00. Law of Chemical Equilibrium and Equilibrium Constant
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Chemical Equilibrium

228923 The equilibrium constant of the following are
$\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; \mathrm{K}_{1}$
$\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons2 \mathrm{NO} ; \mathrm{K}_{2}$
$\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons\mathrm{H}_{2} \mathrm{O} ; \quad \mathrm{K}_{3}$
The equilibrium constant $(K)$ of the reaction:
$2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2}$

1 $\mathrm{K}_{2} \mathrm{~K}_{3}^{3} / \mathrm{K}_{1}$
2 $\mathrm{K}_{2} \mathrm{~K}_{3} / \mathrm{K}_{1}$
3 $\mathrm{K}_{2}^{3} \mathrm{~K}_{3} / \mathrm{K}_{1}$
4 $\mathrm{K}_{1} \mathrm{~K}_{3}^{3} / \mathrm{K}_{2}$
2007
Chemical Equilibrium

228924 For the reaction, $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$, the equilibrium constant is $K_{1}$. The equilibrium constant is $K_{2}$ for the reaction, $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$. What is $\mathrm{K}$ for the reaction, $\mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \frac{1}{\mathbf{2}} \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ ?

1 $1 /\left(4 \mathrm{~K}_{1} \mathrm{~K}_{2}\right)$
2 $\left(1 / \mathrm{K}_{1} \mathrm{~K}_{2}\right)^{1 / 2}$
3 $1 /\left(\mathrm{K}_{1}, \mathrm{~K}_{2}\right)$
4 $1 /\left(2 \mathrm{~K}_{1}, \mathrm{~K}_{2}\right)$
[JIMPER-2014]
Chemical Equilibrium

228926 In the reaction,
$\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})$
if the concentration of $\mathrm{OH}^{-}$ions is decreased $\frac{1}{4}$ by times, then the equilibrium concentration of $\mathrm{Fe}^{3+}$ will increase by

1 8 times
2 16 times
3 64 times
4 4 times
[Karnataka-CET-2016]
Chemical Equilibrium

228927 If $K_{1}$ and $K_{2}$ are respective equilibrium constants for two reactions
$\mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \quad \mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})$
$\mathrm{XeO}_{4}(\mathrm{~g})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons \quad \mathrm{XeOF}_{4}(\mathrm{~g})+\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})$
the equilibrium constant for the reaction
$\mathrm{XeO}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \quad \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
will be

1 $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}^{2}}$
2 $\mathrm{K}_{1} \cdot \mathrm{K}_{2}$
3 $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}$
4 $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$
[BITSAT 2006]
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Chemical Equilibrium

228923 The equilibrium constant of the following are
$\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; \mathrm{K}_{1}$
$\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons2 \mathrm{NO} ; \mathrm{K}_{2}$
$\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons\mathrm{H}_{2} \mathrm{O} ; \quad \mathrm{K}_{3}$
The equilibrium constant $(K)$ of the reaction:
$2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2}$

1 $\mathrm{K}_{2} \mathrm{~K}_{3}^{3} / \mathrm{K}_{1}$
2 $\mathrm{K}_{2} \mathrm{~K}_{3} / \mathrm{K}_{1}$
3 $\mathrm{K}_{2}^{3} \mathrm{~K}_{3} / \mathrm{K}_{1}$
4 $\mathrm{K}_{1} \mathrm{~K}_{3}^{3} / \mathrm{K}_{2}$
2007
Chemical Equilibrium

228924 For the reaction, $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$, the equilibrium constant is $K_{1}$. The equilibrium constant is $K_{2}$ for the reaction, $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$. What is $\mathrm{K}$ for the reaction, $\mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \frac{1}{\mathbf{2}} \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ ?

1 $1 /\left(4 \mathrm{~K}_{1} \mathrm{~K}_{2}\right)$
2 $\left(1 / \mathrm{K}_{1} \mathrm{~K}_{2}\right)^{1 / 2}$
3 $1 /\left(\mathrm{K}_{1}, \mathrm{~K}_{2}\right)$
4 $1 /\left(2 \mathrm{~K}_{1}, \mathrm{~K}_{2}\right)$
[JIMPER-2014]
Chemical Equilibrium

228926 In the reaction,
$\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})$
if the concentration of $\mathrm{OH}^{-}$ions is decreased $\frac{1}{4}$ by times, then the equilibrium concentration of $\mathrm{Fe}^{3+}$ will increase by

1 8 times
2 16 times
3 64 times
4 4 times
[Karnataka-CET-2016]
Chemical Equilibrium

228927 If $K_{1}$ and $K_{2}$ are respective equilibrium constants for two reactions
$\mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \quad \mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})$
$\mathrm{XeO}_{4}(\mathrm{~g})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons \quad \mathrm{XeOF}_{4}(\mathrm{~g})+\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})$
the equilibrium constant for the reaction
$\mathrm{XeO}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \quad \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
will be

1 $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}^{2}}$
2 $\mathrm{K}_{1} \cdot \mathrm{K}_{2}$
3 $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}$
4 $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$
[BITSAT 2006]
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Chemical Equilibrium

