Edexcel CHEMISTRY DOUBTS!!!!

[nid404]

If the reaction is in the form

mA+nB--->pC+qD

R= k [A]^m [ B]ⁿ
a)R=k[NO2][CO]
b) Rate determining step is the slowest step in a mechanism.The second step uses the NO3 produced in the first step. So it is dependent on step 1 for it's NO3. hence Step 1 is the rate determining step.

[nid404]

there is a Question in A2 chemistry ann fullick book page 19 it says

Q: in a gas phase, molecule A decomposes to molecule B and C at high temperature. A chemist suspects that this reaction is first order with respect to A. in an experiment to explore the kinetics of the reaction, the data in the table were obtained for the decomposition at 800K.

then there is a table with time and partial pressure values for A .
example

T            A/KPA
0               1300
20             1051

and so on, then it says by using a graph find out if this reaction is really first order with respect to A.
for the graph i need concentration/time .  but they have given me pressure. so how will i make the graph ? or turn the pressure into conc. ?

Have the given the equation for the reaction?

[helai.j]

thanks for answering the Question and about that previous one i asked, i re-posted it in another post in the science section because no one was replying here and someone answered it for me.

thanks again <3

[Deadly_king]

Well i do NOT have a Q , cuz so far we did NOT use it at ALL .....so idk ....the only thing iwant to know is the new Equations like this and how to use them , Thanks

Well the equation you wrote is very rarely used.

Most of the time the one I gave you is used. SO no need to worry about the other one 

[nid404]

thanks for answering the Question and about that previous one i asked, i re-posted it in another post in the science section because no one was replying here and someone answered it for me.

thanks again <3

You're welcome

[The Golden Girl =D]

Well the equation you wrote is very rarely used.

Most of the time the one I gave you is used. SO no need to worry about the other one 

OMG .. Srsly that is a Relief for me !!

Jazaka Allah Kulla Kair bro

[Deadly_king]

OMG .. Srsly that is a Relief for me !!

Jazaka Allah Kulla Kair bro

You are welcome sis
We all form part of the huge SF family......so it's quite normal to help each other

[The Golden Girl =D]

You are welcome sis
We all form part of the huge SF family......so it's quite normal to help each other

so true

[The Golden Girl =D]

Hey guys i really need this thing it's really Urgent


it is asked to find  enthalpy change of formation for example : of butane the enthalpy changes given to us are of Combustion graphite , hydrogen and butane but when the teacher solved the Questiong he solved it as if it was a combustion reaction


Now i 'm really confused how can we do that !?

PLZ do Help me  

[Deadly_king]

Hey guys i really need this thing it's really Urgent


it is asked to find  enthalpy change of formation for example : of butane the enthalpy changes given to us are of Combustion graphite , hydrogen and butane but when the teacher solved the Questiong he solved it as if it was a combustion reaction


Now i 'm really confused how can we do that !?

PLZ do Help me  

Hmm.....it's not that difficult. Let me get it clear for you.

First you need to write equation indicating the formation of butane.
4C + 5H₂ ----> C₄H₁₀

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. C₄H₁₀ + 13/2 O₂ ----> 4CO₂ + 5H₂O (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 4 moles of carbon and 5 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to butane.

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 = 4(^C) + 5(^H) - (^C₄H₁₀)

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

[nid404]

I was just about to post...Good job DK 

[The Golden Girl =D]

Hmm.....it's not that difficult. Let me get it clear for you.

First you need to write equation indicating the formation of butane.
C + H₂ ----> C₄H₁₀

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. C₄H₁₀ + 13/2 O₂ ----> 4CO₂ + 5H₂O (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 4 moles of carbon and 5 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to butane.

In practise these reactions are not feasible. But in theory we may opt for that methos 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 = 4(^C) + 5(^H) - (^C₄H₁₀)

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

I do Understand it but the true Question that is going on and on in my Mind is that is some Questions we are given the Enthalpy changes of Combustion and are asked to find the enthalpy of change of formation of the product

So is that the only way i can get formation is by finding it in combustion (cuz that's the only route we got obvsly)

Ex of my Qs is when they ask for enthalpy of formation of Methane wen they have given us enthalpy change of combustion of it's compouds , another example would be enthalpy of change of fromation of Ammonia ...etc ?

[The Golden Girl =D]

I do Understand it but the true Question that is going on and on in my Mind is that is some Questions we are given the Enthalpy changes of Combustion and are asked to find the enthalpy of change of formation of the product

So is that the only way i can get formation is by finding it in combustion (cuz that's the only route we got obvsly)

Ex of my Qs is when they ask for enthalpy of formation of Methane wen they have given us enthalpy change of combustion of it's compouds , another example would be enthalpy of change of fromation of Ammonia ...etc ?

Is it really Urgent so Plz do help I have to go cuz of tuition , will be back perhaps late at night if i could iA . 

[Deadly_king]

I do Understand it but the true Question that is going on and on in my Mind is that is some Questions we are given the Enthalpy changes of Combustion and are asked to find the enthalpy of change of formation of the product

So is that the only way i can get formation is by finding it in combustion (cuz that's the only route we got obvsly)

Ex of my Qs is when they ask for enthalpy of formation of Methane wen they have given us enthalpy change of combustion of it's compouds , another example would be enthalpy of change of fromation of Ammonia ...etc ?

Hmm..........yeah, this is the only way to find enthalpy changes of formation when you are given enthalpy changes of combustion.

1. Methane is a hydrocarbon, so same principle used for butane is applied here.

2. For ammonia, you should first write its equation of formation, which is :
N₂ + 3H₂ ----> 2NH₃

Therefore the question needs to provide you with the enthalpy changes of combustion of ammonia, nitrogen and hydrogen as well.

Next, you write the equations indicating the combustion of all the reactants and products separately.
1. N₂ + 2O₂ ---> 2NO₂
2. 2H₂ + O₂ ---> 2H₂O
3. 2NH₃ + 7/2 O₂ ----> 2NO₂ + 3H₂O

You'll note that the combustion of nitrogen and hydrogen together will form the products of the combustion of ammonia.

So we'll take another route to convert nitrogen and hydrogen to ammonia. First we burn both reactants to obtain nitrogen dioxide and water. Then we convert the latter to ammonia.

Let ^H be the enthalpy change of combustion of hydrogen
So enthalpy change of formation of ammonia will be : ^N + 3(^H) - 2(^NH₃)

NOTE : Only one mole of nitrogen and 3 moles of hydrogen are required to form 2 moles of ammonia according to the equation. The number of moles are used as coefficients of enthalpy changes as well as indicated by the bold numbers.
I have applied the same theory as before

[The Golden Girl =D]

Hmm..........yeah, this is the only way to find enthalpy changes of formation when you are given enthalpy changes of combustion.

1. Methane is a hydrocarbon, so same principle used for butane is applied here.

2. For ammonia, you should first write its equation of formation, which is :
N₂ + 3H₂ ----> 2NH₃

Therefore the question needs to provide you with the enthalpy changes of combustion of ammonia, nitrogen and hydrogen as well.

Next, you write the equations indicating the combustion of all the reactants and products separately.
1. N₂ + 2O₂ ---> 2NO₂
2. 2H₂ + O₂ ---> 2H₂O
3. 2NH₃ + 7/2 O₂ ----> 2NO₂ + 3H₂O

You'll note that the combustion of nitrogen and hydrogen together will form the products of the combustion of ammonia.

So we'll take another route to convert nitrogen and hydrogen to ammonia. First we burn both reactants to obtain nitrogen dioxide and water. Then we convert the latter to ammonia.

Let ^H be the enthalpy change of combustion of hydrogen
So enthalpy change of formation of ammonia will be : ^N + 3(^H) - 2(^NH₃)

NOTE : Only one mole of nitrogen and 3 moles of hydrogen are required to form 2 moles of ammonia according to the equation. The number of moles are used as coefficients of enthalpy changes as well as indicated by the bold numbers.
I have applied the same theory as before

Jazaka allahu kulla kari dude for helping me out