Law of equilibrium and Equilibrium constant Questions in English

(B) The given reaction is $2 X_{(g)} + Y_{(g)} \rightleftharpoons 3 Z_{(g)}$.
The equilibrium constant $(k)$ is defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants,each raised to the power of their stoichiometric coefficients.
For the reaction $aA + bB \rightleftharpoons cC + dD$,the equilibrium constant is $k = \frac{[C]^c [D]^d}{[A]^a [B]^b}$.
Applying this to the given reaction,we get $k = \frac{[Z]^3}{[X]^2 [Y]}$.

(C) For the reaction $A + B \rightleftharpoons C + D$,the equilibrium constant expression is given by $K_C = \frac{[C][D]}{[A][B]}$.
Given that $K_C = 10$,we substitute this value into the expression: $10 = \frac{[C][D]}{[A][B]}$.
Rearranging the equation to solve for $[A][B]$,we get $[A][B] = \frac{1}{10}[C][D]$.
Therefore,$[A][B] = 0.1[C][D]$.

(B) Given,the equilibrium constant $K_C = 49$.
The equilibrium concentrations are $[H_2] = 2.0 \times 10^{-2} \ M$ and $[I_2] = 8.0 \times 10^{-2} \ M$.
The expression for the equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $K_C = \frac{[HI]^2}{[H_2][I_2]}$.
Rearranging the formula to solve for $[HI]$,we get $[HI]^2 = K_C \times [H_2] \times [I_2]$.
Substituting the given values: $[HI]^2 = 49 \times (2.0 \times 10^{-2}) \times (8.0 \times 10^{-2}) = 49 \times 16 \times 10^{-4}$.
Taking the square root of both sides: $[HI] = \sqrt{49 \times 16 \times 10^{-4}} = 7 \times 4 \times 10^{-2} = 28 \times 10^{-2} = 0.28 \ mol \ L^{-1}$.

(D) Given reactions:
$I. \frac{1}{2} N_{2(g)} + O_{2(g)} \rightleftharpoons NO_{2(g)}$ with equilibrium constant $X$.
$II. 2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$ with equilibrium constant $Y$.
Target reaction $(III): N_2O_{4(g)} \rightleftharpoons N_{2(g)} + 2 O_{2(g)}$.
To obtain reaction $(III)$,we reverse reaction $(II)$ and add it to twice the reverse of reaction $(I)$ (or simply reverse the sum of $2 \times Eq(I) + Eq(II)$):
$2 \times Eq(I): N_{2(g)} + 2 O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$ with constant $X^2$.
$Eq(II): 2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$ with constant $Y$.
Adding these gives: $N_{2(g)} + 2 O_{2(g)} \rightleftharpoons N_2O_{4(g)}$ with constant $K = X^2Y$.
Since reaction $(III)$ is the reverse of this sum,its equilibrium constant $Z = \frac{1}{K} = \frac{1}{X^2Y}$.

(D) The given reactions are:
(i) $H_3PO_{4\text{(aq)}} \rightleftharpoons H^+{_{\text{(aq)}}} + H_2PO_4^-{_{\text{(aq)}}} ; \ K_1 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]}$
(ii) $H_2PO_4^-{_{\text{(aq)}}} \rightleftharpoons H^+{_{\text{(aq)}}} + HPO_4^{2-}{_{\text{(aq)}}} ; \ K_2 = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]}$
(iii) $HPO_4^{2-}{_{\text{(aq)}}} \rightleftharpoons H^+{_{\text{(aq)}}} + PO_4^{3-}{_{\text{(aq)}}} ; \ K_3 = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$
Adding these three reactions gives the net reaction:
$H_3PO_{4\text{(aq)}} \rightleftharpoons 3H^{+}{_{\text{(aq)}}} + PO_4^{3-}{_{\text{(aq)}}}$
When reactions are added,their equilibrium constants are multiplied.
Therefore,the equilibrium constant $K$ for the net reaction is:
$K = K_1 \times K_2 \times K_3$

(A) For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1 = 5 \times 10^{-2}$.
For the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$,the reaction is the reverse of the first reaction multiplied by $2$.
Therefore,the new equilibrium constant $K_2$ is given by $K_2 = \frac{1}{K_1^2}$.
$K_2 = \frac{1}{(5 \times 10^{-2})^2} = \frac{1}{25 \times 10^{-4}} = \frac{10^4}{25} = 400 \ atm$.

