Transcript THERMO5.76___.ppt

5.76
REVERSE
THIS
REACTION
TO GET B2H6
AS A
REACTANT.
AND TO
CANCEL
B 2 O3
THERE ARE ONLY 2 B
IN THE NET, YOU
MUST MULTIPLY THIS
REACTION BY ½.
THERE ARE 3 H IN
THE NET, YOU MUST
MULTIPLY THIS
REACTION BY 3/2.
4 B + 3 O2  2 B 2 O 3
∆H1=(-2509.1kJ)
O2 + 2 H2  2 H2O
∆H2=(-571.7 kJ)
3O2 + B2H6  B2O3 + 3 H2O
2B + 3H2  B2H6
This is the formation, ∆Hf of B2H6, NET Rx
∆H3=-2147.5
kJ)
2 B + 3/2 O2  B2O3
∆H1=1/2(-2509.1kJ)
3/2O2 + 3 H2  3 H2O
∆H2=3/2(-571.7 kJ)
B2O3 + 3 H2O  B2H6 + 3O2
2B + 3H2  B2H6
∆H3=-(-2147.5 kJ)
∆Hf=+35.4 kJ
This is the formation, ∆Hf of B2H6, NET Rx