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(a)
Interpretation:
The equilibrium constant
Concept introduction:
Equilibrium constant
In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant
Consider the reaction where A reacts to give B.
On rearranging,
Where,
The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.
In same way
Where,
R is gas constant
T is absolute temperature
(b)
Interpretation:
The equilibrium constant
Concept introduction:
Equilibrium constant
In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant
Consider the reaction where A reacts to give B.
On rearranging,
Where,
The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.
In same way
Where,
R is gas constant
T is absolute temperature
(c)
Interpretation:
The given reaction is endothermic or exothermic reaction has to be explained.
Concept introduction:
Le Chatelier's principle states that if a system in equilibrium gets disturbed due to modification of concentration, temperature, volume, and pressure, then it reset to counteract the effect of disturbance.
Effect of change in temperature:
Endothermic reaction: In this reaction increase in temperature which is absorbed in reactant side, the reaction occurs in a way to relieve the stress in reactant side. The reaction occurs along the direction of product side and increases equilibrium constant.
Exothermic reaction: In this reaction increase in temperature which is released in product side, the reaction occurs in a way to relieve the stress in product side. The reaction occurs along the direction of reactant side and decreases equilibrium constant.
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Chapter 13 Solutions
General Chemistry: Atoms First
- Assign these protonarrow_forwardCould you please solve the first problem in this way and present it similarly but color-coded or step by step so I can understand it better? Thank you!arrow_forwardCould you please solve the first problem in this way and present it similarly but color-coded or step by step so I can understand it better? Thank you!arrow_forward
- Could you please solve the first problem in this way and present it similarly but (color-coded) and step by step so I can understand it better? Thank you! I want to see what they are doingarrow_forwardCan you please help mne with this problem. Im a visual person, so can you redraw it, potentislly color code and then as well explain it. I know im given CO2 use that to explain to me, as well as maybe give me a second example just to clarify even more with drawings (visuals) and explanations.arrow_forwardPart 1. Aqueous 0.010M AgNO 3 is slowly added to a 50-ml solution containing both carbonate [co32-] = 0.105 M and sulfate [soy] = 0.164 M anions. Given the ksp of Ag2CO3 and Ag₂ soy below. Answer the ff: Ag₂ CO3 = 2 Ag+ caq) + co} (aq) ksp = 8.10 × 10-12 Ag₂SO4 = 2Ag+(aq) + soy² (aq) ksp = 1.20 × 10-5 a) which salt will precipitate first? (b) What % of the first anion precipitated will remain in the solution. by the time the second anion starts to precipitate? (c) What is the effect of low pH (more acidic) condition on the separate of the carbonate and sulfate anions via silver precipitation? What is the effect of high pH (more basic)? Provide appropriate explanation per answerarrow_forward
- Part 4. Butanoic acid (ka= 1.52× 10-5) has a partition coefficient of 3.0 (favors benzene) when distributed bet. water and benzene. What is the formal concentration of butanoic acid in each phase when 0.10M aqueous butanoic acid is extracted w❘ 25 mL of benzene 100 mL of a) at pit 5.00 b) at pH 9.00arrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 Group of answer choices 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forwardCalculate activation energy (Ea) from the following kinetic data: Temp (oC) Time (s) 23.0 180. 32.1 131 40.0 101 51.8 86.0 choices: 0.0269 kJ/mole 2610 kJ/mole 27.6 kJ/mole 0.215 kJ/mole 20.8 kJ/molearrow_forward
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