(D) The given reaction is: $2 \ AO_{2(g)} + O_{2(g)} \rightleftharpoons 2 \ AO_{3(g)}$ with equilibrium constant $K_{p} = 4 \times 10^{10}$.
We need to find the equilibrium constant $K_{p}^{\prime}$ for the reaction: $3 \ AO_{2(g)} + \frac{3}{2} \ O_{2(g)} \rightleftharpoons 3 \ AO_{3(g)}$.
Notice that the second reaction is obtained by multiplying the first reaction by a factor of $n = \frac{3}{2}$.
According to the properties of equilibrium constants,if a reaction is multiplied by a factor $n$,the new equilibrium constant $K_{p}^{\prime}$ is given by $K_{p}^{\prime} = (K_{p})^n$.
Substituting the values: $K_{p}^{\prime} = (4 \times 10^{10})^{3/2}$.
$K_{p}^{\prime} = (\sqrt{4 \times 10^{10}})^3 = (2 \times 10^5)^3$.
$K_{p}^{\prime} = 8 \times 10^{15}$.

(C) Given reactions:
$(i) \ 2 A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)} ; K_1 = 16$
$(ii) \ 2 B_{(g)} + C_{(g)} \rightleftharpoons 2 D_{(g)} ; K_2 = 25$
We need the equilibrium constant $(K_3)$ for:
$(iii) \ A_{(g)} + \frac{1}{2} B_{(g)} \rightleftharpoons D_{(g)}$
From $(i)$,$K_1 = \frac{[B][C]}{[A]^2} = 16$.
From $(ii)$,$K_2 = \frac{[D]^2}{[C][B]^2} = 25$.
Multiplying $(i)$ and $(ii)$:
$K_1 \times K_2 = \frac{[B][C]}{[A]^2} \times \frac{[D]^2}{[C][B]^2} = \frac{[D]^2}{[A]^2 [B]} = 16 \times 25 = 400$.
Taking the square root of the expression:
$\sqrt{\frac{[D]^2}{[A]^2 [B]}} = \frac{[D]}{[A][B]^{1/2}} = \sqrt{400} = 20$.
Thus,$K_3 = 20$.

(B) The equilibrium constant ($K_c$ or $K_p$) is a function of temperature only for a given reaction.
It remains constant regardless of changes in pressure,volume,or concentration of the reactants or products.
Since the temperature remains constant at $780 \ K$,the equilibrium constant will remain $3.52$.

(B) Given reaction: $\frac{1}{3} N_{2(g)} + H_{2(g)} \rightleftharpoons \frac{2}{3} NH_{3(g)}$,$K_C = 50$.
Multiply the equation by $3$: $N_{2(g)} + 3 H_{2(g)} \rightleftharpoons 2 NH_{3(g)}$. The new equilibrium constant $K_C'' = (K_C)^3 = (50)^3 = 125000$.
Reverse the equation: $2 NH_{3(g)} \rightleftharpoons N_{2(g)} + 3 H_{2(g)}$. The equilibrium constant $K_C' = \frac{1}{K_C''} = \frac{1}{(50)^3}$.
$K_C' = \frac{1}{125000} = 8 \times 10^{-6}$.

(A) For the given reaction: $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$
According to the law of mass action,the equilibrium constant $K_c$ is given by the ratio of the product of concentrations of products to the reactants,each raised to the power of their stoichiometric coefficients.
$K_c = \frac{[P_4O_{10}]}{[P_4][O_2]^5}$
Since $P_{4(s)}$ and $P_4O_{10(s)}$ are pure solids,their active mass (concentration) is taken as $1$.
Substituting these values: $K_c = \frac{1}{1 \times [O_2]^5} = \frac{1}{[O_2]^5}$

(B) Given reaction: $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$,$K = 64$.
Target reaction: $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$,$K' = ?$.
To obtain the target reaction,we first reverse the given reaction: $2 SO_3 \rightleftharpoons 2 SO_2 + O_2$. The equilibrium constant for this reversed reaction is $K_{rev} = \frac{1}{K} = \frac{1}{64}$.
Next,we multiply the coefficients of the reversed reaction by $\frac{1}{2}$: $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$.
When a reaction is multiplied by a factor $n$,the new equilibrium constant is $(K_{original})^n$. Here,$n = \frac{1}{2}$.
Therefore,$K' = (K_{rev})^{1/2} = (\frac{1}{64})^{1/2} = \frac{1}{8} = 0.125$.
Thus,option $(b)$ is the correct answer.

