chem question

ali181 · 7 Apr 2009, 12:40 AM

You can hide this advertisement by registering.

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

**Pwnage101** · 7 Apr 2009, 4:20 PM

which parts dont u get?

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

Fortify · 8 Apr 2009, 11:36 AM

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 · 13 Apr 2009, 6:34 PM

Quote:

Originally Posted by 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 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 · 28 Jul 2009, 12:56 PM

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 · 28 Jul 2009, 1:04 PM

Quote:

Originally Posted by 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.

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 =])

7 Apr 2009, 12:40 AM	#1 (permalink)
ali181 New Member HSC: 2009 Gender: Male Join Date: Mar 2008 Posts: 19 Last Activity: 10 Mar 2010, 9:18 PM	chem question You can hide this advertisement by registering. does any one now how to solve q 24 from 2008 paper section 1

7 Apr 2009, 4:20 PM	#2 (permalink)
Pwnage101 Moderator HSC: N/A Gender: Undisclosed Location: in Pursuit of Happiness. Join Date: May 2008 Posts: 1,453 Last Activity: Yesterday, 11:35 PM	Re: chem question which parts dont u get? surely u have an idea on where to start for some parts...

8 Apr 2009, 11:36 AM	#3 (permalink)
Fortify ♪웨딩드레스 HSC: 2009 Gender: Male Location: In a box under your bed . Join Date: Mar 2007 Posts: 1,319 Last Activity: 13 Mar 2010, 1:20 PM	Re: chem question 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. __________________ Last edited by Fortify; 13 Apr 2009 at 6:34 PM.

28 Jul 2009, 12:56 PM	#5 (permalink)
howlc New Member HSC: 2010 Gender: Male Join Date: Nov 2008 Posts: 5 Last Activity: 30 Oct 2009, 8:47 AM	Re: chem question 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? Last edited by howlc; 28 Jul 2009 at 1:48 PM.

Thread Tools
Show Printable Version Email this Page
Rate This Thread
Rate This Thread: