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some questions
Methanol can be formed on an industrial scale from carbon dioxide and hydrogen by a reversible reaction as shown below.
CO2(g) + 3H2 -><- CH3OH(g) + H20(g)
Under the conditions (700K and 30MPa), equilibrium is reached when 2% of the carbon dioxide has been converted.
(a) If the pressure was reduced but the temperature was kept the some, deduce what would happen to the equilibrium yield of methanol. Explain Your answer.
Yield - Decrease
Explanation - I am not sure about this one.
(b) In the presence of a very efficient copper-based catalyst, this industrial process can be operated at lower temperature of 500K and a pressure of 30MPa. Under these conditions, at equilibrium, more of the carbon dioxide is converted into methanol.
Use this information to deduce the sign of the enthalpy change for the reaction. Explain your deduction.
Sign of enthalpy change - Positive
Explanation - I am not sure about this one.
(c)In the processes above, the equilibrium yield of methanol is low. Suggest what is done with the unreacted carbon dioxide and hydrogen.
CO2(g) + 3H2 -><- CH3OH(g) + H20(g)
Under the conditions (700K and 30MPa), equilibrium is reached when 2% of the carbon dioxide has been converted.
(a) If the pressure was reduced but the temperature was kept the some, deduce what would happen to the equilibrium yield of methanol. Explain Your answer.
Yield - Decrease
Explanation - I am not sure about this one.
(b) In the presence of a very efficient copper-based catalyst, this industrial process can be operated at lower temperature of 500K and a pressure of 30MPa. Under these conditions, at equilibrium, more of the carbon dioxide is converted into methanol.
Use this information to deduce the sign of the enthalpy change for the reaction. Explain your deduction.
Sign of enthalpy change - Positive
Explanation - I am not sure about this one.
(c)In the processes above, the equilibrium yield of methanol is low. Suggest what is done with the unreacted carbon dioxide and hydrogen.
Reply 1
rlagksquf
(a) If the pressure was reduced but the temperature was kept the some, deduce what would happen to the equilibrium yield of methanol. Explain Your answer.
Yield - Decrease
Explanation - I am not sure about this one.
(a) If the pressure was reduced but the temperature was kept the some, deduce what would happen to the equilibrium yield of methanol. Explain Your answer.
Yield - Decrease
Explanation - I am not sure about this one.
Higher pressure would increase yield as equilibrium is reached quicker (side with more moles to the side with less). So an decrease of pressure would decrease yield as it would take longer for equilibrium to be reached.
(This should be correct)
Reply 2
rlagksquf
(c)In the processes above, the equilibrium yield of methanol is low. Suggest what is done with the unreacted carbon dioxide and hydrogen.
(c)In the processes above, the equilibrium yield of methanol is low. Suggest what is done with the unreacted carbon dioxide and hydrogen.
recycled (like the haber process)
Reply 3
12
Well, I think b) should be negative sign for enthalpy change, isnt it?
Reply 4
rlagksquf
(b) In the presence of a very efficient copper-based catalyst, this industrial process can be operated at lower temperature of 500K and a pressure of 30MPa. Under these conditions, at equilibrium, more of the carbon dioxide is converted into methanol.
Use this information to deduce the sign of the enthalpy change for the reaction. Explain your deduction.
Sign of enthalpy change - Positive
Explanation - I am not sure about this one.
(b) In the presence of a very efficient copper-based catalyst, this industrial process can be operated at lower temperature of 500K and a pressure of 30MPa. Under these conditions, at equilibrium, more of the carbon dioxide is converted into methanol.
Use this information to deduce the sign of the enthalpy change for the reaction. Explain your deduction.
Sign of enthalpy change - Positive
Explanation - I am not sure about this one.
Explanation - I am not sure about this one.
I don't see how a catalyst can affect wheather something is endo or exo thermic, all is does is lower the activation energy so that eq. can be reached quicker. If it is positive it means that something has been broken.
I don't know
Reply 5
12
a) As P decreases, CO2 and H2 molecules are further apart -> less reactions/second occur -> less yield
Reply 6
BCHL85
a) As P decreases, CO2 and H2 molecules are further apart -> less reactions/second occur -> less yield
Oh yeah, I forgot to add that. (*Within a fixed volume)
Reply 7
there seems to be a little confusion regarding equilibrium...
catalysts - no effect on equilibrium position (constant) or yield. The equilibrium is merely attained faster from a non-equilibrium position
pressure - increase disturbs the equilibrium and the reactions (forward and reverse) respond by making more of the side with fewer moles re-establishing the equilibrium constant. This can be understood by considering the concentrations of the gases. If you increase Pressure at constant Temperature then you have to decrease volume. The gases increase in concentration. If there are unequal moles on either side then the side with the fewer moles will have a smaller increase in concentration and consequently less of an increase in rate. The reaction responds by making more of this side to re-establish the equilibrium constant.
temperature - this is the ONLY factor that affects the equilibrium constant. Increase the temperature and the equilibrium will shift towards the direction of endothermic change. Ths is because the endothermic rate direction is affected more than the exothermic rate. To fully understand why you have to involve the Maxwell Boltzmann distribution and the activation energy. the endothermic change has a higher activation energy and an increase in temperature will increase the number of particles with this energy by a greater prorportion than for the exothermic change, increaing the endothermic rate more so than the exothermic rate moving the equilibrium in the direction of endothermic change.
catalysts - no effect on equilibrium position (constant) or yield. The equilibrium is merely attained faster from a non-equilibrium position
pressure - increase disturbs the equilibrium and the reactions (forward and reverse) respond by making more of the side with fewer moles re-establishing the equilibrium constant. This can be understood by considering the concentrations of the gases. If you increase Pressure at constant Temperature then you have to decrease volume. The gases increase in concentration. If there are unequal moles on either side then the side with the fewer moles will have a smaller increase in concentration and consequently less of an increase in rate. The reaction responds by making more of this side to re-establish the equilibrium constant.
temperature - this is the ONLY factor that affects the equilibrium constant. Increase the temperature and the equilibrium will shift towards the direction of endothermic change. Ths is because the endothermic rate direction is affected more than the exothermic rate. To fully understand why you have to involve the Maxwell Boltzmann distribution and the activation energy. the endothermic change has a higher activation energy and an increase in temperature will increase the number of particles with this energy by a greater prorportion than for the exothermic change, increaing the endothermic rate more so than the exothermic rate moving the equilibrium in the direction of endothermic change.
Reply 8
charco
there seems to be a little confusion regarding equilibrium...
whats the confusion?
Reply 9
gordon2002
Explanation - I am not sure about this one.
I don't see how a catalyst can affect wheather something is endo or exo thermic, all is does is lower the activation energy so that eq. can be reached quicker. If it is positive it means that something has been broken.
I don't know
I don't see how a catalyst can affect wheather something is endo or exo thermic, all is does is lower the activation energy so that eq. can be reached quicker. If it is positive it means that something has been broken.
I don't know
doesn't sound very clear to me!
Reply 10
charco
doesn't sound very clear to me!
which bit don't you understand?
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