228961 For a reaction,
$\mathbf{A}+\mathbf{B} \rightleftharpoons \mathbf{C}+\mathbf{D},$
at equilibrium, one third of $A$ and $B$ are consumed. The value of equilibrium constant will be

228962 If the equilibrium constant for
$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \boxminus 2 \mathrm{NO}(\mathrm{g})$ is $\mathrm{K}$ the equilibrium constant for $\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \boxminus \mathrm{NO}(\mathrm{g})$ will be

228963 For the reaction, $A(s)+2 B^{+}(a q) \rightarrow A^{2+}(a q)+$ $2 B(s)$, the $E^{0}$ is $1.18 \mathrm{~V}$. Then equilibrium constant for the reaction is

228964 Equilibrium constants are given for the following two equilibria.
$\begin{aligned}
& \text { (i) } \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g}) \text {; } \\
& \mathrm{K}=\mathbf{2} \times 10^{-4} \mathrm{~L} \mathrm{~mol}^{-1}
\end{aligned}$
(ii) $2 \mathrm{AB}(\mathrm{g})+\mathrm{C}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ABC}(\mathrm{g})$;
$\mathrm{K}=\mathbf{2} \times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1}$
Calculate the equilibrium constant for the following equilibrium.
$\mathbf{A B C}(\mathbf{g}) \rightleftharpoons \frac{\mathbf{1}}{\mathbf{2}} \mathbf{A}_{2}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{B}_{\mathbf{2}}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{C}_{\mathbf{2}} \text { (g) }$

228965 HA is a weak acid. At $25^{\circ} \mathrm{C}$, the molar conductivity of $0.02 \mathrm{MHA}$ is $150 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, If its $\Lambda_{\mathrm{m}}^{\mathrm{o}}$ is $300 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ the equilibrium constant of $\mathrm{HA}$ dissociation is

228961 For a reaction,
$\mathbf{A}+\mathbf{B} \rightleftharpoons \mathbf{C}+\mathbf{D},$
at equilibrium, one third of $A$ and $B$ are consumed. The value of equilibrium constant will be

228962 If the equilibrium constant for
$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \boxminus 2 \mathrm{NO}(\mathrm{g})$ is $\mathrm{K}$ the equilibrium constant for $\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \boxminus \mathrm{NO}(\mathrm{g})$ will be

228963 For the reaction, $A(s)+2 B^{+}(a q) \rightarrow A^{2+}(a q)+$ $2 B(s)$, the $E^{0}$ is $1.18 \mathrm{~V}$. Then equilibrium constant for the reaction is

228964 Equilibrium constants are given for the following two equilibria.
$\begin{aligned}
& \text { (i) } \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g}) \text {; } \\
& \mathrm{K}=\mathbf{2} \times 10^{-4} \mathrm{~L} \mathrm{~mol}^{-1}
\end{aligned}$
(ii) $2 \mathrm{AB}(\mathrm{g})+\mathrm{C}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ABC}(\mathrm{g})$;
$\mathrm{K}=\mathbf{2} \times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1}$
Calculate the equilibrium constant for the following equilibrium.
$\mathbf{A B C}(\mathbf{g}) \rightleftharpoons \frac{\mathbf{1}}{\mathbf{2}} \mathbf{A}_{2}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{B}_{\mathbf{2}}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{C}_{\mathbf{2}} \text { (g) }$

228965 HA is a weak acid. At $25^{\circ} \mathrm{C}$, the molar conductivity of $0.02 \mathrm{MHA}$ is $150 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, If its $\Lambda_{\mathrm{m}}^{\mathrm{o}}$ is $300 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ the equilibrium constant of $\mathrm{HA}$ dissociation is

228961 For a reaction,
$\mathbf{A}+\mathbf{B} \rightleftharpoons \mathbf{C}+\mathbf{D},$
at equilibrium, one third of $A$ and $B$ are consumed. The value of equilibrium constant will be

228962 If the equilibrium constant for
$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \boxminus 2 \mathrm{NO}(\mathrm{g})$ is $\mathrm{K}$ the equilibrium constant for $\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \boxminus \mathrm{NO}(\mathrm{g})$ will be

228963 For the reaction, $A(s)+2 B^{+}(a q) \rightarrow A^{2+}(a q)+$ $2 B(s)$, the $E^{0}$ is $1.18 \mathrm{~V}$. Then equilibrium constant for the reaction is

