The equilibrium constant of a reaction doubles on increasing the temperature of the reaction from 25 ° C to 35 ° C . Calculate enthalpy change of the reaction, assuming it to be constant in this temperature range.

Applying equation of temperature dependency of equilibrium constant: log ( K p 2 ) ( K p 1 ) = ΔH ° 2 ⋅ 303 R ( 1 T 1 - 1 T 2 ) Here: K p 2 = 2 K p 1 , R = 8 . 314 JK - 1 mol - 1 , T 1 = 25 ° C = 298 K , T 2 = ( 35 + 273 ) = 308 K and ∆ H ° = Enthalpy change of the reaction. log 2 = ΔH ° 2 ⋅ 303 ( 8 . 314 ) ( 1 298 - 1 308 ) Put log 2 = 0 . 3 , we get ΔH ° = 52 . 898 KJ mol - 1

Determine K C for the reaction, 1 2 N 2 g + 1 2 O 2 g + 1 2 Br 2 g ⇌ NOBr g , from the following data at 298 K . ( i ) 2 NO ⇌ N 2 + O 2 ; K C = 2 . 4 × 10 30 ii NO + 1 2 Br 2 ⇌ NOBr ; K C = 1 . 4

Given, ( i ) 2 NO ⇌ N 2 + O 2 ; K C = 2 . 4 × 10 30 and ii NO + 1 2 Br 2 ⇌ NOBr ; K C = 1 . 4 Reversing reaction i , N 2 + O 2 ⇌ 2 NO ; K C = 1 2 . 4 × 10 30 . . . 1 , equilibrium constant gets inversed. Multiplying equation 1 by 1 2 , we get: 1 2 N 2 + 1 2 O 2 ⇌ NO ; K C = 1 2 . 4 × 10 30 1 2 . . . 2 , equilibrium constant becomes the original constant raised to the power of the multiplying factor. Adding equations ii and 2 , we get the resultant equation: (Their K C are multiplied) 1 2 N 2 g + 1 2 O 2 g + 1 2 Br 2 g ⇌ NOBr g K C = K C ( i ) × K C ( 2 ) = 1 2 . 4 × 10 30 1 2 × 1 . 4 = 9 × 10 - 16

For the reaction, CuSO 4 . 3 H 2 O ( s ) ⇌ CuSO 4 . H 2 O ( s ) + 2 H 2 O ( g ) , the dissociation pressure is 7 × 10 - 3 atm at 25 ° C and ΔH ° = 2750 cal . What will be the dissociation pressure at 127 ° C ?

For the reaction, CuSO 4 . 3 H 2 O ( s ) ⇌ CuSO 4 . H 2 O ( s ) + 2 H 2 O ( g ) , the equilibrium constant is given by K p = ( p ( H 2 O ) ) 2 . At T = 25 ° C = 298 K , K p = ( 7 × 10 - 3 ) 2 = 49 × 10 - 6 . Using Arrhenius equation: log K 2 K 1 = ∆ H 2 . 303 R 1 T 1 - 1 T 2 ⇒ log K 2 49 × 10 - 6 = 2750 2 . 303 × 2 1 298 - 1 400 ⇒ K 2 = 1 . 58 × 10 - 4 Since, at 127 C o , K p = 1 . 58 × 10 - 4 = ( p ( H 2 O ) ) 2 ∴ p ( H 2 O ) = 1 . 26 × 10 - 2 atm .

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${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$
Which of the following sketches may represent the above equilibrium? Assume equilibrium can be achieved from either side and by taking anyone or more components initially.(Given $K_{C}$ for the reaction $< 2$ ).

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Important Questions on Chemical Equilibrium

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 6 - Chemical Equilibrium>DPP 6.2>Q 1

The equilibrium constant of a reaction doubles on increasing the temperature of the reaction from

25 ° C

35 ° C

. Calculate enthalpy change of the reaction, assuming it to be constant in this temperature range.

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 6 - Chemical Equilibrium>DPP 6.2>Q 2

Sulphide ion in an alkaline solution reacts with solid sulphur to form polysulphide ions having formula

S_{2}^{^{2 -}}, S_{3}^{2 -}, S_{4}^{2 -}, etc . If K_{1} = 12 for S + S^{- 2} ⇌ S_{2}^{2 -} and K_{2} = 132 for 2 S + S^{2 -} ⇋ S_{3}^{2 -},

calculate

K_{3}

for

S + S_{2}^{2 -} ⇄ S_{3}^{2 -}

EASY

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 6 - Chemical Equilibrium>DPP 6.2>Q 3

Predict which of the following reaction will have an appreciable concentration of reactants and products:

{Cl}_{2} (g) ⇌ 2 Cl (g)

K_{c} = 5 \times 10^{- 39}

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 6 - Chemical Equilibrium>DPP 6.2>Q 3

Predict which of the following reaction will have an appreciable concentration of reactants and products:

{Cl}_{2} (g) + 2 NO (g) ⇌ 2 NOCl (g)

K_{c} = 3.7 \times 10^{8}

EASY

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 6 - Chemical Equilibrium>DPP 6.2>Q 3

Predict which of the following reaction will have an appreciable concentration of reactants and products:

{Cl}_{2} (g) + 2 {NO}_{2} (g) ⇌ 2 {NO}_{2} Cl (g)

K_{c} = 1.8

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 6 - Chemical Equilibrium>DPP 6.2>Q 4

Determine $K_{C}$ for the reaction, $\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) + \frac{1}{2} {Br}_{2} (g) ⇌ NOBr (g)$ , from the following data at $298 K$ .
$(i) 2 NO ⇌ N_{2} + O_{2}; K_{C} = 2.4 \times 10^{30}$

$(ii) NO + \frac{1}{2} {Br}_{2} ⇌ NOBr; K_{C} = 1.4$

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 6 - Chemical Equilibrium>DPP 6.2>Q 5

For the reaction,

{CuSO}_{4} . 3 H_{2} O (s) ⇌ {CuSO}_{4} . H_{2} O (s) + 2 H_{2} O (g)

, the dissociation pressure is

7 \times 10^{- 3} atm

25 ° C

and

ΔH ° = 2750 cal

. What will be the dissociation pressure at

127 ° C

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 6 - Chemical Equilibrium>DPP 6.2>Q 6

The variation of $logK$ with $\frac{1}{T}$ is shown in the following graph. $ΔH °$ for the reaction will be:

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PCl5(g)⇌PCl3(g)+Cl2(g) Which of the following sketches may represent the above equilibrium? Assume equilibrium can be achieved from either side and by taking anyone or more components initially.(Given KC for the reaction<2 ).

Important Questions on Chemical Equilibrium

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${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$
Which of the following sketches may represent the above equilibrium? Assume equilibrium can be achieved from either side and by taking anyone or more components initially.(Given $K_{C}$ for the reaction $< 2$ ).