Equilibrium reaction of HI with H 2 and I 2 is given, for the given conditions whether the system is at equilibrium or not has to be checked. Concept Introduction: Equilibrium constant ( K c ) : Equilibrium constant ( K c ) is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction . Consider the reaction where A reacts to give B. aA ⇌ bB Rate of forward reaction = Rate of reverse reaction k f [ A ] a =k r [ B ] b On rearranging, [ B ] b [ A ] a = k f k r =K c Where, k f is the rate constant of the forward reaction. k r is the rate constant of the reverse reaction. K c is the equilibrium constant.

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Answer 1

Question

Chapter 12, Problem 62QRT

(a)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions whether the system is at equilibrium or not has to be checked.

Concept Introduction:

Equilibrium constant $(Kc)$ :

Equilibrium constant $(Kc)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

     $aA⇌bB$

     $Rate of forward reaction = Rate of reverse reactionkf[A]a =kr[B]b$

On rearranging,

     $[B]b [A]a=kfkr=Kc$

Where,

     $kf$ is the rate constant of the forward reaction.

     $kr$ is the rate constant of the reverse reaction.

     $Kc$ is the equilibrium constant.

(a)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

Therefore, the given reaction is not at equilibrium for the given conditions.

(b)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions the direction of the reaction has to be given.

Concept Introduction:

Refer part (a)

(b)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

The given reaction is not at equilibrium for the given conditions.

And also $QC > KC$ so the given reaction must proceeds toward reactants to reach equilibrium condition.

(c)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, concentrations of all three substances has to be calculated.

Concept Introduction:

Refer part (a)

(c)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Construct ICE table,

$H2(g) + I2(g) ⇌ 2 HI(g) Initial conc. (M) 0.00125 0.00125 0.0375Change in conc. (M) + x + x - 2 xEquilibrium conc (M) 0.00125 + x 0.00125 + x 0.0375 - 2 x$

At equilibrium,

$KC = [HI]2[H2] [I2] = (0.0375 - 2x)2(0.00125 + x) (0.00125 + x) = 50.0$

Solve for x,

$take square root of each side, (0.0375 - 2x)(0.00125 + x) = 50.0 = 7.07 0.0375 - 2x = 0.0088 + 7.07 x 0.0375 - 0.0088 = 9.07xtherefore, x = 0.0032 M$

Calculate the concentrations,

$[H2] = [I2] = 0.00125 + x = 0.00125 + 0.0032 M = 0.0045 M[HI] = 0.0375 - 2 x = 0.0375 M - 2 (0.0032 M) = 0.0375 M - 0.0064 = 0.031 M$

Therefore, the concentration of $H2 and I2$ is $0.0045 M$ and the concentration of $HI$ is $0.031 M$ .

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Answer 2

Question

Chapter 12, Problem 62QRT

(a)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions whether the system is at equilibrium or not has to be checked.

Concept Introduction:

Equilibrium constant $(Kc)$ :

Equilibrium constant $(Kc)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

     $aA⇌bB$

     $Rate of forward reaction = Rate of reverse reactionkf[A]a =kr[B]b$

On rearranging,

     $[B]b [A]a=kfkr=Kc$

Where,

     $kf$ is the rate constant of the forward reaction.

     $kr$ is the rate constant of the reverse reaction.

     $Kc$ is the equilibrium constant.

(a)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

Therefore, the given reaction is not at equilibrium for the given conditions.

(b)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions the direction of the reaction has to be given.

Concept Introduction:

Refer part (a)

(b)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

The given reaction is not at equilibrium for the given conditions.

And also $QC > KC$ so the given reaction must proceeds toward reactants to reach equilibrium condition.

(c)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, concentrations of all three substances has to be calculated.

