ALL CIE CHEMISTRY DOUBTS HERE !!

[cs]

Yes Deadly King, the examples are exactly what i asked, so water is "eliminated", and the other one is reduced. Thank you so much. +REP as usual. Great help

[Deadly_king]

Yes Deadly King, the examples are exactly what i asked, so water is "eliminated", and the other one is reduced. Thank you so much. +REP as usual. Great help

Yeah......most of the time, water is eliminated but in some case HCl or HBr may also be eliminated, depending on the structure of the reacting compound

Anytime dear 

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Guys, for AS chemistry in practicals, when they tell us to calculate answers in which we have to use q=mc(delta)t, i know that c=4.2, (delta)t is easy to calculate from our results, but m, i have no idea where to get it from :/ All i have are the volumes and not masses. Can i use volumes for the value of m in the equation?

If this isn't clear, i am basically referring to S10 P33 Chemistry (9701) question 1(c)(i).

Thankx

[ashish]

Guys, for AS chemistry in practicals, when they tell us to calculate answers in which we have to use q=mc(delta)t, i know that c=4.2, (delta)t is easy to calculate from our results, but m, i have no idea where to get it from :/ All i have are the volumes and not masses. Can i use volumes for the value of m in the equation?

If this isn't clear, i am basically referring to S10 P33 Chemistry (9701) question 1(c)(i).

Thankx

yes you must take the total volumes as M

[Deadly_king]

yes you must take the total volumes as M

Yupz.......m is taken as the total volume of solutions which you used when measuring the temperature change.

[$!$RatJumper$!$]

Thank you

[moon]

plz can somebody help me in calculation  ppr2   Nov.07 Q1 (e)
thanx in advance.

[$!$RatJumper$!$]

plz can somebody help me in calculation  ppr2   Nov.07 Q1 (e)
thanx in advance.

The general equation for enthalpy change of formation is: (delta)H(formation) = Products - Reactants

Now, the information they have given us for the reaction, is all in (delta)H(combustion). So what you have to do, is simply reverse the signs on each of the values to change them to the formation values. In this case, since they are all negative, we make them positive.

Once done that, now identify your products and your reactants.
Products: C2H4
Reactants: 2C, 2H2

Then plug in their values for formation into the equation above like this:

(delta)H(formation) = (–1411.0) - [(2 * –285.9) + (2 * –393.7)]

Therefore, (delta)H(formation) = 51.8 kJ mol–1

P.S We multiplied them by 2 because the equation said 2C and 2H2. The 2 represents the moles so we take them into account.

Hope this helped

[Deadly_king]

The general equation for enthalpy change of formation is: (delta)H(formation) = Products - Reactants

Now, the information they have given us for the reaction, is all in (delta)H(combustion). So what you have to do, is simply reverse the signs on each of the values to change them to the formation values. In this case, since they are all negative, we make them positive.

Once done that, now identify your products and your reactants.
Products: C2H4
Reactants: 2C, 2H2

Then plug in their values for formation into the equation above like this:

(delta)H(formation) = (–1411.0) - [(2 * –285.9) + (2 * –393.7)]

Therefore, (delta)H(formation) = 51.8 kJ mol–1

P.S We multiplied them by 2 because the equation said 2C and 2H2. The 2 represents the moles so we take them into account.

Hope this helped

Good work buddy 

+rep

[moon]

The general equation for enthalpy change of formation is: (delta)H(formation) = Products - Reactants

Now, the information they have given us for the reaction, is all in (delta)H(combustion). So what you have to do, is simply reverse the signs on each of the values to change them to the formation values. In this case, since they are all negative, we make them positive.

Once done that, now identify your products and your reactants.
Products: C2H4
Reactants: 2C, 2H2

Then plug in their values for formation into the equation above like this:

(delta)H(formation) = (–1411.0) - [(2 * –285.9) + (2 * –393.7)]

Therefore, (delta)H(formation) = 51.8 kJ mol–1

P.S We multiplied them by 2 because the equation said 2C and 2H2. The 2 represents the moles so we take them into account.

Hope this helped

I got the same answer -51.8 KJmol^-1 but the markscheme's answer is +51.8KJmol^-1 How come???

[Deadly_king]

I got the same answer -51.8 KJmol^-1 but the markscheme's answer is +51.8KJmol^-1 How come???

The answer is indeed +51.8 KJmol^-1

Delta H = 2(-393.7) + 2(-285.9) - (-1411) = +51.8


2C + 2H₂ ----> C₂H₄

Since the enthalpy changes of combustion has been given to you, you also need to write the equation relating each substance and its respective combustion.
1. C + O₂ ----> CO₂  (Enthalpy change of combustion of carbon or graphite)
2. H₂ + 0.5O₂ ----> H₂O (Enthalpy change of combustion of hydrogen)
3. C₂H₄ + 3O₂ ----> 2CO₂ + 2H₂O (Enthalpy change of combustion of butane)

As you may observe, the combustion of both hydrogen and carbon leads to carbon dioxide and hydrogen as well as the combustion of butane.

Hence, according to Hess's law, we may opt for another route in the conversion of carbon and hydrogen to butane.

Let's say we'll convert 2 moles of carbon and 2 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to ethane.

In practise these reactions are not feasible. But in theory we may opt for that method to find standard enthalpy changes of a particular reaction.

Let ^C be the enthalpy change of combustion of carbon.
So Enthalpy change of formation of butane = 2(^C) + 2(^H) - (^C₂H₄)

NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to butane whereas the enthalpy change of combustion of butane actually is the energy evolved when 1 mole of butane is completely burnt to carbon dioxide and water. In other words we are doing just the contrary. that's why we need to subtract it.

If you don't understand, let me know  

[$!$RatJumper$!$]

I got the same answer -51.8 KJmol^-1 but the markscheme's answer is +51.8KJmol^-1 How come???

You do get +51.8 when you do the calculation i mentioned

[Deadly_king]

You do get +51.8 when you do the calculation i mentioned

Yeah.......you've done it correctly br0

[$!$RatJumper$!$]

The answer is indeed +51.8 KJmol^-1

Delta H = 2(-393.7) + 2(-285.9) - (-1411) = +51.8


2C + 2H₂ ----> C₂H₄

Since the enthalpy changes of combustion has been given to you, you also need to write the equation relating each substance and its respective combustion.
1. C + O₂ ----> CO₂  (Enthalpy change of combustion of carbon or graphite)
2. H₂ + 0.5O₂ ----> H₂O (Enthalpy change of combustion of hydrogen)
3. C₂H₄ + 13/2 O₂ ----> 2CO₂ + 2H₂O (Enthalpy change of combustion of butane)

As you may observe, the combustion of both hydrogen and carbon leads to carbon dioxide and hydrogen as well as the combustion of butane.

Hence, according to Hess's law, we may opt for another route in the conversion of carbon and hydrogen to butane.

Let's say we'll convert 2 moles of carbon and 2 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to ethane.

In practise these reactions are not feasible. But in theory we may opt for that method to find standard enthalpy changes of a particular reaction.

Let ^C be the enthalpy change of combustion of carbon.
So Enthalpy change of formation of butane = 2(^C) + 2(^H) - (^C₂H₄)

NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to butane whereas the enthalpy change of combustion of butane actually is the energy evolved when 1 mole of butane is completely burnt to carbon dioxide and water. In other words we are doing just the contrary. that's why we need to subtract it.

If you don't understand, let me know 

good one

[moon]

I understood it now thanks Deadly_king and thanks 4 u too $!$RatJumper$!$