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Topic: Tricky Thermodynamics Problem. about calorimetry  (Read 16860 times)

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Offline DrCMS

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #15 on: January 08, 2011, 03:30:20 PM »
hey so how would you work out 1b i, ii and iii please? i do not know how to them. thank you

i) Use the reaction equation and the given heats of formation for the reactants and the products to calculate the heat of reaction.
ii) Use the reaction equation to calculate the volume change for the reaction which you can then use to calculate the work done by the reaction which with the heat of reaction from i above allows you to calculate U.
iii) Use the heats of vaporisation of water and octanol to modify the answer for i to account for the reaction occurring in the vapour phase at 1000K

Offline SAMMY

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #16 on: January 08, 2011, 08:00:28 PM »
Please can someone finish the last bits off:::

II know 3 moles produce 2 moles so how do we work out delta U ???

Offline DrCMS

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #17 on: January 09, 2011, 06:37:20 AM »
@ SAMMY and thegirlbehindtheveil

Go and look at the relationship between H and U.  Then look up how to calculate the work done by a system.

Offline SAMMY

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #18 on: January 09, 2011, 08:07:58 AM »
Yes but what equation do i use??

Delta H = Delta u + Delta (PV) is that correct??


Offline thegirlbehindtheveil

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #19 on: January 09, 2011, 08:13:24 AM »
if we are working out ∆U

the the equation we use is:

∆U= q + w (?)

as we know q is the heat and we have just worked that out and now we need to calculate the work done to find the value of
∆U

but heres where im confused

do we use

w= pex∆v

to work out work done???


Offline SAMMY

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #20 on: January 09, 2011, 08:19:50 AM »
Do we not use this formula??

Calculate Work, W. Use energy balance,
∆U = Q - W, Q = 0
n*Cv*∆T = -W


Calculate ∆U,
∆U = -W

Is this correct??

Offline thegirlbehindtheveil

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #21 on: January 09, 2011, 08:50:36 AM »
n*Cv*∆T = -W????/

i don't know thats why im confused which one do we use  ???

Offline SAMMY

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #22 on: January 09, 2011, 08:56:13 AM »
Does anyone know which formula we use?????


Offline SAMMY

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #23 on: January 09, 2011, 09:25:26 AM »
DrCMS please help

Offline thegirlbehindtheveil

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #24 on: January 09, 2011, 01:45:04 PM »
what equation do we use then???


because this is what ive got so far

 if we are working out ∆U

the the equation we use is:

∆U= q + w (?)

as we know q is the heat and we have just worked that out and now we need to calculate the work done to find the value of
∆U

but heres where im confused

do we use

w= pex∆v

to work out work done???

Offline thegirlbehindtheveil

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #25 on: January 09, 2011, 02:01:07 PM »
im still confused?

Offline SAMMY

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #26 on: January 09, 2011, 02:17:42 PM »
Me2..anyone out there that can help??

Offline DrCMS

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #27 on: January 09, 2011, 04:02:37 PM »
From wikipedia  http://en.wikipedia.org/wiki/Enthalpy

The enthalpy of a system is defined as:

    H = U + pV

where

    H is the enthalpy of the system (in joules),
    U is the internal energy of the system (in joules),
    p is the pressure at the boundary of the system and its environment, (in pascals), and
    V is the volume of the system, (in cubic meters).


For this question the equation to use is ΔH = ΔU + W = ΔU + Δ(pV).  Taking 1 mole of starting material you can calculate Δ(pV) and using the value for ΔH calculated from part i solve for ΔU.



Offline thegirlbehindtheveil

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #28 on: January 09, 2011, 04:39:35 PM »
Taking 1 mole of starting material- do you mean octane?

i know Δ (pV) = (pV) products − (pV) reactants.

is this right?

Offline SAMMY

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Re: Tricky Thermodynamics Problem. about calorimetry
« Reply #29 on: January 09, 2011, 04:54:34 PM »
Hi..Im struggling here...when you figure it then please can you post it on here..

I think its no of moles for reactant minus no of moles products?

does that make sense??

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