Chemical Equilibrium

Calculate $ K_C $:

The equilibrium constant in terms of concentrations ($ K_C $) is given by the ratio of the forward reaction rate constant ($ k_f $) to the backward reaction rate constant ($ k_b $):

$ K_C = \frac{k_f}{k_b} = \frac{1}{2500} $

Convert $ K_C $ to $ K_P $:

To convert the concentration equilibrium constant ($ K_C $) to the pressure equilibrium constant ($ K_P $), use the following relation:

$ K_P = K_C (RT)^{\Delta n_g} $

where:

$ R = 0.0831 \, \mathrm{L \, atm \, mol^{-1} \, K^{-1}} $

$ T = 1000 \, \mathrm{K} $

$\Delta n_g = 2 - 1 = 1$ (change in the number of moles of gas, $ \Delta n_g $, from reactants to products)

Substituting these values in, we get:

$ K_P = \frac{1}{2500} \times 0.0831 \times 1000 $

Calculate $ K_P $:

Solving the above expression gives:

$ K_P = 0.033 $

Therefore, the equilibrium constant $ K_P $ for the reaction at 1000 K is 0.033.

Yield of the product generally depends on

$\bullet$ Temperature

$\bullet$ Concentration of reactant(s) and product(s)

$\bullet$ Pressure

As this is an endothermic reaction ( $\Delta \mathrm{H}=+180.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ), so, increase in temperature will shift equilibrium in forward direction to increase yield of NO.

Increase in concentration of reactants ( $\mathrm{N}_2$ and $\mathrm{O}_2$ ) also shifts the equilibrium in forward direction and increase the yield of NO.

Hence, (A), (C) and (D) only will increase yield of NO.

To find the relationship between $\mathrm{K}_1$ and $\mathrm{K}_2$, let's carefully analyze the given equilibrium reactions and their constants.

First, let's write down the equilibrium reactions clearly:

Reaction 1: $3 \mathrm{~A}_2+\mathrm{~B}_2 \rightleftharpoons 2 \mathrm{~A}_3 \mathrm{~B}$ with equilibrium constant $\mathrm{K}_1$.

Reaction 2: $\mathrm{~A}_3 \mathrm{~B} \rightleftharpoons \frac{3}{2} \mathrm{~A}_2 + \frac{1}{2} \mathrm{~B}_2$ with equilibrium constant $\mathrm{K}_2$.

Now, understand that $\mathrm{K}_2$ represents the equilibrium constant of the reverse reaction of Reaction 1 but with both sides divided by 2. We need to relate these constants. Here is the step-by-step process:

We know Reaction 1 is: $3 \mathrm{~A}_2 + \mathrm{~B}_2 \rightleftharpoons 2 \mathrm{~A}_3 \mathrm{~B}$.

The equilibrium constant for Reaction 1 is: $\mathrm{K}_1 = \frac{[\mathrm{A}_3 \mathrm{~B}]^2}{[\mathrm{A}_2]^3[\mathrm{B}_2]}$.

For Reaction 2: $\mathrm{~A}_3 \mathrm{~B} \rightleftharpoons \frac{3}{2} \mathrm{~A}_2 + \frac{1}{2} \mathrm{~B}_2$, the equilibrium constant $\mathrm{K}_2$ can be expressed as the equilibrium constant of the reverse of Reaction 1, with adjusted coefficients.

The equilibrium constant for the reverse reaction of Reaction 1 would be $\frac{1}{\mathrm{K}_1}$. Since Reaction 2 includes halving the coefficients, the equilibrium constant should be adjusted by taking the square root:

$\mathrm{K}_2 = \left(\frac{1}{\mathrm{K}_1}\right)^{1/2} = \frac{1}{\sqrt{\mathrm{K}_1}}$.

Therefore, the relationship between $\mathrm{K}_1$ and $\mathrm{K}_2$ is:

Option D: $\mathrm{K}_2 = \frac{1}{\sqrt{\mathrm{K}_1}}$.

The given chemical reaction is an example of the synthesis of ammonia, commonly known as the Haber process:

$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}), \Delta \mathrm{H}=-\mathrm{Q}$

This reaction is exothermic, as indicated by the negative enthalpy change ($\Delta \mathrm{H} = -\mathrm{Q}$). In order to determine which conditions favor the forward reaction, we need to consider Le Chatelier's principle, which states that the system will adjust to counteract any changes imposed upon it.

