Chemistry: Post your doubts here!

[anastasia grey113]

Ans :21/9

Look here's the thing
The ratio we already have for tertiary : primary is 21 : 1.
However, as you can see the number of atoms of hydrogen which can be substituted by Cl to make primary halogenoalkane are 9
while in tertiary there is only 1.
Thus, 9 kinds of primary products can be formed but only one kind of tertiary
thus the ratio of products formed tertiary : primary would be 1 : 9
thus the chances of primary product being formed are 9 times greater than tertiary and so products made will be 9 times higher in concentration IF RATE OF REACTION IS NOT CONSIDERED
however, as the reaction RATE is slower, the ratio will be somewhat different

this can be calculated by multiplication
21 x 1 : 1 x 9
thus the ratio would be 21 : 9

 

[hamadh]

Sulfur can be oxidised in two ways.
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 2S(s) + 3O2(g) → 2SO3(g) ∆H o = –791.4 kJ mol–1

Sulfur trioxide can be made from sulfur dioxide and oxygen.

2SO2(g) + O2(g) → 2SO3(g)

What is the standard enthalpy change for this reaction? A –1384.4 kJ mol–1 B –989.8 kJ mol–1 C –494.9 kJ mol–1 D –198.4 kJ mol–1
please somebody explain me this twist in the hess's law and the point of doing it. i would be grateful.

 

[Hamnah Zahoor]

Sulfur can be oxidised in two ways.
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 2S(s) + 3O2(g) → 2SO3(g) ∆H o = –791.4 kJ mol–1

Sulfur trioxide can be made from sulfur dioxide and oxygen.

2SO2(g) + O2(g) → 2SO3(g)

What is the standard enthalpy change for this reaction? A –1384.4 kJ mol–1 B –989.8 kJ mol–1 C –494.9 kJ mol–1 D –198.4 kJ mol–1
please somebody explain me this twist in the hess's law and the point of doing it. i would be grateful.

First of all reverse the first reaction and multiply it by 2 as we do in writing equations for contact process this will also change the sign of enthalpy change required for this reaction i.e 2(296.5)=+593 make an overall equation of the both equations given to get 2SO2(g) + O2(g) → 2SO3(g)
then
-791.4+593= -198.4 thus D is correct option.

 

[anastasia grey113]

Sulfur can be oxidised in two ways.
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 2S(s) + 3O2(g) → 2SO3(g) ∆H o = –791.4 kJ mol–1

Sulfur trioxide can be made from sulfur dioxide and oxygen.

2SO2(g) + O2(g) → 2SO3(g)

What is the standard enthalpy change for this reaction? A –1384.4 kJ mol–1 B –989.8 kJ mol–1 C –494.9 kJ mol–1 D –198.4 kJ mol–1
please somebody explain me this twist in the hess's law and the point of doing it. i would be grateful.

The reaction
S(s) + O2(g) → SO2(g) ∆H o = –296.5 kJ mol–1 is basically the Δ H f ∘ of SO2 while
S(s) + 1.5O2(g) → 2 SO3(g) ∆H o = –791.4 kJ mol–1 is of (2 x Δ H f ∘ SO3).
You can see in the 2nd reaction, I have multiplied the enthalpy by 2 to find enthalpy for formation of only two moles of SO2.
because as you can see, in the given reaction the product is made from 2 moles of SO2.

This is the Hess's law diagram.
The arrows both moving upwards are showing the two reactions stated.
As you can see, the two moles of SO3 can be made by two routes. Either directly, or by doing the 1st reaction followed by oxidation of SO2.
The indirect route is shown with the green arrow.
So the equation for the reaction 2SO2 + O2 -> 2SO3 will be as follows.
2 x Δ H f ∘ of SO2 + ΔHr⊖ = 2 x Δ H f ∘ SO3

 

[hamadh]

First of all reverse the first reaction and multiply it by 2 as we do in writing equations for contact process this will also change the sign of enthalpy change required for this reaction i.e 2(296.5)=+593 make an overall equation of the both equations given to get 2SO2(g) + O2(g) → 2SO3(g)
then
-791.4+593= -198.4 thus D is correct option.

Thank you so much for the reply. I wanted to ask why don't we multiply enthalpy change for sulfurtrioxide with 2 .

 

[anastasia grey113]

Thank you so much for the reply. I wanted to ask why don't we multiply enthalpy change for sulfurtrioxide with 2 .

cuz the reaction is like 2SO2 + O" -> 2SO3
so that means the enthalpy change already shows 2 moles being made n in oxidation the same no. of moles are made.
But in SO2 formation, only one mole is made in the reaction.
But in oxidation of SO2, 2 moles are used. So to equal the moles, it was multiplied.

 

[hamadh]

a big big big thanks to you from my side. but isn't it a bit misleading to write -781.4 per mole. They could have just written enthalpy change = 781.4 . i hope i am not annoying you. this per mole thing is what is causing me the whole confusion anastasia grey113

 

[anastasia grey113]

a big big big thanks to you from my side. but isn't it a bit misleading to write -781.4 per mole. They could have just written enthalpy change = 781.4 . i hope i am not annoying you. this per mole thing is what is causing me the whole confusion anastasia grey113

its okay
okay let me explain
We have one goal and that is MAKING 2 MOLES OF SO3.
let's start from the oxidation reaction
to make 2 moles of SO3 you need 2 moles of SO2 right?
So let's say we are given 1 mole of O2 and one mole of S and we make one mole of SO2 and we get an energy change of -791.4
Now we are asked to make 2 moles of SO3 from the product we got.
But we only have 1 mole of SO2 when we actually need 2 so do you think it's possible for the reaction to occur with just 1 mole of SO2?
According to the equation we need 2 moles of it.
So we will need to do the reaction twice to have two moles of SO2 for oxidation.
So we do the reaction twice. And the enthalpy becomes -791.4 x 2 right?

