The standard molar enthalpy of formation, of diborane cannot be determined direc

Question

The standard molar enthalpy of formation, of diborane cannot be determined directly because the compound cannot be prepared by reaction of boron and hydrogen. However, the value can be calculated. Calculate the standard enthalpy of formation of gaseous diborane(B2H6) using the following thermochemical information:

i) 4B(s) + 3O2(g) --> WB2O3(S) = -2509.1kJ

ii) 2H2(g) + O2(g) --> 2H2O(l) = - 571.7kJ

iii) B2H6(g) + 3O2(g) --> B2O3(s) + 3H2O(l) = -2147.5 kJ

The standard molar enthalpy of formation, of diborane cannot be determined directly because the compound cannot be prepared by reaction of boron and hydrogen. However, the value can be calculated. Calculate the standard enthalpy of formation of gaseous diborane(B2H6) using the following thermochemical information: i) 4B(s) + 3O2(g) --> WB2O3(S) = -2509.1kJ ii) 2H2(g) + O2(g) --> 2H2O(l) = - 571.7kJ iii) B2H6(g) + 3O2(g) --> B2O3(s) + 3H2O(l) = -2147.5 kJ

Explanation / Answer

ANSWER:

Devide Reaction (i) by 2 and rewrite equation (i) as

i) 2B(s) + 3O2 (g) ---------> B2O3 (s)   ?Ho = -2509.1 = -1254.55KJ

When we devide areaction by anumber we have to devide the ?Ho as well

Multiply reaction (ii) by 3 and rewrite as

ii) 3H2(g) + 3O2 (g) ------------> 3H2O (l)    ?Ho = - 3 X 571.7 = 857.55KJ

Add (i) and (ii) to get the following net equation

2B(s) + 3H2(g) + 3O2 (g) ---------> B2O3 (s) + 3H2O (l)   ?Ho = -2112.1KJ   ?Ho of the two reactions will add up.

Now subtract (iii) From the above net reaction.

2B(s) + 3H2(g) + 3O2 (g) ---------> B2O3 (s) + 3H2O (l)   ?Ho = -2112.1KJ

B2O3(s) + 3H2O(l) ---------------> B2H6 + 3O2 (g) ?Ho = +2147.5KJ

Net reaction is given below

2B(s) + 3H2(g) ---------------> B2H6 (g) ?Ho =-2112.1 + 2147.5KJ = 35.4KJ

Therefore ?Ho = 35.4KJ

NOTE: subtracting a reaction as

(I) simply do the reverse the reaction.

(II) change the sign of ?Ho as well

(III)Cancel the same reactants on opposite sides.

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