228923 The equilibrium constant of the following are
$\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; \mathrm{K}_{1}$
$\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons2 \mathrm{NO} ; \mathrm{K}_{2}$
$\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons\mathrm{H}_{2} \mathrm{O} ; \quad \mathrm{K}_{3}$
The equilibrium constant $(K)$ of the reaction:
$2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2}$

1 $\mathrm{K}_{2} \mathrm{~K}_{3}^{3} / \mathrm{K}_{1}$
2 $\mathrm{K}_{2} \mathrm{~K}_{3} / \mathrm{K}_{1}$
3 $\mathrm{K}_{2}^{3} \mathrm{~K}_{3} / \mathrm{K}_{1}$
4 $\mathrm{K}_{1} \mathrm{~K}_{3}^{3} / \mathrm{K}_{2}$
2007
Chemical Equilibrium

228924 For the reaction, $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$, the equilibrium constant is $K_{1}$. The equilibrium constant is $K_{2}$ for the reaction, $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$. What is $\mathrm{K}$ for the reaction, $\mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \frac{1}{\mathbf{2}} \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ ?

1 $1 /\left(4 \mathrm{~K}_{1} \mathrm{~K}_{2}\right)$
2 $\left(1 / \mathrm{K}_{1} \mathrm{~K}_{2}\right)^{1 / 2}$
3 $1 /\left(\mathrm{K}_{1}, \mathrm{~K}_{2}\right)$
4 $1 /\left(2 \mathrm{~K}_{1}, \mathrm{~K}_{2}\right)$
[JIMPER-2014]
Chemical Equilibrium

228926 In the reaction,
$\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})$
if the concentration of $\mathrm{OH}^{-}$ions is decreased $\frac{1}{4}$ by times, then the equilibrium concentration of $\mathrm{Fe}^{3+}$ will increase by

1 8 times
2 16 times
3 64 times
4 4 times
[Karnataka-CET-2016]
Chemical Equilibrium

228927 If $K_{1}$ and $K_{2}$ are respective equilibrium constants for two reactions
$\mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \quad \mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})$
$\mathrm{XeO}_{4}(\mathrm{~g})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons \quad \mathrm{XeOF}_{4}(\mathrm{~g})+\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})$
the equilibrium constant for the reaction
$\mathrm{XeO}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \quad \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
will be

1 $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}^{2}}$
2 $\mathrm{K}_{1} \cdot \mathrm{K}_{2}$
3 $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}$
4 $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$
[BITSAT 2006]
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Chemical Equilibrium

228923 The equilibrium constant of the following are
$\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; \mathrm{K}_{1}$
$\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons2 \mathrm{NO} ; \mathrm{K}_{2}$
$\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons\mathrm{H}_{2} \mathrm{O} ; \quad \mathrm{K}_{3}$
The equilibrium constant $(K)$ of the reaction:
$2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2}$

1 $\mathrm{K}_{2} \mathrm{~K}_{3}^{3} / \mathrm{K}_{1}$
2 $\mathrm{K}_{2} \mathrm{~K}_{3} / \mathrm{K}_{1}$
3 $\mathrm{K}_{2}^{3} \mathrm{~K}_{3} / \mathrm{K}_{1}$
4 $\mathrm{K}_{1} \mathrm{~K}_{3}^{3} / \mathrm{K}_{2}$
2007
Chemical Equilibrium

228924 For the reaction, $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$, the equilibrium constant is $K_{1}$. The equilibrium constant is $K_{2}$ for the reaction, $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$. What is $\mathrm{K}$ for the reaction, $\mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \frac{1}{\mathbf{2}} \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ ?

1 $1 /\left(4 \mathrm{~K}_{1} \mathrm{~K}_{2}\right)$
2 $\left(1 / \mathrm{K}_{1} \mathrm{~K}_{2}\right)^{1 / 2}$
3 $1 /\left(\mathrm{K}_{1}, \mathrm{~K}_{2}\right)$
4 $1 /\left(2 \mathrm{~K}_{1}, \mathrm{~K}_{2}\right)$
[JIMPER-2014]
Chemical Equilibrium

228926 In the reaction,
$\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})$
if the concentration of $\mathrm{OH}^{-}$ions is decreased $\frac{1}{4}$ by times, then the equilibrium concentration of $\mathrm{Fe}^{3+}$ will increase by

1 8 times
2 16 times
3 64 times
4 4 times
[Karnataka-CET-2016]
Chemical Equilibrium

228927 If $K_{1}$ and $K_{2}$ are respective equilibrium constants for two reactions
$\mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \quad \mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})$
$\mathrm{XeO}_{4}(\mathrm{~g})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons \quad \mathrm{XeOF}_{4}(\mathrm{~g})+\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})$
the equilibrium constant for the reaction
$\mathrm{XeO}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \quad \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
will be

1 $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}^{2}}$
2 $\mathrm{K}_{1} \cdot \mathrm{K}_{2}$
3 $\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}$
4 $\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$
[BITSAT 2006]
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