Equilibrium Test

Directions For Questions

The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion of ammonia molecules reacts with

{H}_{2}O

to form

{NH}_{4}OH

molecules.

{NH}_{4}OH

again dissociates into

{NH}_{4}^{+}

and

{OH}^{-}

ions. In solution therefore, we have

{NH}_{3}

molecules,

{NH}_{4}OH

molecules and

{NH}_{4}^{+}

ions and the following equilibrium exist:

{NH}_{3}(g)

\rightleftharpoons{NH}_{3}(l)+{H}_{2}O{(l)}\rightleftharpoons  {NH}_{4}OH{(aq)}\rightleftharpoons  {NH}_{4}^{+}{(aq)}+{OH}^{-}{(aq)}

Let,

{c}_{1}

{(mol/L)}

{NH}_{3}

pass in liquid state which on dissolution in water forms

{c}_{2}

{(mol/L)}

{NH}_{4}OH

. The solution contains

{c}_{3}

{(mol/L)}

{NH}_{4}^{+}

ions

...view full instructions

Equilibrium Tes...

Question 1

ΔH\Delta HΔH is dependent on TTT

KKK is independent of the initial amount of CaCO3CaC{O}_{3}CaCO3​

KKK is dependent on the pressure of CO2C{O}_{2}CO2​ at a given TTT

ΔH\Delta HΔH is independent of the catalyst, if any

Question 2

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Assertion is incorrect but Reason is correct

Both Assertion and Reason are incorrect

Question 3

Fe2O3+6HCl→2FeCl3+3H2OF{e}_{2}{O}_{3} + 6HCl \rightarrow 2FeC{l}_{3} + 3{H}_{2}OFe2​O3​+6HCl→2FeCl3​+3H2​O

SnCl4+Hg2Cl2→SnCl2+2HgCl2SnC{l}_{4} + H{g}_{2}C{l}_{2} \rightarrow SnC{l}_{2} + 2HgC{l}_{2}SnCl4​+Hg2​Cl2​→SnCl2​+2HgCl2​

NH3+H2O+NaCl→NH4Cl+NaOHN{H}_{3} + {H}_{2}O + NaCl \rightarrow N{H}_{4}Cl + NaOHNH3​+H2​O+NaCl→NH4​Cl+NaOH

2CuI+I2+4K+→2Cu2++3KI2CuI + {I}_{2} + 4{K}^{+} \rightarrow 2{Cu}^{2+} + 3KI2CuI+I2​+4K+→2Cu2++3KI

Question 4

Kb=(c3)2(c2−c3){ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) } Kb​=(c2​−c3​)(c3​)2​

Kb=(c3)2c2{ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ { c }_{ 2 } } Kb​=c2​(c3​)2​

Kb=c3(c2−c3){ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) } Kb​=(c2​−c3​)c3​​

Kb=c3(c1−c2){ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 1 }-{ c }_{ 2 } \right) } Kb​=(c1​−c2​)c3​​

Question 5

4×10−54\times{10}^{-5}4×10−5

4×10−44\times{10}^{-4}4×10−4

4×10−34\times{10}^{-3}4×10−3

2×10−42 \times 10^{-4}2×10−4

Question 6

Increase in temperature

Removal of Cl2C{l}_{2}Cl2​

Increase in volume of container

Addition of F2{F}_{2}F2​

Question 7

the mass of SO2(g)S{O}_{2}\left(g\right)SO2​(g) will decrease

the mass of SO2Cl2(g)S{O}_{2}C{l}_{2}\left(g\right)SO2​Cl2​(g) will increase

the mass of Cl2(g)C{l}_{2}\left(g\right)Cl2​(g) will increase

the mass of Cl2(g)C{l}_{2}\left(g\right)Cl2​(g) will remain unchanged

Question 8

favours backward reaction

favours forward reaction

have no effect

increases the reaction rate

Question 9

1.61×10−9M1.61\times 10^{-9}M1.61×10−9M

1.61×10−7M1.61\times 10^{-7}M1.61×10−7M

1.61×10−3M1.61\times 10^{-3}M1.61×10−3M

1.61×10−5M1.61\times 10^{-5}M1.61×10−5M

Question 10

1.4×10−12,1.6×10−7,2.2×10−2,H2PO4⊝1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}1.4×10−12,1.6×10−7,2.2×10−2,H2​PO4⊝​ is acidic and HPO42−HPO_4^{2-}HPO42−​ is basic

2.2×10−12,1.6×10−7,1.4×10−2,H2PO4⊝2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}2.2×10−12,1.6×10−7,1.4×10−2,H2​PO4⊝​ is basic and HPO42−HPO_4^{2-}HPO42−​ is basic

2.2×10−12,1.6×10−7,1.4×10−2,H2PO4⊝2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}2.2×10−12,1.6×10−7,1.4×10−2,H2​PO4⊝​ is acidic and HPO42−HPO_4^{2-}HPO42−​ is basic

1.4×10−12,1.6×10−7,2.2×10−2,H2PO4⊝1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}1.4×10−12,1.6×10−7,2.2×10−2,H2​PO4⊝​ is basic and HPO42−HPO_4^{2-}HPO42−​ is acidic

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Question Analysis

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$\Delta H$ is dependent on $T$

$K$ is independent of the initial amount of $CaC{O}_{3}$

$K$ is dependent on the pressure of $C{O}_{2}$ at a given $T$

$\Delta H$ is independent of the catalyst, if any

$F{e}_{2}{O}_{3} + 6HCl \rightarrow 2FeC{l}_{3} + 3{H}_{2}O$

$SnC{l}_{4} + H{g}_{2}C{l}_{2} \rightarrow SnC{l}_{2} + 2HgC{l}_{2}$

$N{H}_{3} + {H}_{2}O + NaCl \rightarrow N{H}_{4}Cl + NaOH$

$2CuI + {I}_{2} + 4{K}^{+} \rightarrow 2{Cu}^{2+} + 3KI$

${ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) }$

${ K }_{ b }=\cfrac { { \left( { c }_{ 3 } \right) }^{ 2 } }{ { c }_{ 2 } }$

${ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 2 }-{ c }_{ 3 } \right) }$

${ K }_{ b }=\cfrac { { c }_{ 3 } }{ \left( { c }_{ 1 }-{ c }_{ 2 } \right) }$

$4\times{10}^{-5}$

$4\times{10}^{-4}$

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

$2 \times 10^{-4}$

Removal of $C{l}_{2}$

Addition of ${F}_{2}$

the mass of $S{O}_{2}\left(g\right)$ will decrease

the mass of $S{O}_{2}C{l}_{2}\left(g\right)$ will increase

the mass of $C{l}_{2}\left(g\right)$ will increase

the mass of $C{l}_{2}\left(g\right)$ will remain unchanged

$1.61\times 10^{-9}M$

$1.61\times 10^{-7}M$

$1.61\times 10^{-3}M$

$1.61\times 10^{-5}M$

$1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}$ is acidic and $HPO_4^{2-}$ is basic

$2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}$ is basic and $HPO_4^{2-}$ is basic

$2.2\times 10^{-12}, 1.6\times 10^{-7}, 1.4\times 10^{-2}, H_2PO_4^{\circleddash}$ is acidic and $HPO_4^{2-}$ is basic

$1.4\times 10^{-12}, 1.6\times 10^{-7}, 2.2\times 10^{-2}, H_2PO_4^{\circleddash}$ is basic and $HPO_4^{2-}$ is acidic