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One method to produce hydrogen on an ind...

One method to produce hydrogen on an industrial on an industrial scale is the reaction of methane with overheated water vapour at 1100 K to form hydrogen and carbon monoxide . The reaction is known as steam reforming
Calcuate the pressure in the vessel at 1100K and the % conversion of methane.

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The correct Answer is:
Total pressure at 1100K =6.550 bar
Conversion (methene) `~~ 49%`
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One method to produce hydrogen on an industrial on an industrial scale is the reaction of methane with overheated water vapour at 1100 K to form hydrogen and carbon monoxide . The reaction is known as steam reforming Write the balanced equations for the steam reforming of methane .

It we see the reaction of methane with halogen, the rate determining step for chlorination is, endothermic reaction of the chlorine atom with methane to form methyl radical and a molecule of HCl. So free radical is the intermediate of the reaction. Formation of free radical depends upon the energy required to break a bond between a hydrogen atom and a carbon atom. Chlorination of propane and Bromination of propane. when compared it is found that bromination is more selective than chlorination. The probability factor for 3^(@),2^(@),1^(@)H atom is 5.0:3.8:1.0 at 25^(@)C for chlorination. Isobutane when reacts with chlorine in presence of ultra violet radiations yield 2 products primary hydrogen substituted and 3^(@) hydrogen substituted Find their % in product mixture

Large quantities of hydrogen are required these days in industry for various purposes. Water is found to be the industry raw material from which hydrogen can be obtained . One of the important methods for the commerical production of hyrogen in Lane's process in which steam is passes over hot iron. Iron decomposes steam with formation of magnetic oxide (Fe_(3)O_(4)) and hydrogen. (i) {:(3Fe,+,4H_(2)O,rarr,Fe_(3)O_(4),+, 4H_(2)),(,,"steam",,"magnetic ",,),(,,,,"oxide",,):} Iron is regenerated by reducing magnetic oxide with water gas (it is mixture of CO and H_(2) in 1:1 mole ratio). {:((ii),Fe_(3)O_(4),+,4CO,rarr,3Fe ,+,4CO_(2)uarr),((iii),Fe_(3)O_(4),+,4H_(2) ,rarr,3Fe,+,3H_(2)O),(,,,,,,,("Steam recovered")):} [equal amount sof magnetic oxide react in steps (ii) and (iii)] Steam so produced is again used in the previous step in which magnitic oxide is produced : [At . wt. implies Fe=56] How many grams of steam is consumed in order to produce 11.35 litres H_(2) gas at STP?

Large quantities of hydrogen are required these days in industry for various purposes. Water is found to be the industry raw material from which hydrogen can be obtained . One of the important methods for the commerical production of hyrogen in Lane's process in which steam is passes over hot iron. Iron decomposes steam with formation of magnetic oxide (Fe_(3)O_(4)) and hydrogen. (i) {:(3Fe,+,4H_(2)O,rarr,Fe_(3)O_(4),+, 4H_(2)),(,,"steam",,"magnetic ",,),(,,,,"oxide",,):} Iron is regenerated by reducing magnetic oxide with water gas (it is mixture of CO and H_(2) in 1:1 mole ratio). {:((ii),Fe_(3)O_(4),+,4CO,rarr,3Fe ,+,4CO_(2)uarr),((iii),Fe_(3)O_(4),+,4H_(2) ,rarr,3Fe,+,3H_(2)O),(,,,,,,,("Steam recovered")):} [equal amount sof magnetic oxide react in steps (ii) and (iii)] Steam so produced is again used in the previous step in which magnitic oxide is produced : [At . wt. implies Fe=56] if iron is recovered in (ii) an (iii) with 80% efficiency , then how many gram iron is recovered if initially 56 gm iron is taken?

Large quantities of hydrogen are required these days in industry for various purposes. Water is found to be the industry raw material from which hydrogen can be obtained . One of the important methods for the commerical production of hyrogen in Lane's process in which steam is passes over hot iron. Iron decomposes steam with formation of magnetic oxide (Fe_(3)O_(4)) and hydrogen. (i) {:(3Fe,+,4H_(2)O,rarr,Fe_(3)O_(4),+, 4H_(2)),(,,"steam",,"magnetic ",,),(,,,,"oxide",,):} Iron is regenerated by reducing magnetic oxide with water gas (it is mixture of CO and H_(2) in 1:1 mole ratio). {:((ii),Fe_(3)O_(4),+,4CO,rarr,3Fe ,+,4CO_(2)uarr),((iii),Fe_(3)O_(4),+,4H_(2) ,rarr,3Fe,+,3H_(2)O),(,,,,,,,("Steam recovered")):} [equal amount sof magnetic oxide react in steps (ii) and (iii)] Steam so produced is again used in the previous step in which magnitic oxide is produced : [At . wt. implies Fe=56] How many grams of iron are needed in order to produce 11.35 litre fo H_(2) gas at STP?

Large quantities of hydrogen are required these days in industry for various purposes. Water is found to be the industry raw material from which hydrogen can be obtained . One of the important methods for the commerical production of hyrogen in Lane's process in which steam is passes over hot iron. Iron decomposes steam with formation of magnetic oxide (Fe_(3)O_(4)) and hydrogen. (i) {:(3Fe,+,4H_(2)O,rarr,Fe_(3)O_(4),+, 4H_(2)),(,,"steam",,"magnetic ",,),(,,,,"oxide",,):} Iron is regenerated by reducing magnetic oxide with water gas (it is mixture of CO and H_(2) in 1:1 mole ratio). {:((ii),Fe_(3)O_(4),+,4CO,rarr,3Fe ,+,4CO_(2)uarr),((iii),Fe_(3)O_(4),+,4H_(2) ,rarr,3Fe,+,3H_(2)O),(,,,,,,,("Steam recovered")):} [equal amount sof magnetic oxide react in steps (ii) and (iii)] Steam so produced is again used in the previous step in which magnitic oxide is produced : [At . wt. implies Fe=56] How many litre H_(2) gas is produced in (i) at STP in the above problem?

Large quantities of hydrogen are required these days in industry for various purposes. Water is found to be the industry raw material from which hydrogen can be obtained . One of the important methods for the commerical production of hyrogen in Lane's process in which steam is passes over hot iron. Iron decomposes steam with formation of magnetic oxide (Fe_(3)O_(4)) and hydrogen. (i) {:(3Fe,+,4H_(2)O,rarr,Fe_(3)O_(4),+, 4H_(2)),(,,"steam",,"magnetic ",,),(,,,,"oxide",,):} Iron is regenerated by reducing magnetic oxide with water gas (it is mixture of CO and H_(2) in 1:1 mole ratio). {:((ii),Fe_(3)O_(4),+,4CO,rarr,3Fe ,+,4CO_(2)uarr),((iii),Fe_(3)O_(4),+,4H_(2) ,rarr,3Fe,+,3H_(2)O),(,,,,,,,("Steam recovered")):} [equal amount sof magnetic oxide react in steps (ii) and (iii)] Steam so produced is again used in the previous step in which magnitic oxide is produced : [At . wt. implies Fe=56] How many gram water gas is needed in order to recover all the iron in previous problem?

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 :

RESONANCE-THERMODYNAMICS-exercise-3 part-III Advanced level Solutions (STAGE-II)
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