Calculate the enthalpy of the reaction 4B(s)+3O2(g)?2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)?3O2(g)+B2H6(g), ?H?A=+2035 kJ 2B(s)+3H2(g)?B2H6(g), &nbsp

Calculate the enthalpy of the reaction

4B(s)+3O2(g)?2B2O3(s)

given the following pertinent information:

B2O3(s)+3H2O(g)?3O2(g)+B2H6(g), ?H?A=+2035 kJ

2B(s)+3H2(g)?B2H6(g), ?H?B=+36 kJ

H2(g)+1/2O2(g)?H2O(l), ?H?C=?285 kJ

H2O(l)?H2O(g), ?H?D=+44 kJ

Expert Solution

I assume that the reaction for which you need the enthalpy change is actually

4.B (s) + 3.O2 (g) ---> 2.B2O3 (s)

since there is no reference to boron in the gaseous state in any of the equations supplied.

The combination of reactions required is

6.O2 (g) + 2.B2H6 (g) ---> 2.B2O3 (s) + 6.H2O (g) . . ?H = - 4070 kJ
. . . 4.B (s) + 6.H2 (g) ---> 2.B2H6 ((g) . . . . . . . . . . ?H = + 72 kJ
. . . . . . . . . 6.H2O (l) ---> 6.H2 (g) + 3.O2 (g). . . . . .?H = + 1710 kJ
. . . . . . . . .6.H2O (g) ---> 6.H2O (l) . . . . . . . . . . . . ?H = - 264 kJ
--------------------------------------...
. . .4.B (s) + 3.O2 (g) ---> 2.B2O3 (s) . . . . . . . . . . . ?H = - 2552 kJ

First equation we flip backwards because we see that 2B2O3 is on products side and multiply by 2 making
2B2H6+3O2->2B2O3+6H2O(g) Since we flipped it we make the sign opposite and since we multiplied by 2 we mulitply by 2 so (-2035)x2
Reaction 2
Next equation multiply by 2 so we get 4B like in the original equation
4B+3H2->2B2H6 so all we do to the number is multiply by 2 so (36)x2
Reaction 3
Next equation there are different ways to view on how to do this
1) you can see that you need a total of 3O2 in the reactant side so you know you need 3O2 in the product side because in the reaction 1 we have 6O2 and once we cancel out at the end we will get 3O2 in the reactant side.
2)notice in reaction 1 that H2O(g) is in the product side so we know that H2O(l) is going to have to be opposite of it as seen in reaction 4
From this we flip it and multiply by 6
6H2O(l)->6H2+3O2 so we flip the sign and multiply by 6 (285)x6

Reaction 4 we Flip it to cancel out the H2O on both sides this is why the state is important and we have 6 of each so multiply by 6
(-44)x6

Now add up the total from the reactions
(-2035)x2=-4070
(36)x2= 72
(285)x6= 1710
(-44)x6= -264

-2552kJ is the answer or -2550kJ with sig figs

queen_honey_blossom answered 3 days ago

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