Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

answer:- \(\text{Given:}\) Given equations are, \[ \begin{align*} 1. & \quad 2H_{3}BO_{3(aq)} \rightarrow B_{2}O_{3(s)} + 3H_{2}O_{(l)}, \Delta_{r}H^{\circ} = +14.4 \, \text{kJ} \quad \text{(i)} \\ 2. & \quad H_{3}BO_{3(aq)} \rightarrow HBO_{2(aq)} + H_{2}O_{(l)}, \Delta_{r}H^{\circ} = -0.02 \, \text{kJ} \quad \text{(ii)} \\ 3. & \quad H_{2}B_{4}O_{7(s)} \rightarrow 2B_{2}O_{3(s)} + H_{2}O_{(l)}, \Delta_{r}H^{\circ} = +17.3 \, \text{kJ} \quad \text{(iii)} \end{align*} \] \(\text{To find:}\) The standard enthalpy of the given reaction (\(\Delta_{r}H^{\circ}\)) \(\text{Calculation:}\) Reverse equation (i) and multiply by 2, \[ 2B_{2}O_{3(s)} + 6H_{2}O_{(l)} \rightarrow 4H_{3}BO_{3(aq)}, \Delta_{r}H^{\circ} = -28.8 \, \text{kJ} \quad \text{(iv)} \] Multiply equation (ii) by 4, \[ 4H_{3}BO_{3(aq)} \rightarrow 4HBO_{2(aq)} + 4H_{2}O_{(l)}, \Delta_{r}H^{\circ} = -0.08 \, \text{kJ} \quad \text{(v)} \] Add equations (iv), (v), and (iii), \[ \begin{align*} 2B_{2}O_{3(s)} + 6H_{2}O_{(l)} & \rightarrow 4H_{3}BO_{3(aq)}, \Delta_{r}H^{\circ} = -28.8 \, \text{kJ} \\ 4H_{3}BO_{3(aq)} & \rightarrow 4HBO_{2(aq)} + 4H_{2}O_{(l)}, \Delta_{r}H^{\circ} = -0.08 \, \text{kJ} \\ H_{2}B_{4}O_{7(s)} & \rightarrow 2B_{2}O_{3(s)} + H_{2}O_{(l)}, \Delta_{r}H^{\circ} = +17.3 \, \text{kJ} \end{align*} \] \(\begin{align*} H_{2}B_{4}O_{7(s)} + H_{2}O_{(l)} & \rightarrow 4HBO_{2(aq)} \\ \Delta_{r}H^{\circ} & = -28.8 + (-0.08) + 17.3 = -11.58 \, \text{kJ} \end{align*}\) The standard enthalpy (\(\Delta_{r}H^{\circ}\)) of the given reaction is \(-11.58 \, \text{kJ}\).