1
$\begingroup$

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).

$\endgroup$
6
  • $\begingroup$ Have you tried using the heat capacity of liquid water over the 50 C temperature range. $\endgroup$ Dec 4, 2023 at 13:04
  • $\begingroup$ I am sorry for the late reply everyone, I was stuck in Cyclone Michaung for the past few days and could not check on anything. I had no internet for the past few days. $\endgroup$ Dec 6, 2023 at 2:10
  • $\begingroup$ Regarding @ChetMiller's comment: I have not done that. Maybe I'll try and get back to you $\endgroup$ Dec 6, 2023 at 2:11
  • $\begingroup$ @ChetMiller, nope I could not get the answer. $\endgroup$ Dec 7, 2023 at 2:34
  • 2
    $\begingroup$ It seems there is an error in the task description. $\ce{75 J K-1 mol-1}$ is the approximate molar heat capacity of $\ce{H2O(l)}$ ( $\pu{4.18 J K-1 g-1 \cdot 18 g mol-1 \approx 75.2 J K-1 mol-1}$ ). The molar or specific heat capacity of $\ce{H2O(g)}$ is about a half of that. Considering the reaction related to the considered formation enthalpy involved $\ce{H2O(l)}$, it all points out some error. $\endgroup$
    – Poutnik
    Dec 7, 2023 at 8:57

1 Answer 1

2
+50
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.