Concept Introduction:

Refer part (a)

(c)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Construct ICE table,

$H2(g) + I2(g) ⇌ 2 HI(g) Initial conc. (M) 0.00125 0.00125 0.0375Change in conc. (M) + x + x - 2 xEquilibrium conc (M) 0.00125 + x 0.00125 + x 0.0375 - 2 x$

At equilibrium,

$KC = [HI]2[H2] [I2] = (0.0375 - 2x)2(0.00125 + x) (0.00125 + x) = 50.0$

Solve for x,

$take square root of each side, (0.0375 - 2x)(0.00125 + x) = 50.0 = 7.07 0.0375 - 2x = 0.0088 + 7.07 x 0.0375 - 0.0088 = 9.07xtherefore, x = 0.0032 M$

Calculate the concentrations,

$[H2] = [I2] = 0.00125 + x = 0.00125 + 0.0032 M = 0.0045 M[HI] = 0.0375 - 2 x = 0.0375 M - 2 (0.0032 M) = 0.0375 M - 0.0064 = 0.031 M$

Therefore, the concentration of $H2 and I2$ is $0.0045 M$ and the concentration of $HI$ is $0.031 M$ .

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Answer 3

Question

Chapter 12, Problem 62QRT

(a)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions whether the system is at equilibrium or not has to be checked.

Concept Introduction:

Equilibrium constant $(Kc)$ :

Equilibrium constant $(Kc)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

     $aA⇌bB$

     $Rate of forward reaction = Rate of reverse reactionkf[A]a =kr[B]b$

On rearranging,

     $[B]b [A]a=kfkr=Kc$

Where,

     $kf$ is the rate constant of the forward reaction.

     $kr$ is the rate constant of the reverse reaction.

     $Kc$ is the equilibrium constant.

(a)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

Therefore, the given reaction is not at equilibrium for the given conditions.

(b)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions the direction of the reaction has to be given.

Concept Introduction:

Refer part (a)

(b)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

The given reaction is not at equilibrium for the given conditions.

And also $QC > KC$ so the given reaction must proceeds toward reactants to reach equilibrium condition.

(c)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, concentrations of all three substances has to be calculated.

Concept Introduction:

Refer part (a)

(c)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Construct ICE table,

$H2(g) + I2(g) ⇌ 2 HI(g) Initial conc. (M) 0.00125 0.00125 0.0375Change in conc. (M) + x + x - 2 xEquilibrium conc (M) 0.00125 + x 0.00125 + x 0.0375 - 2 x$

At equilibrium,

$KC = [HI]2[H2] [I2] = (0.0375 - 2x)2(0.00125 + x) (0.00125 + x) = 50.0$

Solve for x,

$take square root of each side, (0.0375 - 2x)(0.00125 + x) = 50.0 = 7.07 0.0375 - 2x = 0.0088 + 7.07 x 0.0375 - 0.0088 = 9.07xtherefore, x = 0.0032 M$

Calculate the concentrations,

$[H2] = [I2] = 0.00125 + x = 0.00125 + 0.0032 M = 0.0045 M[HI] = 0.0375 - 2 x = 0.0375 M - 2 (0.0032 M) = 0.0375 M - 0.0064 = 0.031 M$

Therefore, the concentration of $H2 and I2$ is $0.0045 M$ and the concentration of $HI$ is $0.031 M$ .

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Answer 4

Question

Chapter 12, Problem 62QRT

(a)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions whether the system is at equilibrium or not has to be checked.

Concept Introduction:

Equilibrium constant $(Kc)$ :

Equilibrium constant $(Kc)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

     $aA⇌bB$

     $Rate of forward reaction = Rate of reverse reactionkf[A]a =kr[B]b$

On rearranging,

     $[B]b [A]a=kfkr=Kc$

Where,

     $kf$ is the rate constant of the forward reaction.

     $kr$ is the rate constant of the reverse reaction.

     $Kc$ is the equilibrium constant.

(a)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

Therefore, the given reaction is not at equilibrium for the given conditions.

(b)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions the direction of the reaction has to be given.

Concept Introduction:

Refer part (a)

(b)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

The given reaction is not at equilibrium for the given conditions.

And also $QC > KC$ so the given reaction must proceeds toward reactants to reach equilibrium condition.

(c)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, concentrations of all three substances has to be calculated.

