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what's the implication when Qc=Kc is not = to value of one1 ?
e.g. ka of ethanoic acid is 1.74x10^-5 moldm-3, hence there's a net forward rxn? but surely, the system will reach a state of dynamic equilibrium, right?
Please, if you would, look at page 41 of this thread, thank you !
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Look, you have too many little sub qns liao (as expected). Anything I reply to these sub qns, you'll ask even more sub sub qns. Go ask your school teacher or private tutor, to one shot clarify everything. Edart sterces, rmbr?
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Kc, the equilibrium constant, is a measure of the extent to which the reactants are converted to products before equilibrium is reached
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Hi, I have a question regarding selection of equations from data booklet.
"Transition metals are often good homogeneous and heterogeneous catalysts.
a)The decomposition of hydrogen peroxide is very slow and can be catalysed by homogeneous catalyst. By considering relevant E° values from the Data Booklet,
show by means of equations, how Mn2+ (aq) ions can act as a homogeneous catalyst
in this reaction"
For this question, the equations that I selected are the H2O2/H2O (+1.77V) and MnO4/Mn2+(+1.52V) ones. However, I am wrong because it's not supposed to be MnO4. Please help me understand how to choose the right equations. In the Data Booklet, there are so many equations of the same element (Mn related ones have 4, for example). How do I decide which one is "out"?
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Originally posted by supercat:
Hi, I have a question regarding selection of equations from data booklet.
"Transition metals are often good homogeneous and heterogeneous catalysts.
a)The decomposition of hydrogen peroxide is very slow and can be catalysed by homogeneous catalyst. By considering relevant E° values from the Data Booklet,
show by means of equations, how Mn2+ (aq) ions can act as a homogeneous catalyst
in this reaction"
For this question, the equations that I selected are the H2O2/H2O (+1.77V) and MnO4/Mn2+(+1.52V) ones. However, I am wrong because it's not supposed to be MnO4. Please help me understand how to choose the right equations. In the Data Booklet, there are so many equations of the same element (Mn related ones have 4, for example). How do I decide which one is "out"?
PAGING FOR FLYING GRENADE!!! SHOW ME THAT U CAN CORRECTLY ANSWER SUPERCAT'S QN!!! OTHERWISE (IF U CAN'T), U GOT NO HOPE OF GETTING A GRADE FOR 2016 A LEVELS!!!If within 24 hours, Flying Grenade has shown no hope of getting A grade for 2016 A levels, then I will intervene to answer SuperCat's question.
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Standard cell potentials, for use from 2017 onwards
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Originally posted by UltimaOnline:
PAGING FOR FLYING GRENADE!!! SHOW ME THAT U CAN CORRECTLY ANSWER SUPERCAT'S QN!!! OTHERWISE (IF U CAN'T), U GOT NO HOPE OF GETTING A GRADE FOR 2016 A LEVELS!!!If within 24 hours, Flying Grenade has shown no hope of getting A grade for 2016 A levels, then I will intervene to answer SuperCat's question.
the decomposition eqn of H2O2 : 2H2O2(aq) -> 2H2O(l) + O2(g)
° means standard
[R] : H2O2 + 2H+ + 2e- -> H2O , E°redn = +1.77V[O] : Mn2+ -> Mn3+ e- , E°Oxdn = -1.54VE°Cell = E°Redn@Cathode + E°Oxidn@Anode = (+1.77) + (-1.54) = +0.23VSince the (Standard) Cell Potential is positive, the redox reaction between Mn2+ and H2O2 to generate Mn3+ and H2O is thermodynamically feasible under standard conditions.
Standard Reduction Potential of Mn3+ to Mn2+ is +1.54V
Standard Oxidation Potential of H2O2 to O2 is -0.68V
E°Cell = E°Redn@Cathode + E°Oxidn@Anode = (+1.54) + (-0.68) = +0.81VSince the (Standard) Cell Potential is positive, the redox reaction between Mn3+ and H2O2 to generate Mn2+ and O2 is thermodynamically feasible under standard conditions.Conclusion : Mn2+/Mn3+ has the capacity to function as a homogenous catalyst in the decomposition of H2O2, under standard conditions.if [O] Mn2+/MnO4- , E°Oxdn = -1.52V was used, E°cell would still be positive for the above two aforementioned reactions.in order to know which equations to use when solving questions, you'll need to tactfully guess, then check if its correct. use logic, and chemical knowledge.i think the reason that eqns involving MnO4- cannot be used is because, usage of redn eqn (MnO4-/Mn2+) is invalid as you start of with Mn2+, you don't have MnO4- as a reactant.Is this correct, Ultima?my own doubt : will Mn2+ be chemically unchanged and regenerated at the end of the rxn? -
Originally posted by Flying grenade:Is this correct, Ultima?my own doubt : will Mn2+ be chemically unchanged and regenerated at the end of the rxn?
