Chemistry: Post your doubts here!

[hela]

the reactions that take place in the catalylic converter:
*oxidation of CO to form CO2
*the reduction of nitrogen oxides to form nitrogen
*the oxidation of unburnt hydrocarbon to produce CO2 n H2O
HOPE THIS HELPS

thanks a lot
how to explain the solubility of the group 2 hydroxide

 

[littlecloud11]

i think that ΔH(sol) =ΔH(hyd) _ ΔH(latt) and not the opposit

how to explain the solubility of the group 2 hydroxide
t

It is. Apologies, I reversed the energies in the first line.
corrected now.

Hydroxide ion is a relatively small anion compared to the sizes of the group 2 cations, so it has less contribution towards the lattice enthalpy than the cations. As the size of the cation increases down the group the lattice enthalpy decreases down the group (becomes less exothermic) quite drastically as there is usually very little polarisation of the hydroxide anion.

For the hydration enthalpy, the small size of the OH- ion contributes significantly towards the overall energy released. The enthalpy of hydration remains constant for the hydroxide ion. As the cation size increases the overall hydration enthalpy decreases (becomes less exothermic), but only to a small amount as the OH- ion still releases a large amount of energy.

ΔH(sol) =ΔH(hyd) - ΔH(latt)
As the decrease in lattice enthalpy (lattice enthalpy becomes more positive) is more than the hydration enthalpy, ΔH(sol) becomes more exothermic down the group and hence the hydroxides become more soluble.

 

[littlecloud11]

Jazak Allah khiaran....thank you soo much !!!!!!!!!!! Really wonderful explanation ! May Allah reward you and your family with Jannah tul firdous in hereafter. Ameen

You are the nicest well-wisher on this forum!

 

[Joeylicious =)]

Go To Xtreme papers and choose your Level (O OR A)
The syllabus will be Qouted like:9701_y13_sy.pdf
Sy=Syllabus

Thank you!!

 

[Manobilly]

http://www.thestudentroom.co.uk/attachment.php?attachmentid=95974&d=1295962248
http://www.thestudentroom.co.uk/attachment.php?attachmentid=95975&d=1295962248

JazakAllah

 

[hela]

It is. Apologies, I reversed the energies in the first line.
corrected now.

Hydroxide ion is a relatively small anion compared to the sizes of the group 2 cations, so it has less contribution towards the lattice enthalpy than the cations. As the size of the cation increases down the group the lattice enthalpy decreases down the group (becomes less exothermic) quite drastically as there is usually very little polarisation of the hydroxide anion.

For the hydration enthalpy, the small size of the OH- ion contributes significantly towards the overall energy released. The enthalpy of hydration remains constant for the hydroxide ion. As the cation size increases the overall hydration enthalpy decreases (becomes less exothermic), but only to a small amount as the OH- ion still releases a large amount of energy.

ΔH(sol) =ΔH(hyd) - ΔH(latt)
As the decrease in lattice enthalpy (lattice enthalpy becomes more positive) is more than the hydration enthalpy, ΔH(sol) becomes more exothermic down the group and hence the hydroxides become more soluble.

PLEASE IS ΔH(hyd) IS ALWAYS NEGATIVE
thanks

 

[littlecloud11]

PLEASE IS ΔH(hyd) IS ALWAYS NEGATIVE
thanks

Yes. Always negative. It involves the formation of attractive bonds and all bond forming processes are exothermic.

 

[hela]

Yes. Always negative. It involves the formation of attractive bonds and all bond forming processes are exothermic.

can you solve
2 Ammonia is manufactured by the Haber Process, in an exothermic reaction.
Assuming that the amount of catalyst remains constant, which change will not bring about an
increase in the rate of the forward reaction?
A decreasing the size of the catalyst pieces
B increasing the pressure
C increasing the temperature
D removing the ammonia as it is formed

14 Use of the Data Booklet is relevant to this question.
The reaction between aluminium powder and anhydrous barium nitrate is used as the propellant
in some fireworks. The metal oxides and nitrogen are the only products.
Which volume of nitrogen, measured under room conditions, is produced when 0.783 g of
anhydrous barium nitrate reacts with an excess of aluminium?
A 46.8cm3
B 72.0cm3
C 93.6cm3
D 144cm3

