quad \quad \quad \quad0$ $\quad$- initial concentration
- $ {1-\beta/2} \quad \quad \quad \quad β_{eq}$ $\quad$- equilibrium concentration
- 2 mole of X formed = 1 mole of $X_2$ used
- 1 mole of X formed = 1/2 mole of $X_2$ used
- $\beta$ mole of X formed = $\beta/2$ mole of $X_2$ used

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Problem with solution $\rightarrow$ Problem with solution </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **Solution**
- $X_2 (g) \rightleftharpoons 2X (g)$
- $ {1-\beta/2} \quad \quad  β_{eq}$ $\quad$- equilibrium concentration
- To calculate $K_p, P_{X{_2}} and P_X $ should be calculated first
- $n_T =1-\beta / 2+\beta$
- $n_T=1+\beta / 2$
- $Px_2 = x_{X_2} \times P$
- $Px_2=\frac{n_{x_2}}{n_T} \times P$
- $Px_2=\frac{1-\beta / 2}{1+\beta / 2} \times P$

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### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
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- **Solution**
- $P_x=\frac{n_x}{n_T} \times P=\frac{\beta}{1+\beta / 2} \times P$
- $K_p=\frac{(P_X)^2}{P_{X_2}}$
- $K_p=\frac{[(\frac{\beta}{1+\beta / 2}) \times p]^{2}}{[(\frac{1-\beta / 2}{1+\beta / 2})] p}$
- $K_p = \frac{(\frac{\beta}{\frac{2 + \beta}{2}})^2 \times P^2}{\frac{(2- \beta)/2}{(2+ \beta)/2} \times P}$
- $K_p =\frac{\beta^2 \times 4 \times P(2+\beta)}{(2+\beta)^2 \times(2-\beta)}$
- $K_p =\frac{4 P \beta^2}{4-\beta^2}=\frac{8 \beta^2}{4-\beta^2}$

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### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **Question** The equilibrium concentration of A, B and C for the reaction $A \leftrightarrow B+C$ are 4.6, 2.3 and 2.3 moles/L respectively at $25^{\circ} C$ If two moles per litre of A are removed, calculate the equilibrium  concentrations of A, B and C at the same temperature

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### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Problem with solution </primary>
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- **Solution**
- $A \rightleftharpoons B + C$
- $4.6 \quad 2.3 \quad 2.3$ $\quad$- equilibrium concentrations
- 4.6-2 $\quad 2.3 \quad 2.3$ $\quad$- no longer at equilibrium
- 2.6
- 2.6+x $\quad$ 2.3-x $\quad$ 2.3-x $\quad$-  new equilibrium
- $K_C=\frac{[(B)[C]}{[A]}$
- $K_c=\frac{(2.3-x)^2}{2 \cdot 6+x}=\frac{(2.3)^2}{4 \cdot 6}$

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<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Problem with solution </span> $\rightarrow$ Reaction quotient $\rightarrow$ Reaction quotient </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Reaction quotient </primary>
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- $A\rightarrow B$
- $Q=\frac{[B]}{[A]}$
- $B=\frac{1}{5}$
- $\frac{[B]_ {eq}}{[A]_ {e q}}=\frac{3}{3}$=1

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<footer style = "font-size: 18px">Problem with solution $\rightarrow$ Problem with solution $\rightarrow$ <span style = "color:blue"> Reaction quotient </span> $\rightarrow$ Reaction quotient $\rightarrow$ Reaction quotient </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Reaction quotient </primary>
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- Pure B
- $Q=\frac{[B]}{[A]}=\frac{5}{1}$
- $Q>K$
- Reverse reaction will take place

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<footer style = "font-size: 18px">Reaction quotient $\rightarrow$ Reaction quotient $\rightarrow$ <span style = "color:blue"> Reaction quotient </span> $\rightarrow$ Reaction quotient $\rightarrow$ Reaction quotient </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Reaction quotient </primary>
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- $Q=\frac{[P]}{[R]}$
- Reverse reaction is taking place

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### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Direction of the reaction by Q and K </primary>
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<footer style = "font-size: 18px">Reaction quotient $\rightarrow$ Reaction quotient $\rightarrow$ <span style = "color:blue"> Direction of the reaction by Q and K </span> $\rightarrow$ Direction of the reaction by Q and K $\rightarrow$ Direction of the reaction by Q and K </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Direction of the reaction by Q and K </primary>
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- $A\rightarrow B$
- $Q= \frac{[B]}{[A]}$
- $Q= \frac{2}{4}$
- $Q$ = 2
- $Q >K_c$- reaction will proceed in reverse direction
- $A\rightleftharpoons B$
- $2+x \quad  4-x$
- $K_c=\frac{4-x}{2+x} = 1/2$

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<footer style = "font-size: 18px">Direction of the reaction by Q and K $\rightarrow$ Direction of the reaction by Q and K $\rightarrow$ <span style = "color:blue"> Direction of the reaction by Q and K </span> $\rightarrow$ Direction of the reaction by Q and K $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Chemical Equilibrium L-2 </Success>
### <primary style="font-size:24px"> Direction of the reaction by Q and K </primary>
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- It can be simply known now if we mix A and B in which direction the reaction will proceed
- $B \rightarrow A$  or $A \rightarrow B$
- This can be known by comparing the value of K and Q
- If the  value of K and concentration of A and B is known,the it can be identified that how much A goes to B or how much B goes to A if reverse reaction is taking place if reverse reaction is taking place 
- By $K_c, Q$
- Direction of reaction and how much reaction will go in forward direction or reverse direction can be easily known

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