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###### chem help

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12 years ago

http://www.chemsheets.co.uk/AS/CHEM2.1/AS.CHEM2.1.004.pdf

dont give answer but how do i do 1)D

dont give answer but how do i do 1)D

Reply 1

12 years ago

Original post by chignesh10

1. Write out the equation for the combustion of methane.

2. Add up the bond energies of the reactants and then subtract those of the products.

Reply 2

12 years ago

Original post by charco

1. Write out the equation for the combustion of methane.

2. Add up the bond energies of the reactants and then subtract those of the products.

2. Add up the bond energies of the reactants and then subtract those of the products.

Sorry bout this, I finally understood it...but I am now stuck on 4, if i can do 4 i can do 5 and 6. so can i have help on 4?

Reply 3

12 years ago

Original post by chignesh10

Sorry bout this, I finally understood it...but I am now stuck on 4, if i can do 4 i can do 5 and 6. so can i have help on 4?

Calculate the average S-F bond energy in SF

S(s) → S(g) ΔHº = 223 kJ mol-1

The average S-F bond energy is one sixth of the process:

SF

Solution

----------

You are given:

1. F

and you are told that:

2. S(s) → S(g) ΔHº = 223 kJ mol-1

Enthalpy of formation is represented by the equation:

3. S(s) + 3F

So you use equations 1, 2 & 3 to construct the equation you need.

This is really application of Hess's law.

You can't go from:

SF

directly, so you have to construct a way starting with SF

Reverse equation 3:

SF

Then use equation 2.

SF

Then use 3x equation 1: 3F

SF

So, you have got an alternative route from SF

Then just divide the sum by 6 ... chachi

Reply 4

12 years ago

Original post by charco

Calculate the average S-F bond energy in SF_{6}(g) using the bond data in the table, the ΔHf of SF_{6}(g) which is -1100 kJ mol-1, and

S(s) → S(g) ΔHº = 223 kJ mol-1

The average S-F bond energy is one sixth of the process:

SF_{6}(g) --> S(g) + 6F(g)

Solution

----------

You are given:

1. F_{2} --> 2F(g) in the table (+158 kJ)

and you are told that:

2. S(s) → S(g) ΔHº = 223 kJ mol-1

Enthalpy of formation is represented by the equation:

3. S(s) + 3F_{2}(g) --> SF_{6}(g) ........... -1100 kJ

So you use equations 1, 2 & 3 to construct the equation you need.

This is really application of Hess's law.

You can't go from:

SF_{6}(g) --> S(g) + 6F(g)

directly, so you have to construct a way starting with SF_{6}(g).

Reverse equation 3:

SF_{6}(g) --> S(s) + 3F_{2}(g) ..... +1100 kJ

Then use equation 2.

SF_{6}(g) --> S(s) + 3F_{2}(g) --- > S(s) → S(g) + 3F_{2}(g) ....... ΔHº = (223 + 1100) kJ

Then use 3x equation 1: 3F_{2} --> 6F(g) ........ +474 kJ

SF_{6}(g) --> S(s) + 3F_{2}(g) --- > S(s) → S(g) + 3F_{2}(g) --> S(g) + 6F(g) ....... ΔHº = (223 + 1100 + 474 ) kJ

So, you have got an alternative route from SF_{6}(g) to S(g) + 6F(g)

Then just divide the sum by 6 ... chachi

S(s) → S(g) ΔHº = 223 kJ mol-1

The average S-F bond energy is one sixth of the process:

SF

Solution

----------

You are given:

1. F

and you are told that:

2. S(s) → S(g) ΔHº = 223 kJ mol-1

Enthalpy of formation is represented by the equation:

3. S(s) + 3F

So you use equations 1, 2 & 3 to construct the equation you need.

This is really application of Hess's law.

You can't go from:

SF

directly, so you have to construct a way starting with SF

Reverse equation 3:

SF

Then use equation 2.

SF

Then use 3x equation 1: 3F

SF

So, you have got an alternative route from SF

Then just divide the sum by 6 ... chachi

Ah ok thanks a lot of that method! helps a lot! So using your method is the answer to 5., 1206? if u can tlel me?

Reply 5

12 years ago

Original post by chignesh10

Ah ok thanks a lot of that method! helps a lot! So using your method is the answer to 5., 1206? if u can tlel me?

I don't get that (and your answer isn't very realistic) ...

... show your working

(edited 12 years ago)

Reply 6

12 years ago

Original post by charco

I don't get that (and your answer isn't very realistic) ...

... show your working

... show your working

715+185+ 4(hydrogen bond enthalpy) = the bond enthalpy of propyne

Reply 7

12 years ago

Original post by chignesh10

715+185+ 4(hydrogen bond enthalpy) = the bond enthalpy of propyne

You haven't thought it out very carefully:

Calculate the C≡C bond enthalpy in the gas propyne using the bond enthalpy values in the table above and the enthalpy changes for the reactions shown below.

C(s) → C(g) ........... ΔHº = +715 kJ mol-1

CH3C≡CH(g)......... ΔHºf = +185 kJ mol-1

You only want the C≡C bond enthalpy, which can be obtained by subtracting all of the other bond enthalpies from the total bond enthalpy of the process:

1. CH

So, THIS is the equation you want to get to...

You know that:

2. 3C(s) + 2H

So the reverse is:

CH

Now 3 x C(s) → C(g) ......... 2145 kJ gives...

CH

And now you have to change hydrogen molecules to atoms using the hydrogen bond enthalpy: H

CH

Now you have to subtract all of the bonds in propyne that you don't want = (4 x C-H) + (1 x C-C) = 1648 + 348 = 1996

Hence C≡C = 2832 - 1996 = 836 kJ

Reply 8

3 years ago

Original post by charco

Calculate the average S-F bond energy in SF_{6}(g) using the bond data in the table, the ΔHf of SF_{6}(g) which is -1100 kJ mol-1, and

S(s) → S(g) ΔHº = 223 kJ mol-1

The average S-F bond energy is one sixth of the process:

SF_{6}(g) --> S(g) + 6F(g)

Solution

----------

You are given:

1. F_{2} --> 2F(g) in the table (+158 kJ)

and you are told that:

2. S(s) → S(g) ΔHº = 223 kJ mol-1

Enthalpy of formation is represented by the equation:

3. S(s) + 3F_{2}(g) --> SF_{6}(g) ........... -1100 kJ

So you use equations 1, 2 & 3 to construct the equation you need.

This is really application of Hess's law.

You can't go from:

SF_{6}(g) --> S(g) + 6F(g)

directly, so you have to construct a way starting with SF_{6}(g).

Reverse equation 3:

SF_{6}(g) --> S(s) + 3F_{2}(g) ..... +1100 kJ

Then use equation 2.

SF_{6}(g) --> S(s) + 3F_{2}(g) --- > S(s) → S(g) + 3F_{2}(g) ....... ΔHº = (223 + 1100) kJ

Then use 3x equation 1: 3F_{2} --> 6F(g) ........ +474 kJ

SF_{6}(g) --> S(s) + 3F_{2}(g) --- > S(s) → S(g) + 3F_{2}(g) --> S(g) + 6F(g) ....... ΔHº = (223 + 1100 + 474 ) kJ

So, you have got an alternative route from SF_{6}(g) to S(g) + 6F(g)

Then just divide the sum by 6 ... chachi

S(s) → S(g) ΔHº = 223 kJ mol-1

The average S-F bond energy is one sixth of the process:

SF

Solution

----------

You are given:

1. F

and you are told that:

2. S(s) → S(g) ΔHº = 223 kJ mol-1

Enthalpy of formation is represented by the equation:

3. S(s) + 3F

So you use equations 1, 2 & 3 to construct the equation you need.

This is really application of Hess's law.

You can't go from:

SF

directly, so you have to construct a way starting with SF

Reverse equation 3:

SF

Then use equation 2.

SF

Then use 3x equation 1: 3F

SF

So, you have got an alternative route from SF

Then just divide the sum by 6 ... chachi

omg charco, you helped me so much thank you

Reply 9

3 years ago

Original post by reinainoue

omg charco, you helped me so much thank you

This was 8 years ago!

Reply 10

1 year ago

10 years ago now! And still doing the same questions

Can you please help me with 6?? Thanks so much!!

Can you please help me with 6?? Thanks so much!!

Reply 11

11 months ago

I think I can help with this

C(s) ----> C(g) ΔH = 715 kJmol-1

Br2(l) -----> Br2(g) ΔH = 15 kJmol-1

Bond Enthalpies:

C-H = 412 kJmol-1

Br-Br = 276 kJmol-1

H-H = 436 kJmol-1

2C(g) + 3H2(g) + Br2(g) -----> 2CH3Br

(Energy required to change the state of 2 moles of carbon) + (bond enthalpy of two moles of hydrogen gas) + (energy required to change the state of liquid bromine to gas bromine AND the bond enthalpy of Br2)* ------> (bond enthalpies of two moles of three carbon-hydrogen bonds) + (bond enthalpies of two moles of one carbon-bromine bond)

(2 x 715) + (3 x 436) + (193 + 15) -----> 2 x [(3 x 412) + (276)]

