Question

Calculate ∆H0for the following reaction at 298 K H2B4O7(s) + H2O(l) ---> 4HBO2 (aq) i. 2H3BO3(aq) --->B2O3(s) + 3H2O(l) ∆H0 = 14.4 kJ mol-1 ii. H3BO3(aq) --->HBO2(aq) + H2O(l) ∆H0 = -0.02 kJ mol-1 iii. H2B4O7(s) --->2B2O3(s) + H2O(l) ∆H0 =17.3 kJ mol-1

Answer 1

Answer:

The question was unable to match

Answer 2

The required equation is :$H_{2}B_{4} O_{7} (aq)+H_{2}O(l) \rightarrow 4HBO_{2}(aq)$

The given equations are:

$i)2H_{3}BO_{3} (aq)\rightarrow B_{2}O_{3}(s)+3H_{2}O(l)\ \Delta H \textdegree=14.4kJmol^{-1}$

$ii)H_{3}BO_{3} (aq)\rightarrow HBO_{2}(aq) +H_{2}O(l)\ \Delta H \textdegree=-0.02kJmol^{-1}$

$iii)H_{2}B_{4}O_{7}(s) \rightarrow 2B_{2}O_{3}(s)+H_{2}O(l)\ \Delta H \textdegree=17.3kJmol^{-1}$

The above equation can be obtained by:$(iii)+4 (ii)-2(i)$

The $\Delta H$ for the reaction would be⇒$\Delta H \textdegree (iii)+4\Delta H \textdegree (ii)-2\Delta H \textdegree (i)$

$\rightarrow [17.3+4(-0.02)-2(14.4)]kJmol^{-1} =-11.58kJmol^{-1}$

Th $\Delta H$ of the reaction is -11.58 kJmol⁻¹.