But for the formation of SO3 (i'm talking abt formation) 2 moles are already being made.

Now to answer your question.
PER MOLE MEANS the enthalpy change when one mole of the product is made.
Let's suppose we have 2 moles of S and two moles of O2, then obvio 2 moles of SO2 will be formed and the enthalpy change will be different right?
Even though the reaction is same.
So to clarify, they wrote per mole.

 

[Mstudent]

Hey Guys! Long time no see. I hope your studies are going great this year!

Could an A level student help me out here. Is there any easy way to memorize all the Transition metal complex reactions regarding Cobalt, Manganese, copper etc., and their products?

 

[anastasia grey113]

Hey Guys! Long time no see. I hope your studies are going great this year!

Could an A level student help me out here. Is there any easy way to memorize all the Transition metal complex reactions regarding Cobalt, Manganese, copper etc., and their products?

You don't need to revise all of them.Just go with cobalt and copper.

 

[Mstudent]

You don't need to revise all of them.Just go with cobalt and copper.

Thanks, but is there any easy way to memorize the products and the reactions for cobalt and Copper ? Im having a really hard time with Ligands this year!

 

[anastasia grey113]

Thanks, but is there any easy way to memorize

Not really. You just have to memorise them as they are. There's no concept.
I hope you remember the colours of Cu on adding NH3.
They will help you memorise the colours of two reactions and that's that.
Otherwise it's all memorisation.

 

[Holmes]

Hey Guys! Long time no see. I hope your studies are going great this year!

Could an A level student help me out here. Is there any easy way to memorize all the Transition metal complex reactions regarding Cobalt, Manganese, copper etc., and their products?

Hey where have you been all the time. Seeing you almost after a year though Welcome back to the XPC!
Yes unfortunately you will have to memorize these reactions.

 

[Hinafatima]

May june 2016/12, q#7 why the answer is D instead of C, any comments please?

 

[Hamnah Zahoor]

May june 2016/12, q#7 why the answer is D instead of C, any comments please?

Are you sure you are substituting the values correctly.

 

[mehran46]

T is an alcohol, CxHyO. A gaseous sample of T occupied a volume of 20cm3 at 120°C and 100kPa. The sample was completely burned in 200cm3 of oxygen (an excess). The final volume, measured under the same conditions as the gaseous sample, was 250cm3 . Under these conditions, all water present is vaporised. Removal of the water vapour from the gaseous mixture decreased the volume to 170cm3 . Treating the remaining gaseous mixture with concentrated alkali, to absorb carbon dioxide, decreased the volume to 110cm3 . The equation for the complete combustion of T can be represented as shown. CxHyO + zO2 xCO2 + y/ 2 H2O (i) Use the data given to calculate the value of x ..


the answer states that x= 3
How?????!!!!!!!!

Thx for help

 

[anastasia grey113]

T is an alcohol, CxHyO. A gaseous sample of T occupied a volume of 20cm3 at 120°C and 100kPa. The sample was completely burned in 200cm3 of oxygen (an excess). The final volume, measured under the same conditions as the gaseous sample, was 250cm3 . Under these conditions, all water present is vaporised. Removal of the water vapour from the gaseous mixture decreased the volume to 170cm3 . Treating the remaining gaseous mixture with concentrated alkali, to absorb carbon dioxide, decreased the volume to 110cm3 . The equation for the complete combustion of T can be represented as shown. CxHyO + zO2 xCO2 + y/ 2 H2O (i) Use the data given to calculate the value of x ..


the answer states that x= 3
How?????!!!!!!!!

Thx for help

Well first write down the volumes of each gas.
1- The initial mixture with H2O + CO2 + O2 has a total volume of 250cm3 but removing H2O gave a volume of 170cm3.
So VOLUME OF H2O = 250cm3 - 170cm3 = 80cm3.

2- The remaining mixture of CO2 + O2 = 170cm3. Removing CO2 with alkali gives 110cm3.
So VOLUME OF CO2 = 170cm3 - 110cm3 = 60cm3.

3- The final mixture only contains O2 which is 110cm3. So out of 2o0cm3, 110cm3 did not react.
so VOLUME of reacted O2 = 200cm3 - 110cm3 = 90cm3.

4- Volume of alcohol = 20cm3.

Now since all are gases (remember water is actually water vapour so its a gas too) , we can take their ratios to find no. of moles.
so
CxHyO + zO2 -> xCO2 + y/2H2O
20 : 90 : 60 : 80
Now in order to find the value of x, we are supposed to take only 1 mole of the alcohol. So we will reduce the ratios down in such a way that the no. of moles of alcohol are 1.
So it'll be like
CxHyO : O2 : CO2 : H2O
1 : 4.5 : 3 : 4
The no. of moles of CO2 r 3 so x = 3.

 

[Fareez]

Hi. I just want to ask.

In the data booklet there's so many standard electrode potential equation specifically for Mn04-. So which one should I use?

 

[anastasia grey113]

Hi. I just want to ask.

In the data booklet there's so many standard electrode potential equation specifically for Mn04-. So which one should I use?

When the question is related to reduction of acidified KMnO4, use
MnO4 – + 8H+ + 5e ---> Mn2+ + 4H2O

 

[Thought blocker]

Hi. I just want to ask.

In the data booklet there's so many standard electrode potential equation specifically for Mn04-. So which one should I use?

These are the three electrode potentials given to us in data booklet for MnO4-.
You can see on RHS of each equation, there are different species mentioned, namely MnO4(-2), MnO2 and Mn(+2), so now depending on to the data given to you in the question, you use respective electrode potentials.. I hope that made some sense.