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In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:
At constant temperature and constant pressure for a fixed mass of gaseous mixtures,
PV = nRT Volume ∝ number of moles
Where Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200°C and 252°C are 70.2 and 57.2 respectively at one atmosphere
Q.
This observation shows that
  • a)
    There action is endothermic
  • b)
    The reaction is exothermic
  • c)
    The reaction is not affected by change in pressure
  • d)
    The reaction is not affected by adding inert gas at constant pressure.
Correct answer is option 'A'. Can you explain this answer?
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In gaseous dissociation reactions, the total mass remains unchanged an...
As Kp value increases with increase in temperature, this implies the reaction is endothermic.
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In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.The observation shows that the dissociation constants (KP) for the reaction are

In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.From the given data, it can be interpreted that .

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculatedfrom their molecular formula. The observed densities decrease towards a limit as thetemperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.Examples :With increase in the number of molecules, the volume increases (pressure remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,t = 0 at = teq a x nx Total no. of moles Observed molecular weight or molar mass of the mixtureQ.The K for the reaction is 640 mm at 775 K. The percentage dissociation of N2O4 at equilibrium pressure of 160 mm is

For certain substances such as ammonium chloride, nitrogen peroxide, phosphorus pentachloride, etc. the measured densities are found to be less than those calculatedfrom their molecular formula. The observed densities decrease towards a limit as thetemperature is raised. This is due to the splitting of the molecules into simpler ones. The process is reversible and is called thermal dissociation.Examples :With increase in the number of molecules, the volume increases (pressure remaining constant) and, in consequence, the density decreases. As the temperature rises, more and more dissociation takes place, and when practically complete dissociation occursthe density reaches its lowest limit.The extent of issociation, ice., the fraction of the total number of molecules which suffers dissociation is called the degree of dissociation. Gas density measurements can be used to determine the degree of dissociation. Let us take by general case where one molecule of a substance A splits up into n molecule of B on heating; i.e.,t = 0 at = teq a x nx Total no. of moles Observed molecular weight or molar mass of the mixtureQ.A sample of mixture of A(g), B(g) and C(g) under equilibrium has a mean molecular weight (observe d) is 80.The equilibrium is Find the degree of dissociation for A(g).

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In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer?
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In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer?.
Solutions for In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer?, a detailed solution for In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice In gaseous dissociation reactions, the total mass remains unchanged and the number of moles increases as a result of the reaction. Thus the average molecular weight and hence the vapour density decreases. In other words, the volume increases at constant temperature and pressure and hence the vapour density decreases.The relationship between vapour density in the beginning and vapour density at equilibrium can be found as follows:At constant temperature and constant pressure for a fixed mass of gaseous mixtures,PV = nRT Volume number of molesWhere Do and D are the vapour densities in the beginning and at equilibrium and no and n are the number of moles in the beginning and at equilibrium.The vapour density of PCI5 at 200C and 252C are 70.2 and 57.2 respectively at one atmosphereQ.This observation shows thata)There action is endothermicb)The reaction is exothermicc)The reaction is not affected by change in pressured)The reaction is not affected by adding inert gas at constant pressure.Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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