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 13.128MP

(a)

Interpretation Introduction

Interpretation:

The ratio of dimers to monomers for acetic acid in benzene has to be calculated.

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,

[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.

(b)

Interpretation Introduction

Interpretation:

The ratio of dimers to monomers for acetic acid in water has to be calculated.

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,

[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.

(c)

Interpretation Introduction

Interpretation:

The reason of less Kc value for water solution than benzene solution has to be explained.

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,

[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.

Hydrogen Bond:

A bond between hydrogen and most electronegative atom is known as hydrogen bond.

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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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