Q.xvii. Calculate $\Delta H^0$ for the following reaction at 298 K $\mathrm{H_2B_4O_7(s) + H_2O}(l) \rightarrow \mathrm{4HBO_2\,(aq)}$ i. $\mathrm{2H_3BO_3(aq) \rightarrow B_2O_3(s) + 3H_2O}(l)$, $\Delta H^0$ = 14.4 kJ mol$^{-1}$ ii. $\mathrm{H_3BO_3(aq) \rightarrow HBO_2(aq) + H_2O,}(l)$ $\Delta H^0$ = -0.02 kJ mol$^{-1}$ iii. $\mathrm{H_2B_4O_7(s) \rightarrow 2P_2O_3(s) + H_2O}(l)$, $\Delta H^0$ =17.3 kJ mol$^{-1}$ Ans. : (- 11.58 kJ)

[!NOTE]
Equation iii is transcribed exactly as the textbook prints it, including the product "2P$_2$O$_3$(s)" -- an apparent misprint for 2B$_2$O$_3$(s) (boron trioxide), the only reading consistent with the problem's boron chemistry and with the printed answer. Equation ii's "H$_2$O,($l$)" comma placement is also the book's own.

Combine the three given combustion enthalpies via Hess's law to build the target formation reaction C(graphite) + 2H2(g) -> CH4(g) — the sum works out to -74.8 kJ/mol.

We want the enthalpy of formation of methane:

Target: C(graphite) + 2H2(g) -> CH4(g), Delta_f H = ?

Given combustion reactions:

(1) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), Delta H1 = -890.3 kJ/mol

(2) C(graphite) + O2(g) -> CO2(g), Delta H2 = -393.5 kJ/mol

(3) H2(g) + 1/2 O2(g) -> H2O(l), Delta H3 = -285.8 kJ/mol

By Hess's law, construct the target equation from these three: Target = (2) + 2x(3) - (1)

Check the atoms cancel correctly:

(2): C + O2 -> CO2

2x(3): 2H2 + O2 -> 2H2O

Sum: C + 2H2 + 2O2 -> CO2 + 2H2O …

Using Hess's law, the enthalpy of formation of methane equals the combustion enthalpy of carbon plus twice that of hydrogen, minus the combustion enthalpy of methane itself.

Target formation reaction: C(graphite) + 2 H2(g) -> CH4(g), delta Hf = ?

Given combustion enthalpies (delta Hc):

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), delta Hc(CH4) = -890.3 kJ/mol

C(graphite) + O2(g) -> CO2(g), delta Hc(C) = -393.5 kJ/mol

H2(g) + 1/2 O2(g) -> H2O(l), delta Hc(H2) = -285.8 kJ/mol

…

Hess's Law states that the total enthalpy change for a reaction is the same whether it occurs in a single step or in a series of steps, since enthalpy is a state function.

Statement: If a reaction takes place in several steps, the standard enthalpy change of the overall reaction is equal to the sum of the standard enthalpy changes of the individual steps, regardless of the path or number of steps taken to go from reactants to products.

…

Hess's Law: total enthalpy change of a reaction is independent of the path taken, depending only on the initial and final states.

Because enthalpy (H) is a state function, delta-H for converting reactants into products has one fixed value, regardless of whether the reaction happens directly in one step or through a series of intermediate steps. If a reaction can be written as the sum of two or more other reactions, its enthalpy change equals the sum of the enthalpy changes of those reactions. This principle lets chemists calculate delta-H for reactions that are hard to measure directly, by …

Hess's Law: total delta-H for a reaction is path-independent, depending only on initial and final states.

Step 1 -- statement: Hess's Law of Constant Heat Summation states that if a chemical reaction takes place in several steps, the standard enthalpy change (heat of reaction) for the overall reaction is equal to the sum of the standard enthalpy changes for the individual steps, and this total is the SAME regardless of the actual path or number of steps by which the reaction is carried out -- it depends only on the initial reactants and final products.

Step 2 -- basis: This follows because enthalpy (H) is a state function: delta-H for any process depends only on the initial and final states of the system, not on the route taken to get there. …