General Chemistry: Atoms First
General Chemistry: Atoms First
2nd Edition
ISBN: 9780321809261
Author: John E. McMurry, Robert C. Fay
Publisher: Prentice Hall
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Chapter 13, Problem 13.56SP

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant expression (KcandKp) for the given reactionhas to be written.

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.

aAbB

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

On rearranging,

[A]a[B]b=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.

Equilibrium constant(Kp):

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant (Kp) is defined as ratio of partial pressure of products to partial pressure of reactants. Each partial pressure 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.

aAbB

Rate of forward reaction = Rate of reverse reactionkfPAa=krPBa

On rearranging,

PAaPBb=kfkr=Kp

Where,

kf is the rate constant of the forward reaction.

kr is the rate constant of the reverse reaction.

Kp is the equilibrium constant.

The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.

PAV = nARTPA = nAVRT=[A]RT

In same way PB =[B]RT. Thus, equilibrium constant Kp is given as

Kp=[B]b[A]a×(RT)b-aKp = Kc(RT)Δn

Where,

R is gas constant

T is absolute temperature

Δn is difference in moles of products to reactants in gas phase.

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant expression (KcandKp ) for the given reaction has to be written.

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.

aAbB

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

On rearranging,

[A]a[B]b=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.

Equilibrium constant(Kp):

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant (Kp) is defined as ratio of partial pressure of products to partial pressure of reactants. Each partial pressure 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.

aAbB

Rate of forward reaction = Rate of reverse reactionkfPAa=krPBa

On rearranging,

PAaPBb=kfkr=Kp

Where,

kf is the rate constant of the forward reaction.

kr is the rate constant of the reverse reaction.

Kp is the equilibrium constant.

The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.

PAV = nARTPA = nAVRT=[A]RT

In same way PB =[B]RT. Thus, equilibrium constant Kp is given as

Kp=[B]b[A]a×(RT)b-aKp = Kc(RT)Δn

Where,

R is gas constant

T is absolute temperature

Δn is difference in moles of products to reactants in gas phase.

(c)

Interpretation Introduction

Interpretation:

The equilibrium constant expression (KcandKp ) for the given reaction has to be written.

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.

aAbB

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

On rearranging,

[A]a[B]b=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.

Equilibrium constant(Kp):

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant (Kp) is defined as ratio of partial pressure of products to partial pressure of reactants. Each partial pressure 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.

aAbB

Rate of forward reaction = Rate of reverse reactionkfPAa=krPBa

On rearranging,

PAaPBb=kfkr=Kp

Where,

kf is the rate constant of the forward reaction.

kr is the rate constant of the reverse reaction.

Kp is the equilibrium constant.

The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.

PAV = nARTPA = nAVRT=[A]RT

In same way PB =[B]RT. Thus, equilibrium constant Kp is given as

Kp=[B]b[A]a×(RT)b-aKp = Kc(RT)Δn

Where,

R is gas constant

T is absolute temperature

Δn is difference in moles of products to reactants in gas phase.

(d)

Interpretation Introduction

Interpretation:

The equilibrium constant expression (KcandKp ) for the given reaction has to be written.

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.

aAbB

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

On rearranging,

[A]a[B]b=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.

Equilibrium constant(Kp):

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant (Kp) is defined as ratio of partial pressure of products to partial pressure of reactants. Each partial pressure 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.

aAbB

Rate of forward reaction = Rate of reverse reactionkfPAa=krPBa

On rearranging,

PAaPBb=kfkr=Kp

Where,

kf is the rate constant of the forward reaction.

kr is the rate constant of the reverse reaction.

Kp is the equilibrium constant.

The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.

PAV = nARTPA = nAVRT=[A]RT

In same way PB =[B]RT. Thus, equilibrium constant Kp is given as

Kp=[B]b[A]a×(RT)b-aKp = Kc(RT)Δn

Where,

R is gas constant

T is absolute temperature

Δn is difference in moles of products to reactants in gas phase.

