Calculation of formation heat given molar heat capacities and reaction enthalpy

Question

Consider the following reaction for the formation of liquid water at $\pu{298 K}:$

$$\ce{H2(g) + 1/2 O2(g) -> H2O (l)} \quad \Delta_\mathrm rH = \pu{-286 kJ/mol}$$

If molar heat capacities (at constant pressure) of $\ce{H2(g)},$ $\ce{O2(g)},$ $\ce{H2O(g)}$ are given as $39,$ $29$ and $\pu{75 J K^-1 mol^-1},$ calculate the heat of formation of $\ce{H2O(l)}$ at $\pu{348 K}.$ (Assume that molar heat capacities are independent of temperature in given range)

---

I tried solving this problem by taking $\Delta H = nC_p\Delta T$ and I added but I am getting a wrong answer. The data given is bit unusual (in thermodynamics we frequently work with enthalpy more). Any hints to the answer?

My problem is sourced from my prep material for JEE(Main).

Answer 1

You can apply the Hess law (Wikipedia, Libretexts), mere saying the change of enthalpy (or more generally any state variable) depends only on the starting and ending state, not on the path how the end state has been reached.

Applying it on the water formation enthalpies at different temperatures:

The enthalpy of liquid water formation at $\pu{298 K}$ is equal to

• the enthalpy change to warm up respective elements from $\pu{298 K}$ to $\pu{348 K}$
• plus the enthalpy of water formation at $\ce{348 K}$
• plus the enthalpy change to cool down water back from $\pu{348 K}$ to $\pu{298 K}$.

Calculation:

\begin{align} \Delta_\text{f} H_\ce{product}^{T_1} &= \Delta H_{\text{elements},T_1 \to T_2} + \Delta_\text{f} H_\ce{product}^{T_2} + \Delta H_{\text{product},T_2 \to T_1}\\ \Delta_\text{f} H_\ce{product}^{T_1} &= C_\text{elements} \cdot (T_2 - T_1) + \Delta_\text{f} H_\ce{product}^{T_2} + C_\text{product} (T_1 - T_2)\\ \Delta_\text{f} H_\ce{H2O(l)}^{298} &= (C_\ce{m,H2} + \frac 12 C_\ce{m,O2})(\pu{348 K} - \pu{298 K}) \\ &+ \Delta_\text{f} H_\ce{H2O(l)}^{348} + C_\ce{m,H2O(l)} (\pu{298 K} - \pu{348 K})\\ - \pu{286 kJ mol-1} &= (\pu{39 J K-1 mol-1} + \frac 12 \pu{29 J K-1 mol-1})(\pu{348 K} - \pu{298 K}) + \Delta_\text{f} H_\ce{H2O(l)}^{348} \\ &+ (\pu{75 J K-1 mol-1}) (\pu{298 K} - \pu{348 K})\\ - \pu{286 kJ mol-1} &= (\pu{53.5 J K-1 mol-1} - \pu{75 J K-1 mol-1})(\pu{50 K} ) + \Delta_\text{f} H_\ce{H2O(l)}^{348}\\ \Delta_\text{f} H_\ce{H2O(l)}^{348} &= -\pu{286 kJ mol-1} - \pu{1.075 kJ K-1 mol-1} \approx \pu{-287.1 kJ mol-1} \end{align}