Find the value of delta H net for the following equation:
SnBr2(s) + TiCl4(l) -> TiBr2(s) + SnCl4(l)
I know that I already posted this once but I am still lost on what to do.
I know that TiCl4=-804.2 and SnCl4=-511.3 but I thought that the values for SnBr2 is 0 as well as for TiBr2.
Chemistry - DrBob222, Wednesday, February 1, 2012 at 10:00pm
SnBr2 is not zero.
TiBr2 is not zero.
I don't know what they are but they aren't zero. The only DHf values that are zero are elements which appear in an equation in their normal state. O2(g), H2(g), etc.
Chemistry - Hannah, Thursday, February 2, 2012 at 11:55pm
How should I go about solving this problem? Am I missing information?
I don't know what to tell you. I can't find that in any of the texts I have nor can I find it on the web. I'm sure it's somewhere on the web; I've just not found it.
Ok well thank you for trying!
I don't know if this will help but I realized there was more information for the problem.
1) SnCl2(s) + TiBr2(s) -> SnBr2(s) + Ticl2(s) Delta H=+4.2
2) SnCl2(s) + Cl2(g) -> TiCl4(l) deltaH=-195
3) TiCl2(s) + Cl2(g) ->TiCl4(l) deltaH=-273
SnCl2(s) + Cl2(g) -> SnCl4(l) deltaH=-195
TiCl4(l) -> TiCl2(s) + Cl2(g) deltaH=+273
SnCl2(s) + TiCl4(l) -> TiCl2(s) + SnCl4(l) deltaH net = 78
The directions for the original question that I posted said to use the reactions above to find the net value for the new equation.
Hannah, will you PLEASE check your post. I've already spent (wasted may be a better word) hours on this thing only to find that you didn't post the entire problem. I MAY (and I emphasize MAY) be able to do something with that; however, I do not believe equation 2. Finally you don't have any units. Start over, post the entire problem and question (as a new post), and all of the data, including units. Proof it thoroughly before posting. I assume "net value" means to find the delta H for the reaction. Is that right?
To solve this problem, you need to use the concept of Hess's Law. Hess's Law states that the enthalpy change for a reaction is the same whether it occurs in one step or multiple steps.
First, you need to determine the enthalpy changes for the individual reactions involved:
1. SnBr2(s) → Sn(s) + Br2(g)
2. TiBr2(s) → Ti(s) + Br2(g)
3. TiCl4(l) → Ti(s) + 4Cl(g)
4. SnCl4(l) → Sn(s) + 4Cl(g)
The enthalpy changes for reactions 1 and 2 can be obtained from standard enthalpy of formation values (ΔHf). However, you mentioned that you do not have these values, so I'll assume you are given the ΔHf values for SnBr2 and TiBr2.
Let's denote the enthalpy changes for the reactions as follows:
ΔH1: SnBr2(s) → Sn(s) + Br2(g)
ΔH2: TiBr2(s) → Ti(s) + Br2(g)
ΔH3: TiCl4(l) → Ti(s) + 4Cl(g)
ΔH4: SnCl4(l) → Sn(s) + 4Cl(g)
Next, you need to determine the net equation by combining the individual reactions in a way that cancels out the intermediate compounds (Sn(s) and Br2(g)):
ΔHnet: SnBr2(s) + TiCl4(l) → TiBr2(s) + SnCl4(l)
Since ΔH1 and ΔH2 both produce Br2(g) and ΔH3 and ΔH4 both consume Cl(g), you can add these reactions together to get the net equation.
ΔHnet = ΔH1 + ΔH3 + (ΔH2 + ΔH4)
Now, substitute the given enthalpy change values for TiCl4 and SnCl4:
ΔHnet = ΔH1 + (-804.2 kJ) + (ΔH2 + (-511.3 kJ))
Keep in mind that the ΔH values are usually given in kJ/mol, so you'll need to multiply them by the appropriate stoichiometric coefficients to obtain the overall equation's ΔH value.
Unfortunately, without the values for ΔH1 and ΔH2, we cannot provide a specific numerical answer. You'll need to refer to a reliable source such as a textbook or online database to find the specific ΔHf values for SnBr2 and TiBr2.