Find the value of delta H net for the following equation:
SnBr2(s) + TiCl4(l) -> TiBr2(s) + SnCl4(l)
Use the following reactions to help solve for the net value:
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 = +78kJ
Everything that I have posted is exactly what was stated in the problem so I guess everything is in kJ.
You have an error on the product side of the third reaction. The reaction produces SnCl4.
I was wondering what was wrong Haha.
Anyway...
Hess's Law...Oh Joy!
If we look at the given reactions we see that we must flip three of the reactions to produce the overall reaction that you stated above. So you just fiddle around with it to cancel out compounds. And you end up flipping the first, third, and sixth reaction. You change the delta H value to the opposite sign and add all the enthalphys. So you get.... +73.8 Kj for your answer!
I mean an error on the second reaction.
To find the value of ΔH net for the given equation, we need to use Hess's Law.
First, let's write the overall equation for the reaction:
SnBr2(s) + TiCl4(l) → TiBr2(s) + SnCl4(l)
Now, let's use the given reactions and their enthalpy changes to manipulate their equations in order to cancel out the intermediates:
1) SnCl2(s) + TiBr2(s) → SnBr2(s) + TiCl2(s) ΔH = +4.2 kJ
2) SnCl2(s) + Cl2(g) → SnCl4(l) ΔH = -195 kJ
3) TiCl2(s) + Cl2(g) → TiCl4(l) ΔH = -273 kJ
TiCl4(l) → TiCl2(s) + Cl2(g) ΔH = +273 kJ (inverted sign)
Next, we manipulate the given reactions to cancel out the intermediates:
Multiply reaction 1 by 1 to have the same amount of SnBr2 on both sides.
Multiply reaction 2 by 1 to have the same amount of SnCl2 on both sides.
Add reaction 1 and reaction 2 to eliminate SnCl2 and SnBr2.
SnCl2(s) + Cl2(g) → SnCl4(l) ΔH = -195 kJ
SnBr2(s) + Cl2(g) → SnCl4(l) ΔH = +4.2 kJ
------------------
2SnCl2(s) + 2Cl2(g) → 2SnCl4(l) ΔH = -190.8 kJ
Now, substitute reaction 3 with the modified reaction above:
2SnCl2(s) + 2Cl2(g) → 2SnCl4(l) ΔH = -190.8 kJ
TiCl2(s) + 2Cl2(g) → TiCl4(l) ΔH = -273 kJ
SnCl2(s) + TiCl4(l) → TiCl2(s) + SnCl4(l) ΔH_net = -190.8 kJ + (-273 kJ)
ΔH_net = -463.8 kJ
Therefore, the value of ΔH_net for the given equation is -463.8 kJ.
To find the value of delta H net for the given equation, we can use Hess's Law and the given reactions.
Hess's Law states that if a reaction can be expressed as the sum of two or more reactions, the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual reactions.
Start with the given equation:
SnBr2(s) + TiCl4(l) -> TiBr2(s) + SnCl4(l)
We can break this reaction down into the following steps:
Step 1:
SnBr2(s) + TiCl4(l) -> SnCl2(s) + TiBr2(s)
This is given in reaction 1:
Delta H1 = +4.2 kJ
Step 2:
SnCl2(s) + TiBr2(s) -> TiCl2(s) + SnBr2(s)
To get this step, reverse reaction 1:
Delta H2 = - Delta H1 = - (+4.2 kJ) = -4.2 kJ
Step 3:
SnCl2(s) + Cl2(g) -> SnCl4(l)
This is given in reaction 4:
Delta H3 = -195 kJ
Step 4:
TiCl2(s) + Cl2(g) -> TiCl4(l)
This is given in reaction 3:
Delta H4 = -273 kJ
Now, we can add these steps together to get the overall reaction:
Step 2 + Step 3 + Step 4:
Delta Hnet = Delta H2 + Delta H3 + Delta H4
= (-4.2 kJ) + (-195 kJ) + (-273 kJ)
= -472.2 kJ
Therefore, the value of delta Hnet for the given equation is -472.2 kJ.