2946 - 3024 = -78 kJmol-1

* I am having trouble with this question myself - I don't know whether I should use both of these values for bromine, seeing bromine changes shape and has a bond enthalpy as gaseous bromine is diatomic, or I need to use one value (if so, I don't know which value)

Please let me know what you think, I just thought I'd share my process

C(s) ----> C(g) ΔH = 715 kJmol-1

Br2(l) -----> Br2(g) ΔH = 15 kJmol-1

Bond Enthalpies:

C-H = 412 kJmol-1

Br-Br = 276 kJmol-1

H-H = 436 kJmol-1

2C(g) + 3H2(g) + Br2(g) -----> 2CH3Br

(Energy required to change the state of 2 moles of carbon) + (bond enthalpy of two moles of hydrogen gas) + (energy required to change the state of liquid bromine to gas bromine AND the bond enthalpy of Br2)* ------> (bond enthalpies of two moles of three carbon-hydrogen bonds) + (bond enthalpies of two moles of one carbon-bromine bond)

(2 x 715) + (3 x 436) + (193 + 15) -----> 2 x [(3 x 412) + (276)]

2946 - 3024 = -78 kJmol-1

* I am having trouble with this question myself - I don't know whether I should use both of these values for bromine, seeing bromine changes shape and has a bond enthalpy as gaseous bromine is diatomic, or I need to use one value (if so, I don't know which value)

Please let me know what you think, I just thought I'd share my process

Reply 12

11 months ago

dude this thread is awesome

Reply 13

11 months ago

Original post by reinainoue

dude this thread is awesome

Yeah my teacher uses chemsheets too but doesn't upload the answers, and it's hard to find workings online!

Original post by drunkcucombers

I think I can help with this

C(s) ----> C(g) ΔH = 715 kJmol-1

Br2(l) -----> Br2(g) ΔH = 15 kJmol-1

Bond Enthalpies:

C-H = 412 kJmol-1

Br-Br = 276 kJmol-1

H-H = 436 kJmol-1

2C(g) + 3H2(g) + Br2(g) -----> 2CH3Br

(Energy required to change the state of 2 moles of carbon) + (bond enthalpy of two moles of hydrogen gas) + (energy required to change the state of liquid bromine to gas bromine AND the bond enthalpy of Br2)* ------> (bond enthalpies of two moles of three carbon-hydrogen bonds) + (bond enthalpies of two moles of one carbon-bromine bond)

(2 x 715) + (3 x 436) + (193 + 15) -----> 2 x [(3 x 412) + (276)]

2946 - 3024 = -78 kJmol-1

* I am having trouble with this question myself - I don't know whether I should use both of these values for bromine, seeing bromine changes shape and has a bond enthalpy as gaseous bromine is diatomic, or I need to use one value (if so, I don't know which value)

Please let me know what you think, I just thought I'd share my process

C(s) ----> C(g) ΔH = 715 kJmol-1

Br2(l) -----> Br2(g) ΔH = 15 kJmol-1

Bond Enthalpies:

C-H = 412 kJmol-1

Br-Br = 276 kJmol-1

H-H = 436 kJmol-1

2C(g) + 3H2(g) + Br2(g) -----> 2CH3Br

(Energy required to change the state of 2 moles of carbon) + (bond enthalpy of two moles of hydrogen gas) + (energy required to change the state of liquid bromine to gas bromine AND the bond enthalpy of Br2)* ------> (bond enthalpies of two moles of three carbon-hydrogen bonds) + (bond enthalpies of two moles of one carbon-bromine bond)

(2 x 715) + (3 x 436) + (193 + 15) -----> 2 x [(3 x 412) + (276)]

2946 - 3024 = -78 kJmol-1

* I am having trouble with this question myself - I don't know whether I should use both of these values for bromine, seeing bromine changes shape and has a bond enthalpy as gaseous bromine is diatomic, or I need to use one value (if so, I don't know which value)

Please let me know what you think, I just thought I'd share my process

i did the same method almost but i got half your answer (-39 kjmol-1) bc i halved everything - i think when you calculate enthalpy of formation you should only produce 1 mol of desired product i.e. the thing being formed so you half all your values

not too sure about my answer but it would make sense to include both values for bromine since the energy has to be taken into account to convert it into a gas from standard states

Reply 15

9 months ago

omg yeah I totally forgot about that, thank you!

Original post by masa_

i did the same method almost but i got half your answer (-39 kjmol-1) bc i halved everything - i think when you calculate enthalpy of formation you should only produce 1 mol of desired product i.e. the thing being formed so you half all your values

not too sure about my answer but it would make sense to include both values for bromine since the energy has to be taken into account to convert it into a gas from standard states

not too sure about my answer but it would make sense to include both values for bromine since the energy has to be taken into account to convert it into a gas from standard states

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