Home
Class 12
CHEMISTRY
The dissociation of (molecular) chlorine...

The dissociation of (molecular) chlorine is an endothermic process, `DeltaH=243.6 kJ mol^(-1)`. The disoociation can also attained by the effect of light.
How can the value obtained be (qualitatovely) explained? Describe the reaction mechanism.

Text Solution

Verified by Experts

The correct Answer is:
The observed quantum yield is based on a chain mechanism.
The start of reaction chain : `Cl_(2)+hv rarr 2 Cl*`
The propagation of the chain: `2 Cl *+H_(2) rarr HCl+2 H*`
`H*+Cl_(2) rarr HCl +Cl*`
The chain termination mainly by: `2 H* rarr H_(2)`
`2 Cl* rarr Cl_(2)`
`H*+Cl* rarr HCl`
Promotional Banner

Similar Questions

Explore conceptually related problems

The dissociation of (molecular) chlorine is an endothermic process, DeltaH=243.6 kJ mol^(-1) . The disoociation can also attained by the effect of light. At what wavelength can the dissociating effect of light be expected?

The dissociation of (molecular) chlorine is an endothermic process, DeltaH=243.6 kJ mol^(-1) . The disoociation can also attained by the effect of light. What is the energy of the photon with the critical wavelength?

The dissociation of (molecular) chlorine is an endothermic process, DeltaH=243.6 kJ mol^(-1) . The disoociation can also attained by the effect of light. Can this effect also be obtained with light whose wavelength is smailer or larger than the calculayed critical wavelength?

The dissociation of (molecular) chlorine is an endothermic process, DeltaH=243.6 kJ mol^(-1) . The disoociation can also attained by the effect of light. How large is the quantum yield (=the number of product molecules per absorbed photons)?

Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. A reaction has a value of DeltaH =-40 Kcal at 400 k cal mol^(-1) . The reaction is spontaneous, below this temperature , it is not . The values fo DeltaG and DeltaS at 400 k are respectively

The thermodynamic property that measures the extent of molecular disorder is called entropy. Entropy change of phase transformation can be calculated using Trouton's formula (DeltaS = DeltaH//T) . In the reversible adiabatic process, however, DeltaS will be zero. the rise in temperature in isobaric and isochoric conditions is found to increase the randomness or entropy of the system. DeltaS = 2.303 C log (T_(1)//T_(2)) When 1 mol of an ideal gas is compressed to half of its volume, its temperature becomes half. Then the change in entropy (DeltaS) would be

The thermodynamic property that measures the extent of molecular disorder is called entropy. Entropy change of phase transformation can be calculated using Trouton's formula (DeltaS = DeltaH//T) . In the reversible adiabatic process, however, DeltaS will be zero. the rise in temperature in isobaric and isochoric conditions is found to increase the randomness or entropy of the system. DeltaS = 2.303 C log (T_(1)//T_(2)) The melting point of a solid is 200K and its latent heat of fusion is 400cal mol^(-1) . The entropy changes for the fusion of 1 mole of the solid (in cal K^(-1)) at the same temperature would be

