Question

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), ΔH∘A=+2035 kJ
2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36 kJ
H2(g)+12O2(g)→H2O(l), ΔH∘C=−285 kJ
H2O(l)→H2O(g), ΔH∘D=+44 kJ

Answer 1

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$    $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$     $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$    $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$    $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$    $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole