chem question (1 Viewer)

[ali181]

does any one now how to solve q 24 from 2008 paper section 1

 

[Pwnage101]

which parts dont u get?

surely u have an idea on where to start for some parts...

 

[Fortify]

a)
Petrol - 5460/114 = 47.89KJ/g
Kerosene - 10000/210 = 47.62KJ/g
Hydrogen - 285/2 = 142.5KJ/g
Ethanol - 1370/46 = 29.78KJ/g

Therefore Hydrogen provides the greatest amount of energy per gram.

b)
Mass of Petrol combusted = 0.69 * 1000 * 80 = 55200
n(Petrol) = m/mm
n(Petrol) = 55200/114
n(Petrol) = 484.2105263moles

Heat of combustion = Delta H / n = 5460

Delta H = 5460 * n
Delta H = 5460 * 484.2105263
Delta H = 2643789.474KJ

Therefore energy produced by Petrol is 2643789.474KJ

c)
1 mol of hydrogen gas at 25 degress C at 100 kPa occupies 24.79 L and releases 285 kJ of energy

Therefore amount needed to release 22643789.474 kJ = (2643789.474)/285 = 9276.454294 moles

V (hydrogen gas) = 9276.454294 x 24.79 = 229963.3019 L = 2 x 10^5 L (1 sig fig)

I'm not sure if I'm 100% correct. Hope it helps.

 

[annabackwards]

a)
Petrol - 5460/114 = 47.89KJ/g
Kerosene - 10000/210 = 47.62KJ/g
Hydrogen - 285/2 = 142.5KJ/g
Ethanol - 1370/46 = 29.78KJ/g

Therefore Hydrogen provides the greatest amount of energy per gram.

b)
Mass of Petrol combusted = 0.69 * 1000 * 80 = 55200
n(Petrol) = m/mm
n(Petrol) = 55200/114
n(Petrol) = 484.2105263moles

Heat of combustion = Delta H / n = 5460

Delta H = 5460 * n
Delta H = 5460 * 484.2105263
Delta H = 2643789.474KJ

Therefore energy produced by Petrol is 2643789.474KJ

c)
1 mol of hydrogen gas at 25 degress C at 100 kPa occupies 24.79 L and releases 285 kJ of energy

Therefore amount needed to release 2.64 x 10^6 kJ = (2.64 x 10^6)/285 = 9276.4 moles

V (hydrogen gas) = 9276.4 x 24.79 = 2.3 x 10^5 L = 2 x 10^5 L (1 sig fig)

I'm not sure if I'm 100% correct. Hope it helps.

They're correct

 

[howlc]

is the reason u divide heat of combustion by average molar mass is so u get kj per gram?

hey soz haha but im really unsure why in part b you got:
heat of combustion= delta H/n =5460
i dont understand how you got this?

 

[howlc]

a)
Petrol - 5460/114 = 47.89KJ/g
Kerosene - 10000/210 = 47.62KJ/g
Hydrogen - 285/2 = 142.5KJ/g
Ethanol - 1370/46 = 29.78KJ/g

Therefore Hydrogen provides the greatest amount of energy per gram.

b)
Mass of Petrol combusted = 0.69 * 1000 * 80 = 55200
n(Petrol) = m/mm
n(Petrol) = 55200/114
n(Petrol) = 484.2105263moles

Heat of combustion = Delta H / n = 5460

Delta H = 5460 * n
Delta H = 5460 * 484.2105263
Delta H = 2643789.474KJ

Therefore energy produced by Petrol is 2643789.474KJ

c)
1 mol of hydrogen gas at 25 degress C at 100 kPa occupies 24.79 L and releases 285 kJ of energy

Therefore amount needed to release 22643789.474 kJ = (2643789.474)/285 = 9276.454294 moles

V (hydrogen gas) = 9276.454294 x 24.79 = 229963.3019 L = 2 x 10^5 L (1 sig fig)

I'm not sure if I'm 100% correct. Hope it helps.

Small error in part (c) Therefore amount needed to release 22643789.474 kJ = (2643789.474)/285 = 9276.454294 moles (the 22643789.474 should be 2643789.474 there was just an extra 2 just a typo =])