Consider a solution of CH_3COONH_4 which is a salt weak acid and weak base. The equilibrium involved in the solutions are : CH_3COO^(-)+H_2OhArrCH_3COOH+OH^(-) " ........"(i) NH_(4)^(+)+H_(2)hArrNH_(4)OH+H^(+)" ........"(ii) H^(+)+OH^(-)hArrH_(2)O" ........"(iii) If we add these reactions, then the net reaction is : CH_(3)COO^(-)+H_(2)^(+)+H_(2)OhArrCH_(3)COOH+NH_(4)OH" ........"(iv) Both CH_(3)COO^(-) and NH_(4)^(+) get hydrolysed independently and their hydrolysis depends on : (a) their initial concentration (b) The value of K_(h) which is (K_(w))/(K_(a)) for CH_(3)COO^(-) and (K_(w))/(K_(b)) for NH_(4)^(+) . Since both of the ions were produced form the salt, their initial concertration are same. Therefore, unless and until the value of (K_(w))/(K_(a)) or K_(a) and K_(b) is same, the degree of hydrogen of ions can't be same. To explain why we assume that degree of hydrolysis of cation and anion is same, we needed to now look at the third reaction i.e.,combination of H^(+) and OH^(-) ions. It is obvious that this reaction happens only because one reaction produced H^(+) ion and the other prodcued OH^(-) ions. We can also note that this reaction causes both the hydrolysis reaction to occur more since their product ions are being consumed. Keep this in mind that the equilibrium which has smaller value of the equilibrium constant is affected more by the common ion effect. For the same reason if for any reson a reaction is made to occur to a greater extent by the consumption of any of the prodcut ion, the reaction with the smaller value of equilibrium constant tends to get affected more. Therefore, we conclude that firstly the hydroylsis of both the ions occurs more in the presence of each other (due to consumption of the product ions) than in each other's absence. Secondly,the hydroylsis of the ion which occurs to a lesser extent (due to smaller value of K_(h)) is affected more than the one whole K_(h) is greater. Hence we can see that the degree of hydroylsis of both the ions would be close to each other when they are getting hyderolysed in the presence of each other. In the hydrolysis of salt weak acid and weak base :

Consider a solution of CH_3COONH_4 which is a salt weak acid and weak base. The equilibrium involved in the solutions are : CH_3COO^(-)+H_2OhArrCH_3COOH+OH^(-) " ........"(i) NH_(4)^(+)+H_(2)hArrNH_(4)OH+H^(+)" ........"(ii) H^(+)+OH^(-)hArrH_(2)O" ........"(iii) If we add these reactions, then the net reaction is : CH_(3)COO^(-)+H_(2)^(+)+H_(2)OhArrCH_(3)COOH+NH_(4)OH" ........"(iv) Both CH_(3)COO^(-) and NH_(4)^(+) get hydrolysed independently and their hydrolysis depends on : (a) their initial concentration (b) The value of K_(h) which is (K_(w))/(K_(a)) for CH_(3)COO^(-) and (K_(w))/(K_(b)) for NH_(4)^(+) . Since both of the ions were produced form the salt, their initial concertration are same. Therefore, unless and until the value of (K_(w))/(K_(a)) or K_(a) and K_(b) is same, the degree of hydrogen of ions can't be same. To explain why we assume that degree of hydrolysis of cation and anion is same, we needed to now look at the third reaction i.e.,combination of H^(+) and OH^(-) ions. It is obvious that this reaction happens only because one reaction produced H^(+) ion and the other prodcued OH^(-) ions. We can also note that this reaction causes both the hydrolysis reaction to occur more since their product ions are being consumed. Keep this in mind that the equilibrium which has smaller value of the equilibrium constant is affected more by the common ion effect. For the same reason if for any reson a reaction is made to occur to a greater extent by the consumption of any of the prodcut ion, the reaction with the smaller value of equilibrium constant tends to get affected more. Therefore, we conclude that firstly the hydroylsis of both the ions occurs more in the presence of each other (due to consumption of the product ions) than in each other's absence. Secondly,the hydroylsis of the ion which occurs to a lesser extent (due to smaller value of K_(h)) is affected more than the one whole K_(h) is greater. Hence we can see that the degree of hydroylsis of both the ions would be close to each other when they are getting hyderolysed in the presence of each other. For 0.1 M CH_3COONH_(4) salt solution given, K_(a)(CH_(3)COOH)=K_(b)(NH_(4)OH)=2xx10^(-5) . In the case : degree of hydrolysis of cation and anion are :

An organic compound (A) having C = 16.27%,H = 0.677%, Cl = 72.203% reduces Fehling solution and on oxidation gives an acid (B) having C = 14.679%, H = 0.612% and Cl = 65.137%. (B) on distillation with sodalime gives a sweet smelling liquid (C) which contains 89.12% chlorine. (C) can also be obtained by heating (A) with alkali. (A) can also be obtained by the action of Cl2 on C_2 H_5 OH . Identify (A), (B) and (C) and explain the reactions.