(A) The given reaction is: $N_2 + O_2 \rightleftharpoons 2 NO$
The equilibrium constant expression is given by:
$K_C = \frac{[NO]^2}{[N_2][O_2]}$
Given equilibrium concentrations:
$[N_2] = 4 \times 10^{-3} \ M$
$[O_2] = 3 \times 10^{-3} \ M$
$[NO] = 3 \times 10^{-3} \ M$
Substituting the values into the expression:
$K_C = \frac{(3 \times 10^{-3})^2}{(4 \times 10^{-3})(3 \times 10^{-3})}$
$K_C = \frac{9 \times 10^{-6}}{12 \times 10^{-6}}$
$K_C = \frac{9}{12} = 0.75$

(D) The equilibrium constant expression for the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is given by:
$K_C = \frac{[PCl_3][Cl_2]}{[PCl_5]}$
Given $K_C = 1.79$,$[PCl_5] = 1.41 \ M$,and $[PCl_3] = 1.59 \ M$.
Substituting these values into the expression:
$1.79 = \frac{1.59 \times [Cl_2]}{1.41}$
Solving for $[Cl_2]$:
$[Cl_2] = \frac{1.79 \times 1.41}{1.59} = 1.587 \ M$
Rounding to two decimal places,we get $[Cl_2] \approx 1.59 \ M$.

(A) For the reaction $2AB \rightleftharpoons A_2 + B_2$,the equilibrium constant is $K_c = 49$.
When the coefficients of a balanced chemical equation are multiplied by a factor $n$,the new equilibrium constant $K'_c$ is given by $K'_c = (K_c)^n$.
In this case,the original equation is multiplied by $n = 1/2$.
Therefore,$K'_c = (49)^{1/2} = \sqrt{49} = 7$.

(D) For the reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,the equilibrium constant is $K = \frac{[HI]^2}{[H_2][I_2]}$ ...$(i)$
For the reaction: $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$,the equilibrium constant is $K' = \frac{[H_2]^{1/2} [I_2]^{1/2}}{[HI]}$ ...$(ii)$
Comparing $(ii)$ with $(i)$,we can see that $K' = \sqrt{\frac{1}{K}} = \frac{1}{\sqrt{K}}$.

(C) For the reaction: $N_{2(g)} + 2 O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$,the equilibrium constant is $K_1 = 100$.
$K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2} = 100$ ... $(i)$
For the target reaction: $NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$,the equilibrium constant is $K_2$.
$K_2 = \frac{[N_2]^{1/2} [O_2]}{[NO_2]}$ ... $(ii)$
Comparing $(i)$ and $(ii)$,we observe that $K_2 = \sqrt{\frac{1}{K_1}}$.
$K_2 = \sqrt{\frac{1}{100}} = \frac{1}{10} = 0.1$.

(C) For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1 = 5 \times 10^{-2}$.
For the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$,we observe that the reaction is the reverse of the first reaction multiplied by a factor of $2$.
Therefore,the new equilibrium constant $K_2$ is given by $K_2 = \frac{1}{K_1^2}$.
Substituting the value of $K_1$:
$K_2 = \frac{1}{(5 \times 10^{-2})^2} = \frac{1}{25 \times 10^{-4}} = \frac{10^4}{25} = 400 = 4 \times 10^2$.

(A) Given reactions:
$(i) 2 A \rightleftharpoons B + C, K_{1} = 1$
$(ii) 2 B \rightleftharpoons C + D, K_{2} = 16$
$(iii) 2 C + D \rightleftharpoons 2 P, K_{3} = 25$
We need the equilibrium constant for $P \rightleftharpoons A + \frac{1}{2} B$.
Reverse reaction $(iii)$ and divide by $2$: $P \rightleftharpoons C + \frac{1}{2} D, K' = \sqrt{\frac{1}{K_{3}}} = \frac{1}{5}$.
Reverse reaction $(ii)$ and divide by $2$: $\frac{1}{2} C + \frac{1}{2} D \rightleftharpoons B, K'' = \sqrt{\frac{1}{K_{2}}} = \frac{1}{4}$.
Reverse reaction $(i)$ and divide by $2$: $\frac{1}{2} B + \frac{1}{2} C \rightleftharpoons A, K''' = \sqrt{\frac{1}{K_{1}}} = 1$.
Adding these reactions: $P + \frac{1}{2} C + \frac{1}{2} D + \frac{1}{2} B + \frac{1}{2} C \rightleftharpoons C + \frac{1}{2} D + B + A$.
Simplifying: $P \rightleftharpoons A + \frac{1}{2} B$.
The equilibrium constant $K_{final} = K' \times K'' \times K''' = \frac{1}{5} \times \frac{1}{4} \times 1 = \frac{1}{20}$.

