For a reaction at equilibriumthe relation between dissociation constant (K), degree of dissociation (andinfin;) and equilibrium pressure (p) is given by:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?

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For a reaction at equilibrium

the relation between dissociation constant (K), degree of dissociation (∞) and equilibrium pressure (p) is given by:

Correct answer is option 'B'. Can you explain this answer?

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For a reaction at equilibriumthe relation between dissociation constan...

Now equilibrium pressue (p),

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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.From the given data, it can be interpreted that .

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Paragraph 1Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equation:The standard reaction Gibbs energy, ΔrG0, of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β. Thus, βequilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: R = 0.083 L bar K-1 mol-1)Q.The INCORRECT statement among the following, for this reaction is

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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 that

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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

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For a reaction at equilibriumthe relation between dissociation constant (K), degree of dissociation (∞) and equilibrium pressure (p) is given by:a)b)c)d)Correct answer is option 'B'. Can you explain this answer?

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