## 8.45

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Joined: Wed Sep 21, 2016 2:57 pm

### 8.45

8.45 Carbon disulfi de can be prepared from coke (an impure form of carbon) and elemental sulfur: 4 C1s2  S81s2¡4 CS21l2 ¢H°  358.8 kJ How much heat is absorbed in the reaction of 1.25 mol S8 at constant pressure? (b) Calculate the heat absorbed in the reaction of 197 g of carbon with an excess of sulfur.

I am so confused as to how to do the first part of this problem. I saw the solution but it doesnt make sense as to how they got the equation to solve part a.

Jeffrey_Huang_2K
Posts: 11
Joined: Fri Jul 22, 2016 3:00 am

### Re: 8.45

The reaction enthalpy of deltaH = +358.8 kJ specifically describes the situation where one mol of S8 reacts. When 1.25 mol is reacted, then the reaction enthalpy (notice it is positive so heat is absorbed) is also multiplied by this new amount of reactant.

(358.8 kJ)(1.5 mol S8) = 448.5 kJ

GabrielaGutierrez2A
Posts: 11
Joined: Wed Sep 21, 2016 2:58 pm

### Re: 8.45

Does anyone know how to do part b?

stephanieyang_3F
Posts: 62
Joined: Wed Sep 21, 2016 2:55 pm

### Re: 8.45

Part B is basically asking how much heat is absorbed when that much amount of carbon reacts with S8. Since sulfur is in excess, the amount of carbon in the reaction goes to completion, which means it will determine the reaction's yield of heat. Since the original reaction said 358 kJ is associated with 4 moles of C reacting with 1 mole of S8, (also means 358 kJ is absorbed by that many moles of reactants reacting together), you'd find the amount of heat absorbed when you input 197 g of C by multiplying the moles of C (197g C to moles) by the reaction's (358kJ/4moles of C). The 358kJ is the amount of heat absorbed when there's 4 moles of C and you wanna find out the amount of heat absorbed when there's 197 g of C.

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