Question

Calculate the standard enthalpy of formation of gaseous diborane (B2H6)
using the following thermochemical information:
(a) 4B(s) + 3O2(g) ---> 2B2O3(s) ΔH° = −2509.1 kJ
(b) 2H2(g) + O2(g) ---> 2H2O(l) ΔH° = −571.7 kJ
(c) B2H6(g) + 3O2(g) ----> B2O3(s) + 3H2O(l) ΔH° = −2147.5 kJ

Answer 1

Answer:

hope it helps........

Answer 2

Given: 4 B (s) + 3 O₂ → 2 B₂O₃ (s)      ΔH° = -2509.1 kJ

           2 H₂ (g) + O₂ → 2 H₂O (l)        ΔH° = -571.7 kJ

           B₂H₆ (g) + 3 O₂ → B₂O₃ (s) + 3 H₂O (l)  ΔH° = -2147.5 kJ

To Find: ΔH° for 2 B(s) + 3 H₂(g) → B₂H₆ (g)

Solution:

• For solving this question we will apply Hess's Law which states:

Total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken.

Let us consider equation (a) first

1. 4 B(s) + 3 O₂ (g) → 2 B₂O₃ (s)    ΔH° = -2509.1 kJ

Dividing both sides of equation by 2 we get:

2 B (s) + $\frac{3}{2}$ O₂ (g) → B₂O₃ (s)    ΔH° = - 1254.55 kJ     (1)

Now adding reciprocal of equation (c) to equation (1)

2 B (s) + $\frac{3}{2}$ O₂ (g) → B₂O₃ (s)  ΔH° = -1254.55 kJ

B₂O₃ (s) + 3 H₂O (l) → B₂H₆ (g) + 3 O₂(g)   ΔH° = 2147.5 kJ  

⇒ 2 B (s) + 3 H₂O (l) → B₂H₆ (g) + $\frac{3}{2}$ O₂ (g)  ΔH° = 892.95 kJ       (2) (enthalpy is intensive property)

• Now we will multiply equation (2) by 2 and equation (b) by 3 to get proportionate coefficients and then add them.

4 B (s) + 6 H₂O (l) → 2 B₂H₆ (g) + 3 O₂ (g)        ΔH° = 892.95 ×2= 1785.9 kJ

6 H₂ (g) + 3 O₂ (g) → 6 H₂O (l)                         ΔH° = -571.7 ×3 = -1715.1 kJ

⇒ 4 B(s) + 6 H₂ (g) → 2 B₂H₆ (g)                    ΔH°= 70.8 kJ

Dividing above equation by 2 to get simplest form:

2 B(s) + 3 H₂(g) → B₂H₆ (g)                            ΔH° = 35.4 kJ

Therefore standard enthalpy of formation of gaseous diborane is 35.4 kJ.