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Chapter 13 Solutions

General Chemistry: Atoms First

Ch. 13.6 - Prob. 13.11CPCh. 13.6 - Prob. 13.12PCh. 13.6 - Prob. 13.13PCh. 13.6 - Prob. 13.14PCh. 13.6 - Prob. 13.15PCh. 13.6 - Prob. 13.16PCh. 13.8 - Prob. 13.17PCh. 13.9 - Prob. 13.18PCh. 13.9 - Prob. 13.19CPCh. 13.10 - Prob. 13.20PCh. 13.10 - Prob. 13.21PCh. 13.10 - Prob. 13.22CPCh. 13.11 - Prob. 13.23PCh. 13.11 - Prob. 13.24PCh. 13.11 - Prob. 13.25PCh. 13 - Consider the interconversion of A molecules (red...Ch. 13 - Prob. 13.27CPCh. 13 - Prob. 13.28CPCh. 13 - Prob. 13.29CPCh. 13 - Prob. 13.30CPCh. 13 - Prob. 13.31CPCh. 13 - Prob. 13.32CPCh. 13 - Prob. 13.33CPCh. 13 - Prob. 13.34CPCh. 13 - Prob. 13.35CPCh. 13 - Prob. 13.36CPCh. 13 - The following pictures represent the initial and...Ch. 13 - Prob. 13.38SPCh. 13 - Prob. 13.39SPCh. 13 - Prob. 13.40SPCh. 13 - Prob. 13.41SPCh. 13 - Prob. 13.42SPCh. 13 - Prob. 13.43SPCh. 13 - Prob. 13.44SPCh. 13 - Prob. 13.45SPCh. 13 - Prob. 13.46SPCh. 13 - Prob. 13.47SPCh. 13 - Prob. 13.48SPCh. 13 - Prob. 13.49SPCh. 13 - Prob. 13.50SPCh. 13 - Prob. 13.51SPCh. 13 - Prob. 13.52SPCh. 13 - Prob. 13.53SPCh. 13 - Prob. 13.54SPCh. 13 - Prob. 13.55SPCh. 13 - Prob. 13.56SPCh. 13 - Prob. 13.57SPCh. 13 - Prob. 13.58SPCh. 13 - Prob. 13.59SPCh. 13 - Prob. 13.60SPCh. 13 - Prob. 13.61SPCh. 13 - Prob. 13.62SPCh. 13 - Prob. 13.63SPCh. 13 - Prob. 13.64SPCh. 13 - Prob. 13.65SPCh. 13 - Prob. 13.66SPCh. 13 - Prob. 13.67SPCh. 13 - Prob. 13.68SPCh. 13 - Prob. 13.69SPCh. 13 - Prob. 13.70SPCh. 13 - Prob. 13.71SPCh. 13 - Prob. 13.72SPCh. 13 - Prob. 13.73SPCh. 13 - Gaseous indium dihydride is formed from the...Ch. 13 - Prob. 13.75SPCh. 13 - Prob. 13.76SPCh. 13 - Prob. 13.77SPCh. 13 - Prob. 13.78SPCh. 13 - Prob. 13.79SPCh. 13 - Prob. 13.80SPCh. 13 - Prob. 13.81SPCh. 13 - The value of Kc for the reaction of acetic acid...Ch. 13 - In a basic aqueous solution, chloromethane...Ch. 13 - Prob. 13.84SPCh. 13 - Prob. 13.85SPCh. 13 - Prob. 13.86SPCh. 13 - Prob. 13.87SPCh. 13 - Prob. 13.88SPCh. 13 - Prob. 13.89SPCh. 13 - Prob. 13.90SPCh. 13 - Prob. 13.91SPCh. 13 - Prob. 13.92SPCh. 13 - Consider the endothermic reaction Fe3+ (aq) + Cl...Ch. 13 - Prob. 13.94SPCh. 13 - Prob. 13.95SPCh. 13 - Prob. 13.96SPCh. 13 - Prob. 13.97SPCh. 13 - Prob. 13.98CHPCh. 13 - Prob. 13.99CHPCh. 13 - Prob. 13.100CHPCh. 13 - Prob. 13.101CHPCh. 13 - Prob. 13.102CHPCh. 13 - Prob. 13.103CHPCh. 13 - Prob. 13.104CHPCh. 13 - Prob. 13.105CHPCh. 13 - Refining petroleum involves cracking large...Ch. 13 - Prob. 13.107CHPCh. 13 - Prob. 13.108CHPCh. 13 - Prob. 13.109CHPCh. 13 - Prob. 13.110CHPCh. 13 - At 1000 K, Kp = 2.1 106 and H = 107.7 kJ for the...Ch. 13 - Consider the gas-phase decomposition of NOBr: 2...Ch. 13 - At 100C, Kc = 4.72 for the reaction 2 NO2(g) ...Ch. 13 - Prob. 13.114CHPCh. 13 - Prob. 13.115CHPCh. 13 - Prob. 13.116CHPCh. 13 - Prob. 13.117CHPCh. 13 - Prob. 13.118CHPCh. 13 - Prob. 13.119CHPCh. 13 - Prob. 13.120CHPCh. 13 - Prob. 13.121CHPCh. 13 - Prob. 13.122CHPCh. 13 - Prob. 13.123CHPCh. 13 - Prob. 13.124CHPCh. 13 - Prob. 13.125MPCh. 13 - Prob. 13.126MPCh. 13 - The equilibrium constant Kc for the gas-phase...Ch. 13 - Prob. 13.128MPCh. 13 - Prob. 13.129MPCh. 13 - Prob. 13.130MPCh. 13 - Prob. 13.131MPCh. 13 - Prob. 13.132MPCh. 13 - Consider the sublimation of mothballs at 27C in a...Ch. 13 - Prob. 13.134MPCh. 13 - Prob. 13.135MPCh. 13 - For the decomposition reaction PCl5(g) PCl3(g) +...Ch. 13 - Prob. 13.137MP
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