Concept Introduction:

Refer part (a)

(c)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Construct ICE table,

$H2(g) + I2(g) ⇌ 2 HI(g) Initial conc. (M) 0.00125 0.00125 0.0375Change in conc. (M) + x + x - 2 xEquilibrium conc (M) 0.00125 + x 0.00125 + x 0.0375 - 2 x$

At equilibrium,

$KC = [HI]2[H2] [I2] = (0.0375 - 2x)2(0.00125 + x) (0.00125 + x) = 50.0$

Solve for x,

$take square root of each side, (0.0375 - 2x)(0.00125 + x) = 50.0 = 7.07 0.0375 - 2x = 0.0088 + 7.07 x 0.0375 - 0.0088 = 9.07xtherefore, x = 0.0032 M$

Calculate the concentrations,

$[H2] = [I2] = 0.00125 + x = 0.00125 + 0.0032 M = 0.0045 M[HI] = 0.0375 - 2 x = 0.0375 M - 2 (0.0032 M) = 0.0375 M - 0.0064 = 0.031 M$

Therefore, the concentration of $H2 and I2$ is $0.0045 M$ and the concentration of $HI$ is $0.031 M$ .

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Answer 5

Chapter 12, Problem 62QRT

(a)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions whether the system is at equilibrium or not has to be checked.

Concept Introduction:

Equilibrium constant $(Kc)$ :

Equilibrium constant $(Kc)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

     $aA⇌bB$

     $Rate of forward reaction = Rate of reverse reactionkf[A]a =kr[B]b$

On rearranging,

     $[B]b [A]a=kfkr=Kc$

Where,

     $kf$ is the rate constant of the forward reaction.

     $kr$ is the rate constant of the reverse reaction.

     $Kc$ is the equilibrium constant.

(a)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

Therefore, the given reaction is not at equilibrium for the given conditions.

(b)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions the direction of the reaction has to be given.

Concept Introduction:

Refer part (a)

(b)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

The given reaction is not at equilibrium for the given conditions.

And also $QC > KC$ so the given reaction must proceeds toward reactants to reach equilibrium condition.

(c)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, concentrations of all three substances has to be calculated.

Concept Introduction:

Refer part (a)

(c)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Construct ICE table,

$H2(g) + I2(g) ⇌ 2 HI(g) Initial conc. (M) 0.00125 0.00125 0.0375Change in conc. (M) + x + x - 2 xEquilibrium conc (M) 0.00125 + x 0.00125 + x 0.0375 - 2 x$

At equilibrium,

$KC = [HI]2[H2] [I2] = (0.0375 - 2x)2(0.00125 + x) (0.00125 + x) = 50.0$

Solve for x,

$take square root of each side, (0.0375 - 2x)(0.00125 + x) = 50.0 = 7.07 0.0375 - 2x = 0.0088 + 7.07 x 0.0375 - 0.0088 = 9.07xtherefore, x = 0.0032 M$

Calculate the concentrations,

$[H2] = [I2] = 0.00125 + x = 0.00125 + 0.0032 M = 0.0045 M[HI] = 0.0375 - 2 x = 0.0375 M - 2 (0.0032 M) = 0.0375 M - 0.0064 = 0.031 M$

Therefore, the concentration of $H2 and I2$ is $0.0045 M$ and the concentration of $HI$ is $0.031 M$ .

Answer 6

Chapter 12, Problem 62QRT

(a)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions whether the system is at equilibrium or not has to be checked.

Concept Introduction:

Equilibrium constant $(Kc)$ :

Equilibrium constant $(Kc)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

     $aA⇌bB$

     $Rate of forward reaction = Rate of reverse reactionkf[A]a =kr[B]b$

On rearranging,

     $[B]b [A]a=kfkr=Kc$

Where,

     $kf$ is the rate constant of the forward reaction.

     $kr$ is the rate constant of the reverse reaction.

     $Kc$ is the equilibrium constant.

(a)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

Therefore, the given reaction is not at equilibrium for the given conditions.