24 How many isomeric esters have the molecular formula C4H8O2?
A 2 B 3 C 4 D 5

 

[Gémeaux]

can you solve
2 Ammonia is manufactured by the Haber Process, in an exothermic reaction.
Assuming that the amount of catalyst remains constant, which change will not bring about an
increase in the rate of the forward reaction?
A decreasing the size of the catalyst pieces
B increasing the pressure
C increasing the temperature
D removing the ammonia as it is formed

24 How many isomeric esters have the molecular formula C4H8O2?
A 2 B 3 C 4 D 5

2) C. the forward reaction is exothermic and is favored by a decrease in temperature. Increasing it would increase the rate of the backward reaction.
A is incorrect because decreasng the catalyst's size would favor both the forward and backward reaction. B rejected because an increase in pressure shifts the equilibrium to the side with fewer moles (in this case 2 moles of NH3)
And removing NH3 obviously favors the R.H.S.

24) C
CH3-COO-CH2-CH3

H-COO-CH2-CH2-CH3

CH3-CH2-COO-CH3

H-COO-CH(CH3)-CH3

 

[hela]

2) C. the forward reaction is exothermic and is favored by a decrease in temperature. Increasing it would increase the rate of the backward reaction.
A is incorrect because decreasng the catalyst's size would favor both the forward and backward reaction. B rejected because an increase in pressure shifts the equilibrium to the side with fewer moles (in this case 2 moles of NH3)
And removing NH3 obviously favors the R.H.S.

24) B
CH3-COO-CH2-CH3

H-COO-CH2-CH2-CH3

CH3-CH2-COO-CH3

these answers are wrong
in marking scheme the answer are
for2) D and for 24 C
i don't care if you like me or not, just answer me!

 

[Gémeaux]

these answers are wrong
in marking scheme the answer are
for2) D and for 24 C
i don't care if you like me or not, just answer me!

eeee Sorry.
For 24th I missed the isomer H-COO-CH(CH3)-CH3

 

[Raiyan3]

Can someone please explain me the the following mcqs
o/n 05 Q1
m/j 05 Q1 & 2
o/n 04 Q3
m/j 04 Q3
o/n 03 Q1
:/

 

[hela]

eeee Sorry.
For 24th I missed the isomer H-COO-CH(CH3)-CH3

I BELIEVE WHATEVER DOESN'T KILL YOU, SIMPLY MAKES YOU STRANGER

 

[Gémeaux]

I BELIEVE WHATEVER DOESN'T KILL YOU, SIMPLY MAKES YOU STRANGER

err.. sorry?

 

[Nab900]

ci) The enthalpy change of solution represents the change of state of an ionic compound from solid to aqueous form. This change corresponds to the difference in energy values for ΔHf MgCl2(s) and ΔHf mgCl2 (aq) as it represents the exact same state change.
So, ΔHsol = -801 - (-641) = -160 KJ/mol

ii) Use the answer form (i) for the formula in (ii). The hydration enthalpy of Cl- should be multiplied by 2 as one mole MgCl2 gives 2 mols of Cl- ions.
ΔHsol (MgCl2) = {2*ΔHhyd(Cl-) + ΔHhyd(Mg2+)} - ΔHlatt(MgCl2)
-160 = 2* ΔHhyd(Cl-) + (-1890) - (-2526)
Therefore, ΔHhyd(Cl-) = (-160 + 1890 - 2526)/2 = -398 Kj/mol

but isn't
Δsol = Δhyd +Δlattice ???
instead of
Δsol = Δhyd - Δlattice ??????

 

[Nab900]

Hi everyone, AsSalamoAlaikum Wr Wb..

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your CHEMISTRY doubts in this thread. InshaAllah members around will help you.

Any Chemistry related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites!

Chemistry Notes:


http://www.chemguide.co.uk This is the website, which contains almost everything classified according to the syllabus.

Tips for solving chemistry MCQs 9701/01

Chemistry P5 Tips and Notes

Some links & Notes - by 'destined007'

Chemistry worksheets Link shared by hassam

Chemistry Application Booklet: Mistakes and Corrections!

Calculations for A level Chemistry, author E.N. Ramsden third edition ebook download.