Let's analyze each option:

Option A: Use of catalyst

A catalyst does not favor the forward or reverse direction of a reaction. It only speeds up the rate at which equilibrium is achieved.

Option B: Decreasing concentration of $\mathrm{N}_2$

Decreasing the concentration of $\mathrm{N}_2$ would shift the equilibrium to the left, favoring the reverse reaction to produce more $\mathrm{N}_2$ and $\mathrm{H}_2$.

Option C: Low pressure, high temperature, and high concentration of ammonia

Low pressure and high temperature would favor the reverse reaction. Moreover, a high concentration of ammonia would also shift the equilibrium to the left. Thus, this set of conditions does not favor the forward reaction.

Option D: High pressure, low temperature and higher concentration of $\mathrm{H}_2$

High pressure favors the formation of ammonia because there are fewer moles of gas on the product side (2 moles) as compared to the reactant side (4 moles, i.e., 1 mole of $\mathrm{N}_2$ and 3 moles of $\mathrm{H}_2$). Low temperature favors the exothermic forward reaction. Additionally, a higher concentration of $\mathrm{H}_2$ will shift the equilibrium to the right, forming more ammonia.

Therefore, the correct answer is:

Option D: High pressure, low temperature and higher concentration of $\mathrm{H}_2$

To determine in which of the given equilibria $\mathrm{K}_p$ and $\mathrm{K}_{\mathrm{c}}$ are not equal, it's important to understand the relationship between these two equilibrium constants. This relationship is expressed by the equation:

$\mathrm{K}_p = \mathrm{K}_c (RT)^{\Delta n}$

where $R$ is the gas constant, $T$ is the temperature in Kelvin, and $\Delta n$ is the change in the number of moles of gas (number of moles of gaseous products minus number of moles of gaseous reactants).

If $\Delta n = 0$, then $\mathrm{K}_p$ and $\mathrm{K}_{\mathrm{c}}$ are equal because $(RT)^0 = 1$. However, if $\Delta n \neq 0$, the constants will not be the same, and the degree to which they differ will depend on the temperature and the value of $\Delta n$.

Now, let's analyze each option:

• Option A: $\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}$

Reactant side moles = 1, Product side moles = 2; $\Delta n = 2 - 1 = 1$.

• Option B: $\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}$

Reactant side moles = 2, Product side moles = 2; $\Delta n = 2 - 2 = 0$.

• Option C: $\mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})}$

Reactant side moles = 2, Product side moles = 2; $\Delta n = 2 - 2 = 0$.

• Option D: $2 \mathrm{BrCl}_{(\mathrm{g})} \rightleftharpoons \mathrm{Br}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}$

Reactant side moles = 2, Product side moles = 2; $\Delta n = 2 - 2 = 0$.

From this analysis, it is evident that $\mathrm{K}_p$ and $\mathrm{K}_{\mathrm{c}}$ are not equal in Option A where $\Delta n = 1$. In all other options, since $\Delta n = 0$, $\mathrm{K}_p$ is equal to $\mathrm{K}_{\mathrm{c}}$. Thus, the correct answer is Option A.

To determine which option is correct regarding the reaction state and its direction, we need to calculate the reaction quotient $Q_c$ and compare it to the equilibrium constant $K_c$. The reaction given is:

$ 2 \mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C} $

The equilibrium constant expression $K_c$ for this reaction is:

$ K_c = \frac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}]^2} $

Given that $K_c = 4 \times 10^{-3}$ and the concentrations of A, B, and C at this time are each $2 \times 10^{-3}$ M, we can substitute these values into the expression for $K_c$ to calculate the reaction quotient $Q_c$:

$ Q_c = \frac{(2 \times 10^{-3} \mathrm{M})(2 \times 10^{-3} \mathrm{M})}{(2 \times 10^{-3} \mathrm{M})^2} $

Simplifying, we find:

$ Q_c = \frac{4 \times 10^{-6} \mathrm{M}^2}{4 \times 10^{-6} \mathrm{M}^2} = 1 $

Comparing $Q_c$ with $K_c$:

$ Q_c = 1 $

$ K_c = 4 \times 10^{-3} $

Since $Q_c > K_c$ (1 > 0.004), the reaction quotient is greater than the equilibrium constant. This indicates that the concentration of products (B and C) is too high relative to the concentration of reactants (A) for the system to be at equilibrium under these conditions.

