Manufacture of CCl4(l) includes the reaction

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Homework Statement


One step in the manufacture of CCl4(l) includes the reaction
3 Cl2(g) + CS2(l) [tex]\rightarrow[/tex] CCl4(l) + S2Cl2(l)

[tex]\Delta H^{\circ}_{f}\left(CS_{2(l)}\right) = 88 KJ[/tex]
[tex]\Delta H^{\circ}_{f}\left(CCl_{4(l)}\right) = -139 KJ[/tex]
[tex]\Delta H^{\circ}_{f}\left(S_{2}Cl_{2(l)}\right) = -60 KJ[/tex]

If the reaction takes place inside a reactor which is cooled by water at [tex]25^{\circ}C[/tex], how many kilograms of water at [tex]12^{\circ}C[/tex] must pass through the cooling coils of the reactor for each kilogram of [tex]Cl_{2}[/tex] reacting in order to keep the temperature at [tex]25^{\circ}C[/tex] ?

Homework Equations


The hard part is the language. What is it really asking ... in simple terms ? and this is this a modified version of a large calorimetry question? the calorimter is just supersized... ??

The Attempt at a Solution


All i can do is calculate the heat of raction which turns out to be -287 KJ
 
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Discussion



In simple terms, water at 120C is used to maintain the temperature of the system at 250C. How many kilograms of water per kilogram of Cl2 do you need to dissipate the heat generated in the reaction?
 


okay i calculate the amount of heat given off from the reaction which is -287 kJ now how do i calculate the mass of water @ 12 degrees celsius needed to keep the temperature of the system at 25 degrees celsius ?
 


You have a difference in temperature, an amount of energy/heat that is to be dissipated, can you think of a simple formula that relates these with the mass of water required per kilogram of Cl2 used?
 


WHAT THE...
lemme try this thang.

287=m(13)(4.18)
287=54.34m
m=5.282 kg ?
nah that don't make sense yo

The answer is 24.7 kg
 


What isn't making sense to you? How did you calculate the enthalpy of the reaction?
And please keep your language clean.
 
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okay you told me there's a simple formula that relates change in temperature and heat that dissapates and such right ? that we found out was the delta H = mc delta T and using that equation i plug in the numbers to find mass which gives me 5.282 kg which is wrong because the answer in the back of my page is 24.7 kg
 


Well, I get the same answer as given in the back of your book.
The unit for enthalpy of formation is usually Kj/mol. Convert this to Kj/kg. And remember you are calculating the mass of water per kilogram of Cl2
 


how did u do it then ? because I am not really clear when they mean mass of water per kilogram of Cl2 that reacted
 


Ok, let's start from the beginning,
Hess's Law states that if the equation is multiplied or divided by a certain constant, the enthalpy of the reaction is also multiplied or divided by the same constant.
So -287KJ/mol is the enthalpy when 3 moles of Cl2 reacts. What will be the enthalpy when 1 mole of Cl2 reacts?
Next find out how many kilograms of Cl2 are there in 1 mole of Cl2. Calculate the enthalpy of reaction in KJ/kg.
Now when you plug in these values in the formula,you'll get the answer, because all quantities have the same units.
 


287/3 = 95.667 Kj/mol of Cl2 that reacts. 1 mol of Cl2 has 71 g/mol -->0.071 kg/mol
 


The enthalpy of Cl2 is zero because its a diatomic molecule. Its an element So I am not quite sure where ur coming at
 


Th enthalpy of Cl2 is 0 because it is in its elemental state. True.
But when the reaction occurs, 3 moles of Cl2 reacts to give you -287KJ/mol of energy. So how much energy do you get if 1kg of Cl2 were to react? It's the enthalpy of reaction due to 1kg of Cl2 that is our concern, not the enthalpy of Cl2 (which is 0 here).
 


How?
3 moles of Cl2 give -287 Kj
1 mole of Cl2 gives -95.67 Kj
1 mole of Cl2= 71/1000 Kg of Cl2
or 71/1000 kg of Cl2 gives -95.67 Kj of energy.
1 kg of Cl2 gives 1347.46 KJ of energy
ie delta H= 1347.46 KJ/kg.