Correct, well done, Flying Grenade. And yes of course, that's the definition of a catalyst.SuperCat, you got it?
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Originally posted by UltimaOnline:
Correct, well done, Flying Grenade. And yes of course, that's the definition of a catalyst.SuperCat, you got it?
yes got it now. for the overall redox reaction, Mn2+ is regenerated at the R.H.S of eqn.
from the oxidation of H2O2 to O2, Mn3+ is reduced to Mn2+.
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Thank you very much. So you must look at all the equations that includes the reactant?
In this case, you used both the equations for H2O2. (+1.77V) and (+0.68V). This means that in electrochem questions, we must consider all the equations??
How I see it is I look at the reactants. For example, if there was H2O in the reaction, I would be looking at
E(H2O/O2) (+1.23V)
E(H2O/OH) (+0.40V)
E(H2O/H2) (-0.83V)
I would include the (+0.68V) and (+1.77V) as well. Asked my teacher about it, I was told that I need to look at what could form. If H2O2 cannot be formed, then the equation is "out". But I still have trouble selecting equations. 3 equations, which 1 to choose?
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cs toh advanced guide pg 210 btm of page
'...misconception that an increase in temp will decrease rate of exothermic rxn'
what's the correct effect of increase in temp on rate of an exothermic rxn?
rate forward and rate backward definitely increase, but the rate forward increases by a larger extent(pg 198)
but wouldn't the position of eqm shift left also , increasing [rxt] ??
Help Ultima !!
Cstoh should have included what he felt was correct, at the Common errors section of every topic he wrote lol
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Originally posted by supercat:
Thank you very much. So you must look at all the equations that includes the reactant?
In this case, you used both the equations for H2O2. (+1.77V) and (+0.68V). This means that in electrochem questions, we must consider all the equations??
How I see it is I look at the reactants. For example, if there was H2O in the reaction, I would be looking at
E(H2O/O2) (+1.23V)
E(H2O/OH) (+0.40V)
E(H2O/H2) (-0.83V)
I would include the (+0.68V) and (+1.77V) as well. Asked my teacher about it, I was told that I need to look at what could form. If H2O2 cannot be formed, then the equation is "out". But I still have trouble selecting equations. 3 equations, which 1 to choose?
You're welcome.
As we've already wrote full detailed solution + explanation, please exercise close reading and put in a little more effort to understand and internalise, before asking redundant questions
as we said, yes, need to look at the eqns that includes the reactant u have, in this case u have Mn2+ , not MnO4-.
yes, of course, u also need to consider what could form,
that is what we meant by tactful guess, so u no need guess and check all the redox half equations. you tactfully select the required one and check, if it doesnt work, test and check the next most probable one.
look at the decomposition products I've written for the first sentence in the above post. H2O2 decomposes to O2 and H2O. so u look for the half eqns involving this two.
Tactfully, you should deduce H2O2 disproportionates. i.e. O in H2O2 is oxidised and reduced simultaneously.
for your example regarding H2O in the rxn, where exactly? H2O as a product or reactant? show us the exact question, type out or post a url linking to the pic of the question.
if your example regarding H2O, pertains to the same qn of decomposition of H2O2 above, then obviously you select the half equation involving H2O2 and H2O, and not the rest e.g. H2O/OH- or H2O/H2
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Originally posted by Flying grenade:
cs toh advanced guide pg 210 btm of page
'...misconception that an increase in temp will decrease rate of exothermic rxn'
what's the correct effect of increase in temp on rate of an exothermic rxn?
rate forward and rate backward definitely increase, but the rate forward increases by a larger extent(pg 198)
but wouldn't the position of eqm shift left also , increasing [rxt] ??
Help Ultima !!
Cstoh should have included what he felt was correct, at the Common errors section of every topic he wrote lol
CS Toh's right. And you're also right about shifting of position of equilibrium. So what's your doubt exactly?!?Ok test you : what happens to k-forward and k-backward, and where does the position of equilibrium shift towards, when you decrease temperature for an endothermic reaction? how about for an exothermic reaction?
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Hi, for 2015 tjc paper 3 qn 3i, on why aspirin is insoluble in water, my answer differs fromthe answer scheme. I said that the formation of predominantly instantaneous dipole induced dipole interactions due to large non polar benzene ring between aspirin and h2o molecules releases insufficient energy to overcome hydrogen bonds between h2o and idid interactions between aspirin. Dissolution is energetically unfavourable. is my answer acceptable?
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Originally posted by UltimaOnline:
CS Toh's right. And you're also right about shifting of position of equilibrium. So what's your doubt exactly?!?Ok test you : what happens to k-forward and k-backward, and where does the position of equilibrium shift towards, when you decrease temperature for an endothermic reaction? how about for an exothermic reaction?
so [rxt] increase as a result of shifting of eqm, and rate of product formation also increase (rate of forward rxn) , as a result of increasing temp on a exothermic rxn?
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Originally posted by ArJoe:
Hi, for 2015 tjc paper 3 qn 3i, on why aspirin is insoluble in water, my answer differs fromthe answer scheme.
I said that the formation of predominantly instantaneous dipole induced dipole interactions due to large non polar benzene ring between aspirin and h2o molecules releases insufficient energy to overcome hydrogen bonds between h2o and idid interactions between aspirin. Dissolution is energetically unfavourable.
is my answer acceptable?
No, your answer is not acceptable.1st error : between polar and non-polar molecules, it's not instantaneous dipole - induced dipole London dispersion van der Waals interactions, it's permanent dipole - induced dipole Debye van der Waals interactions.
2nd error : you didn't mention that the COOH and OCO groups of aspirin can and will participate in limited hydrogen bonding with the water solvent, but the solubility of aspirin as a whole is still limited by the weaker permanent dipole - induced dipole Debye van der Waals interactions of the non-polar hydrophobic benzene ring with water solvent.
Lastly, and this is partly the question's fault : if you say "dissolution is energetically unfavourable", it implies aspirin is completely insoluble (eg. like SiO2 or diamond, etc), but aspirin is actually slightly soluble and does dissolve slightly in water under standard conditions. Which is why it's better to state and explain why "the solubility of aspirin is hence limited" in water.
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Originally posted by UltimaOnline:
CS Toh's right. And you're also right about shifting of position of equilibrium. So what's your doubt exactly?!?Ok test you : what happens to k-forward and k-backward, and where does the position of equilibrium shift towards, when you decrease temperature for an endothermic reaction? how about for an exothermic reaction?
when decrease temp for endo rxn, position of eqm shift left (exothermic side). k-forward ( of endo rxn) decreases less extent than k-backward (of exo rxn) ?
when decrease temp for exo rxn, position of eqm shift right (exothermic side). k-forward (of exo rxn) decreases to a larger extent than k-backward (of endo rxn)?
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Originally posted by Flying grenade:
so [rxt] increase as a result of shifting of eqm, and rate of product formation also increase (rate of forward rxn) , as a result of increasing temp on a exothermic rxn?
Yes, but you haven't answered my question on the effect of DECREASING TEMPERATURE, for endothermic reaction, VERSUS (ie. separate question) exothermic reaction. -
Originally posted by UltimaOnline:
Yes, but you haven't answered my question on the effect of DECREASING TEMPERATURE, for endothermic reaction, VERSUS (ie. separate question) exothermic reaction.OHHH COOL !!! Thanks Ultima !!!!
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so [rxt] increase as a result of shifting of eqm left, would [pdt] decrease, contradictory to increase rate of product formation ???
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from :
http://www.chemicalforums.com/index.php?topic=9699.0 thread title solubility of aspirin
Actually, Aspirin is partially soluble in water. The reason why it is only partially soluble in water is because of its hydrophobic portion of the molecule. Also, the hydrogen on the carboxylic acid functional group can hydrogen bond with the oxygen atom of the ester carbonyl group making it even further less soluble in water.
When you treat aspirin with sodium bicarbonate, it undergoes an acid base reaction to give you a sodium carboxylate, which is an ionic species of aspirin, thus making it water soluble. Also, the intermolecular hydrogen bonding affect is now absent because the proton on the carboxylic acid is now gone.aspirin is slightly soluble in water, as COOH and OCO groups of aspirin can and will participate in limited hydrogen bonding with the water solvent, and
made more soluble after rxn with NaHCO3
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Originally posted by Flying grenade:
so [rxt] increase as a result of shifting of eqm left, would [pdt] decrease, contradictory to increase rate of product formation ???
Stop with your silly acronyms, write out [reactants] and [products] in the A level exam.There's nothing contradictory. Position of equilibrium can shift LHS or RHS, [reactants] and [products] can increase or decrease (alternately not together), and k-forward and k-backward can increase or decrease, depending on whether you increase or decrease temperature, and whether forward reaction is endothermic or exothermic.
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Originally posted by UltimaOnline:
Yes, but you haven't answered my question on the effect of DECREASING TEMPERATURE, for endothermic reaction, VERSUS (ie. separate question) exothermic reaction.repost on top, just before u posted, i posted
when decrease temp for endo rxn, position of eqm shift left (exothermic side). k-forward ( of endo rxn) decreases less extent than k-backward (of exo rxn) ?
when decrease temp for exo rxn, position of eqm shift right (exothermic side). k-forward (of exo rxn) decreases to a larger extent than k-backward (of endo rxn)?