0.600mol H₂OSTARTING AMOUNT0.600 mol H₂O x-2035 kJ3 mol B₂H6X= 407 KJ-20353kJWhat is the change in enthalpy when 0.600 moles of H₂O are produced according tothe following balanced chemical reacton:mol B₂H627.67=ADD FACTORx( )407B₂H6(g) +407g B₂O3 mol B₂O369.6232.003g B₂2H6kJO₂(g) → B₂O3(s) + 3 H₂O(g)ΔΗ = -2035| κυDELETEA-2035-407g O₂J1=ANSWERkJ1.22 x 10³ -1.22 x 10³0.600mol B₂H6g H₂O3RESETJmol O₂18.02mol H₂O

Here the given reaction is ; B2H6(g) + 3O2 --> B2O3(g) + 3H2O (g ) Here boron hydride react…

Answered: What is the change in enthalpy when…

What is the change in enthalpy when 0.600 moles of H₂O are produced according to the following balanced chemical reacton: B₂H6(g) + 3 O₂(g) → B₂O3(s) + ΔΗ = -2035 KJ 3 H₂O(g)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Exergonic Reaction

The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.

Question

0.600
mol H₂O
STARTING AMOUNT
0.600 mol H₂O x
-2035 kJ
3 mol B₂H6
X
= 407 KJ
-2035
3
kJ
What is the change in enthalpy when 0.600 moles of H₂O are produced according to
the following balanced chemical reacton:
mol B₂H6
27.67
=
ADD FACTOR
x( )
407
B₂H6(g) +
407
g B₂O3 mol B₂O3
69.62
32.00
3
g B₂2H6
kJ
O₂(g) → B₂O3(s) + 3 H₂O(g)
ΔΗ = -2035| κυ
DELETE
A
-2035
-407
g O₂
J
1
=
ANSWER
kJ
1.22 x 10³ -1.22 x 10³
0.600
mol B₂H6
g H₂O
3
RESET
J
mol O₂
18.02
mol H₂O

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