Regards,
XPC Staff.

http://a-levelchemistry.co.uk/AQA Chemistry/AQA Chemistry home.html
here u go 1 more addition to ur list of chemistry notes

 

[littlecloud11]

but isn't
Δsol = Δhyd +Δlattice ???
instead of
Δsol = Δhyd - Δlattice ??????

The shortest route to calculate enthalpy of solution is simply to move from lattice to ions in solution, i.e. the state change from (s) to (aq)
The longer route is Δsol= -Δlatt + Δhyd. This is simple vector. Note the direction of the arrow. To move along the longer route from lattice to ions in solution the direction is the lattice energy arrow has to be reversed, hence -Δlatt.

Hope this helps!

 

[littlecloud11]

can you solve
2 Ammonia is manufactured by the Haber Process, in an exothermic reaction.
Assuming that the amount of catalyst remains constant, which change will not bring about an
increase in the rate of the forward reaction?
A decreasing the size of the catalyst pieces
B increasing the pressure
C increasing the temperature
D removing the ammonia as it is formed

14 Use of the Data Booklet is relevant to this question.
The reaction between aluminium powder and anhydrous barium nitrate is used as the propellant
in some fireworks. The metal oxides and nitrogen are the only products.
Which volume of nitrogen, measured under room conditions, is produced when 0.783 g of
anhydrous barium nitrate reacts with an excess of aluminium?
A 46.8cm3
B 72.0cm3
C 93.6cm3
D 144cm3

24 How many isomeric esters have the molecular formula C4H8O2?
A 2 B 3 C 4 D 5

2) Ok, the word to emphasize on is RATE of reaction. Increasing the surface area of catalyst, increasing the temp and increasing the pressure all increases the RATE of reaction for the forward reaction but effects the YIELD of forward reaction differently. Increasing the catalyst SA means more molecules of the reactant can bind to the catalyst hence the rate increases, increasing temp means the reactant molecules have more energy, so the rate of successful collision increases and the rate increases (the yield of forward reaction would decrease because the forward reaction is exo. and the equilibrium will shift to the left), increase in pressure also increases the rate of forward reaction because there are greater no. of moles of reactant on the left. Only D, which involves removing the ammonia as it is formed does not effect the rate, but only effects the yield. So the answer is D.

14) the equation would be-
3 Ba(NO3)2 + 10 Al -------> 5Al2O3 + 3BaO + 3N2

Mr of Ba(NO3)2 = 137 + 14*2 + 16* 6 = 261
looking at the stoichiometry of the reaction, 3mols Ba(NO3)2 gives 3 mols N2. So-
3*261 g Ba(NO3)2 = 3*24 dm^3 N2
.783 g Ba(NO3)2 = .072 dm^3 N2 = 72 cm^3 N2
Your answer is B.

Gemeaux solved 24.

 

[hela]

2) Ok, the word to emphasize on is RATE of reaction. Increasing the surface area of catalyst, increasing the temp and increasing the pressure all increases the RATE of reaction for the forward reaction but effects the YIELD of forward reaction differently. Increasing the catalyst SA means more molecules of the reactant can bind to the catalyst hence the rate increases, increasing temp means the reactant molecules have more energy, so the rate of successful collision increases and the rate increases (the yield of forward reaction would decrease because the forward reaction is exo. and the equilibrium will shift to the left), increase in pressure also increases the rate of forward reaction because there are greater no. of moles of reactant on the left. Only D, which involves removing the ammonia as it is formed does not effect the rate, but only effects the yield. So the answer is D.

14) the equation would be-
3 Ba(NO3)2 + 10 Al -------> 5Al2O3 + 3BaO + 3N2

Mr of Ba(NO3)2 = 137 + 14*2 + 16* 6 = 261
looking at the stoichiometry of the reaction, 3mols Ba(NO3)2 gives 3 mols N2. So-
3*261 g Ba(NO3)2 = 3*24 dm^3 N2
.783 g Ba(NO3)2 = .072 dm^3 N2 = 72 cm^3 N2
Your answer is B.

Gemeaux solved 24.

you are really very helpful

 

[hela]

would you please solve

38 A number of alcohols with the formula C4H10O are separately oxidised. Using 70g of the alcohols
a 62% yield of organic product is achieved.
What mass of product could be obtained?
1 42.2g of butanone
2 51.6g of butanoic acid
3 51.6g of 2-methyl propanoic acid