Answer 7

Chapter 12, Problem 62QRT

(a)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions whether the system is at equilibrium or not has to be checked.

Concept Introduction:

Equilibrium constant $(Kc)$ :

Equilibrium constant $(Kc)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

     $aA⇌bB$

     $Rate of forward reaction = Rate of reverse reactionkf[A]a =kr[B]b$

On rearranging,

     $[B]b [A]a=kfkr=Kc$

Where,

     $kf$ is the rate constant of the forward reaction.

     $kr$ is the rate constant of the reverse reaction.

     $Kc$ is the equilibrium constant.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

Therefore, the given reaction is not at equilibrium for the given conditions.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions the direction of the reaction has to be given.

Concept Introduction:

Refer part (a)

(b)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

The given reaction is not at equilibrium for the given conditions.

And also $QC > KC$ so the given reaction must proceeds toward reactants to reach equilibrium condition.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, for the given conditions the direction of the reaction has to be given.

Concept Introduction:

Refer part (a)

Answer 14

(b)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Calculate the value of $QC$ ,

$Conc. of HI = 0.75 mol20.0 L = 0.0375 M ;Conc. of H2 = Conc. of I2 = 0.025 mol20.0 L = 0.00125 M QC = (Conc HI)2(Conc H2) (Conc I2) = (0.0375)2(0.0025) (0.0025) = 900$

Compare the value of $QC$ and $KC$ ,

$QC ≠ KC$

The given reaction is not at equilibrium for the given conditions.

And also $QC > KC$ so the given reaction must proceeds toward reactants to reach equilibrium condition.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, concentrations of all three substances has to be calculated.

Concept Introduction:

Refer part (a)

(c)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Construct ICE table,

$H2(g) + I2(g) ⇌ 2 HI(g) Initial conc. (M) 0.00125 0.00125 0.0375Change in conc. (M) + x + x - 2 xEquilibrium conc (M) 0.00125 + x 0.00125 + x 0.0375 - 2 x$

At equilibrium,

$KC = [HI]2[H2] [I2] = (0.0375 - 2x)2(0.00125 + x) (0.00125 + x) = 50.0$

Solve for x,

$take square root of each side, (0.0375 - 2x)(0.00125 + x) = 50.0 = 7.07 0.0375 - 2x = 0.0088 + 7.07 x 0.0375 - 0.0088 = 9.07xtherefore, x = 0.0032 M$

Calculate the concentrations,

$[H2] = [I2] = 0.00125 + x = 0.00125 + 0.0032 M = 0.0045 M[HI] = 0.0375 - 2 x = 0.0375 M - 2 (0.0032 M) = 0.0375 M - 0.0064 = 0.031 M$

Therefore, the concentration of $H2 and I2$ is $0.0045 M$ and the concentration of $HI$ is $0.031 M$ .

Answer 19

(c)

Interpretation Introduction

Interpretation:

Equilibrium reaction of $HI$ with $H2 and I2$ is given, concentrations of all three substances has to be calculated.

Concept Introduction:

Refer part (a)

Answer 20

(c)

Expert Solution

Explanation of Solution

Given information,

$H2(g) + I2(g) ⇌ 2 HI(g) KC = 50.0 at 745 KH2 = 0.025 mol ; I2 = 0.025 mol and HI = 0.75 molVolume of flask = 20.0 L$

Construct ICE table,

$H2(g) + I2(g) ⇌ 2 HI(g) Initial conc. (M) 0.00125 0.00125 0.0375Change in conc. (M) + x + x - 2 xEquilibrium conc (M) 0.00125 + x 0.00125 + x 0.0375 - 2 x$

At equilibrium,

$KC = [HI]2[H2] [I2] = (0.0375 - 2x)2(0.00125 + x) (0.00125 + x) = 50.0$

Solve for x,

$take square root of each side, (0.0375 - 2x)(0.00125 + x) = 50.0 = 7.07 0.0375 - 2x = 0.0088 + 7.07 x 0.0375 - 0.0088 = 9.07xtherefore, x = 0.0032 M$