For all of you who did AQA gcse chemistry paper 1 higher separate, what did you get for the bond energy calculation??

I got 1033 but idk

I got 1033 but idk

I can't remember exactly my answer but I believe that firstly you had to convert the total final bond of 2219 into a negative as it was an exothermic reaction (which has a negative energy change). Then you had to calculate the bond energies of the molecules and complete the rearranging to find X. I got a number around 380 although I cannot be sure that this is the correct answer. I think that it is unfair that they should give us the bond energy as a positive integer and not as negative as it was not made entirely clear that you must convert it to a negative. I hope this can help.

Original post by jar3d08

i got 392.375, could be very wrong tho

Thanks but I Doubt it is wrong, lots of people got something similar to that dw

Original post by jar3d08

i got 392.375, could be very wrong tho

That is what I got and my friend too!

Original post by Jasmine6789

For all of you who did AQA gcse chemistry paper 1 higher separate, what did you get for the bond energy calculation??

I got 1033 but idk

I got 1033 but idk

392.375

Original post by Jasmine6789

For all of you who did AQA gcse chemistry paper 1 higher separate, what did you get for the bond energy calculation??

I got 1033 but idk

I got 1033 but idk

I got 349 it was something like 8000 - 2300 then you take away the enthalpy for the other two parts and divided by 6 I think to get the answer

Original post by jar3d08

i got 392.375, could be very wrong tho

I got this too!

392 (3sf) is what I got

the real enthalpy change is -2219, not 2219 (exothermic = negative enthalpy change)

Propane + 5 oxygen -> 3 CO2 + 4H2O

[8(c-h) + 2(c-c) ] + 5 (o=o) -> 3(2(c=o)) + 4(2(o-h))

8(X) + 2(347) + 5(498) -> 6(805) + 8(464)

(8X + 694 + 2490) - (4830 + 3712) = -2219

8X + 3184 - 8542 = -2219

8X = 3139

X = 392.375 = 392 (3sf)

the real enthalpy change is -2219, not 2219 (exothermic = negative enthalpy change)

Propane + 5 oxygen -> 3 CO2 + 4H2O

[8(c-h) + 2(c-c) ] + 5 (o=o) -> 3(2(c=o)) + 4(2(o-h))

8(X) + 2(347) + 5(498) -> 6(805) + 8(464)

(8X + 694 + 2490) - (4830 + 3712) = -2219

8X + 3184 - 8542 = -2219

8X = 3139

X = 392.375 = 392 (3sf)

Original post by zmp903

392 (3sf) is what I got

the real enthalpy change is -2219, not 2219 (exothermic = negative enthalpy change)

Propane + 5 oxygen -> 3 CO2 + 4H2O

[8(c-h) + 2(c-c) ] + 5 (o=o) -> 3(2(c=o)) + 4(2(o-h))

8(X) + 2(347) + 5(498) -> 6(805) + 8(464)

(8X + 694 + 2490) - (4830 + 3712) = -2219

8X + 3184 - 8542 = -2219

8X = 3139

X = 392.375 = 392 (3sf)

the real enthalpy change is -2219, not 2219 (exothermic = negative enthalpy change)

Propane + 5 oxygen -> 3 CO2 + 4H2O

[8(c-h) + 2(c-c) ] + 5 (o=o) -> 3(2(c=o)) + 4(2(o-h))

8(X) + 2(347) + 5(498) -> 6(805) + 8(464)

(8X + 694 + 2490) - (4830 + 3712) = -2219

8X + 3184 - 8542 = -2219

8X = 3139

X = 392.375 = 392 (3sf)

nahhh, i put -2219 at first but then i changed it back to normal 🥲

Original post by abdullahawrr

I got 349 it was something like 8000 - 2300 then you take away the enthalpy for the other two parts and divided by 6 I think to get the answer

divide by 8 not 6 there were 8 C-H bonds

Original post by zmp903

392 (3sf) is what I got

the real enthalpy change is -2219, not 2219 (exothermic = negative enthalpy change)

Propane + 5 oxygen -> 3 CO2 + 4H2O

[8(c-h) + 2(c-c) ] + 5 (o=o) -> 3(2(c=o)) + 4(2(o-h))

8(X) + 2(347) + 5(498) -> 6(805) + 8(464)

(8X + 694 + 2490) - (4830 + 3712) = -2219

8X + 3184 - 8542 = -2219

8X = 3139

X = 392.375 = 392 (3sf)

the real enthalpy change is -2219, not 2219 (exothermic = negative enthalpy change)

Propane + 5 oxygen -> 3 CO2 + 4H2O

[8(c-h) + 2(c-c) ] + 5 (o=o) -> 3(2(c=o)) + 4(2(o-h))

8(X) + 2(347) + 5(498) -> 6(805) + 8(464)

(8X + 694 + 2490) - (4830 + 3712) = -2219

8X + 3184 - 8542 = -2219

8X = 3139

X = 392.375 = 392 (3sf)

did you have to round to sig fig? would i lsoe marks for not doing so? the only reason i didnt was because i wanted to energy amounts to be exact - since no energy can be created or destroyed

Original post by aanaya

did you have to round to sig fig? would i lsoe marks for not doing so? the only reason i didnt was because i wanted to energy amounts to be exact - since no energy can be created or destroyed

You definitely didn't have to - I just did it since the other values were to 3sf

Original post by zmp903

You definitely didn't have to - I just did it since the other values were to 3sf

Original post by InmanAerospace

To be fair sometimes they are annoying and take marks away for it- since you don’t actually change the number e.g if it rounded to 393 I don’t think they take marks away but rounded it correctly does sometimes mean you immediately get full marks for the question whereas if you don’t then you still can get full marks it jus relies on you working so more likely for an examiner to make a mistake

i wrote the decimal and then the rounding so they should count either one

Original post by InmanAerospace

To be fair sometimes they are annoying and take marks away for it- since you don’t actually change the number e.g if it rounded to 393 I don’t think they take marks away but rounded it correctly does sometimes mean you immediately get full marks for the question whereas if you don’t then you still can get full marks it jus relies on you working so more likely for an examiner to make a mistake

yes i just didnt wanna take the risk, as techinaclly if you round it, you're cutitng off a few important kJ of energy and as energy cannot be destroyed or created, i thought it wouldn't be appropriate. should be okay anyways if working is there

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