(B) The target reaction is the oxidation of $2 \text{ mol}$ of $NH_{3}$ to $NO$:
$2 NH_{3} + \frac{5}{2} O_{2} \rightleftharpoons 2 NO + 3 H_{2} O$
Given equations:
$(i) N_{2} + 3 H_{2} \rightleftharpoons 2 NH_{3} ; K_{1}$
$(ii) N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$
$(iii) H_{2} + \frac{1}{2} O_{2} \rightleftharpoons H_{2} O ; K_{3}$
To obtain the target reaction:
$1$. Reverse equation $(i)$: $2 NH_{3} \rightleftharpoons N_{2} + 3 H_{2} ; K' = \frac{1}{K_{1}}$
$2$. Keep equation $(ii)$ as it is: $N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$
$3$. Multiply equation $(iii)$ by $3$: $3 H_{2} + \frac{3}{2} O_{2} \rightleftharpoons 3 H_{2} O ; K'' = K_{3}^{3}$
Adding these three equations:
$(2 NH_{3} + N_{2} + O_{2} + 3 H_{2} + \frac{3}{2} O_{2}) \rightleftharpoons (N_{2} + 3 H_{2} + 2 NO + 3 H_{2} O)$
Simplifying gives: $2 NH_{3} + \frac{5}{2} O_{2} \rightleftharpoons 2 NO + 3 H_{2} O$
The equilibrium constant $K$ is:
$K = K' \cdot K_{2} \cdot K'' = \frac{1}{K_{1}} \cdot K_{2} \cdot K_{3}^{3} = K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

(D) For the reaction $A_{(g)} \rightleftharpoons B_{(g)}$,the equilibrium constant $K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75$.
Initial moles of $A$ in $1 \text{ L}$ flask $= 0.5 \text{ mol}$.
Initial moles of $B$ in $1 \text{ L}$ flask $= 0.375 \text{ mol}$.
After adding $0.1 \text{ mol}$ of $A$,total moles of $A = 0.5 + 0.1 = 0.6 \text{ mol}$.
Let $x$ be the amount of $A$ reacted to reach new equilibrium.
New $[A] = 0.6 - x$ and New $[B] = 0.375 + x$.
$K_c = \frac{0.375 + x}{0.6 - x} = 0.75$.
$0.375 + x = 0.45 - 0.75x$.
$1.75x = 0.075 \Rightarrow x = 0.0428$.
New $[A] = 0.6 - 0.0428 = 0.557 \text{ M}$ and New $[B] = 0.375 + 0.0428 = 0.418 \text{ M}$.

(C) The target reaction is: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$.
We are given:
$(1)$ $2xy \rightleftharpoons x_2 + y_2$ with $K_1 = 2.5 \times 10^5$
$(2)$ $xy + \frac{1}{2}z_2 \rightleftharpoons xyz$ with $K_2 = 5 \times 10^{-3}$
To obtain the target reaction,we perform: (Reaction $2$) - $\frac{1}{2} \times$ (Reaction $1$).
This corresponds to: $(xy + \frac{1}{2}z_2) - \frac{1}{2}(2xy) \rightleftharpoons xyz - \frac{1}{2}(x_2 + y_2)$.
Rearranging gives: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$.
The equilibrium constant $K_3$ is given by $K_3 = \frac{K_2}{(K_1)^{1/2}}$.
$K_3 = \frac{5 \times 10^{-3}}{\sqrt{2.5 \times 10^5}} = \frac{5 \times 10^{-3}}{\sqrt{25 \times 10^4}} = \frac{5 \times 10^{-3}}{500} = \frac{5 \times 10^{-3}}{5 \times 10^2} = 1.0 \times 10^{-5}$.

(D) If a chemical equation is multiplied by a factor '$n$',the new equilibrium constant $K'$ is given by $K' = K^{n}$.
Here,the original reaction is $A_{2}(g) + B_{2}(g) \rightleftharpoons C(g)$ with equilibrium constant $K = 2.7 \times 10^{-5}$.
The new reaction is $\frac{1}{3}A_{2}(g) + \frac{1}{3}B_{2}(g) \rightleftharpoons \frac{1}{3}C(g)$,which is obtained by multiplying the original reaction by $n = 1/3$.
Therefore,the new equilibrium constant $K'$ is $K^{1/3}$.
$K' = (2.7 \times 10^{-5})^{1/3} = (27 \times 10^{-6})^{1/3}$.
$K' = (27)^{1/3} \times (10^{-6})^{1/3} = 3 \times 10^{-2}$.