This means that the reaction has a tendency to move in the backward direction to reach equilibrium, reducing the concentration of the products (B and C) and increasing the concentration of the reactant (A). Therefore, the correct answer to the given question is:

Option C: Reaction has a tendency to go in backward direction.

$ \begin{aligned} &2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}\\ &\begin{aligned} \mathrm{K}_c & =\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2} \\ & =\frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}} \\ & =1.607 \end{aligned} \end{aligned} $

$ \mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{(0.1-0.1 \alpha)^2} $

$ \begin{gathered} \mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{0.01(1-\alpha)^2} \\ 1.607=\frac{(0.05)^2 \alpha^2}{0.01(1-\alpha)^2} \\ \frac{\alpha^2}{(1-\alpha)^2}=\frac{1.607 \times(0.1)^2}{(0.05)^2} \\ \frac{\alpha}{1-\alpha}=\frac{1.27 \times 0.1}{0.05} \\ \frac{\alpha}{1-\alpha}=2.54 \\ \alpha=2.54-2.54 \alpha \\ 3.54 \alpha=2.54 \\ \alpha=\frac{2.54}{3.54}=0.717 \end{gathered} $

$\mathrm{K_a=C\alpha^2}$

$\mathrm{K_a=(0.1)\times(0.01)^2}$

$\mathrm{K_a=1\times10^{-5}}$

CO₂(g) + C(s) $\rightleftharpoons$ 2CO(g)

$\Delta$n_g = 2 $-$ 1 = 1

K_p = K_c(RT)^$\Delta$n_g

K_p = K_c(RT)

[$\because$ K_p = 3]

${\mathrm{K_c}} = {{{\mathrm{K_p}}} \over {\mathrm{RT}}} = {3 \over {0.083 \times 1000}}$

= 0.036

= 3.6 $\times$ 10^$-$2

3O₂(g) $\rightleftharpoons$ 2O₃(g)

${K_c} = {{{{[{O_3}]}^2}} \over {{{[{O_2}]}^3}}}$

${[{O_3}]^2} = {K_c}{[{O_2}]^3} = 3 \times {10^{ - 59}} \times {(0.04)^3}$

${[{O_3}]^2} = 1.9 \times {10^{ - 63}} = 19 \times {10^{ - 64}}$

$[{O_3}] = 4.38 \times {10^{ - 32}}$

Concentration of O₃ at equilibrium $ = 4.38 \times {10^{ - 32}}$ M

$\begin{aligned} \text { Given, } \mathrm{N}_2 & =3.0 \times 10^{-3} \mathrm{M} \\ \mathrm{O}_2 & =4.2 \times 10^{-3} \mathrm{M} \\ \mathrm{NO} & =28 \times 10^{-3} \mathrm{M} \end{aligned}$

For the given reaction,

$\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)$

equilibrium constant $K_C$ can be written as

$\begin{aligned} & K_C=\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]} \\ \therefore & \quad K_C=\frac{\left(2.8 \times 10^{-3} \mathrm{M}\right)^2}{\left(3.0 \times 10^{-3} \mathrm{M}\right)\left(4.2 \times 10^{-3} \mathrm{M}\right)}=0.622 \\ \because \quad & K_p=K_C \cdot(R T)^{\Delta n} \end{aligned}$

$\Delta n=$ Number of moles of gaseous products $-$ number of moles of gaseous reactants

$\therefore \quad \begin{aligned} \Delta n & =2-2=0 \\ & K_p=K_C \cdot(R T)^{\circ} \end{aligned}$

or, $\quad K_p=K_C$ or, $K_p=0.622 \mathrm{~atm}$

Both assertion and reason are correct and reason is the correct explanation of assertion.

$K_p=K_C(R T)^{\Delta n}$

$\Delta n$ = number of moles of gaseous products $-$ number of moles of gaseous reactants

When, $\Delta n>1, K_p>K_C$

When, $\Delta n<1, K_p< K_C$

When, $\Delta n=0, K_p=K_C$