228964 Equilibrium constants are given for the following two equilibria.
$\begin{aligned}
& \text { (i) } \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g}) \text {; } \\
& \mathrm{K}=\mathbf{2} \times 10^{-4} \mathrm{~L} \mathrm{~mol}^{-1}
\end{aligned}$
(ii) $2 \mathrm{AB}(\mathrm{g})+\mathrm{C}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ABC}(\mathrm{g})$;
$\mathrm{K}=\mathbf{2} \times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1}$
Calculate the equilibrium constant for the following equilibrium.
$\mathbf{A B C}(\mathbf{g}) \rightleftharpoons \frac{\mathbf{1}}{\mathbf{2}} \mathbf{A}_{2}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{B}_{\mathbf{2}}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{C}_{\mathbf{2}} \text { (g) }$

228965 HA is a weak acid. At $25^{\circ} \mathrm{C}$, the molar conductivity of $0.02 \mathrm{MHA}$ is $150 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, If its $\Lambda_{\mathrm{m}}^{\mathrm{o}}$ is $300 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ the equilibrium constant of $\mathrm{HA}$ dissociation is

228961 For a reaction,
$\mathbf{A}+\mathbf{B} \rightleftharpoons \mathbf{C}+\mathbf{D},$
at equilibrium, one third of $A$ and $B$ are consumed. The value of equilibrium constant will be

228962 If the equilibrium constant for
$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \boxminus 2 \mathrm{NO}(\mathrm{g})$ is $\mathrm{K}$ the equilibrium constant for $\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \boxminus \mathrm{NO}(\mathrm{g})$ will be

228963 For the reaction, $A(s)+2 B^{+}(a q) \rightarrow A^{2+}(a q)+$ $2 B(s)$, the $E^{0}$ is $1.18 \mathrm{~V}$. Then equilibrium constant for the reaction is

228964 Equilibrium constants are given for the following two equilibria.
$\begin{aligned}
& \text { (i) } \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g}) \text {; } \\
& \mathrm{K}=\mathbf{2} \times 10^{-4} \mathrm{~L} \mathrm{~mol}^{-1}
\end{aligned}$
(ii) $2 \mathrm{AB}(\mathrm{g})+\mathrm{C}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ABC}(\mathrm{g})$;
$\mathrm{K}=\mathbf{2} \times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1}$
Calculate the equilibrium constant for the following equilibrium.
$\mathbf{A B C}(\mathbf{g}) \rightleftharpoons \frac{\mathbf{1}}{\mathbf{2}} \mathbf{A}_{2}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{B}_{\mathbf{2}}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{C}_{\mathbf{2}} \text { (g) }$

228965 HA is a weak acid. At $25^{\circ} \mathrm{C}$, the molar conductivity of $0.02 \mathrm{MHA}$ is $150 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, If its $\Lambda_{\mathrm{m}}^{\mathrm{o}}$ is $300 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ the equilibrium constant of $\mathrm{HA}$ dissociation is

228961 For a reaction,
$\mathbf{A}+\mathbf{B} \rightleftharpoons \mathbf{C}+\mathbf{D},$
at equilibrium, one third of $A$ and $B$ are consumed. The value of equilibrium constant will be

228962 If the equilibrium constant for
$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \boxminus 2 \mathrm{NO}(\mathrm{g})$ is $\mathrm{K}$ the equilibrium constant for $\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \boxminus \mathrm{NO}(\mathrm{g})$ will be

228963 For the reaction, $A(s)+2 B^{+}(a q) \rightarrow A^{2+}(a q)+$ $2 B(s)$, the $E^{0}$ is $1.18 \mathrm{~V}$. Then equilibrium constant for the reaction is

228964 Equilibrium constants are given for the following two equilibria.
$\begin{aligned}
& \text { (i) } \mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g}) \text {; } \\
& \mathrm{K}=\mathbf{2} \times 10^{-4} \mathrm{~L} \mathrm{~mol}^{-1}
\end{aligned}$
(ii) $2 \mathrm{AB}(\mathrm{g})+\mathrm{C}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ABC}(\mathrm{g})$;
$\mathrm{K}=\mathbf{2} \times 10^{-2} \mathrm{~L} \mathrm{~mol}^{-1}$
Calculate the equilibrium constant for the following equilibrium.
$\mathbf{A B C}(\mathbf{g}) \rightleftharpoons \frac{\mathbf{1}}{\mathbf{2}} \mathbf{A}_{2}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{B}_{\mathbf{2}}(\mathbf{g})+\frac{\mathbf{1}}{\mathbf{2}} \mathbf{C}_{\mathbf{2}} \text { (g) }$

228965 HA is a weak acid. At $25^{\circ} \mathrm{C}$, the molar conductivity of $0.02 \mathrm{MHA}$ is $150 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, If its $\Lambda_{\mathrm{m}}^{\mathrm{o}}$ is $300 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ the equilibrium constant of $\mathrm{HA}$ dissociation is