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Find the value of delta H net for the following equation: SnBr2(s) + TiCl4(l) -> TiBr2(s) + SnCl4(l) Use the following reaction...Asked by Hannah
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.
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.
Answers
Answered by
Declan
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 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!
Answered by
Declan
I mean an error on the second reaction.
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