Question

The standard enthalpy change of combustion [to (CO₂(g) and H₂O(l)at 25⁰ Cof the organic liquid cyclohexane, C₆H₁₂(l), is -3923.7kJmol^-1. Determine the $\Delta H_f^0$of C₆H₁₂(l). Use data from Appendix D.

Short answer

$\Delta H_f^{°}Valueof{C}_6{H}_{12}is-152.3kJmo{l}^{-1}.$

Step-by-step solution

Given data

$\Delta H^{°}ValueofCyclohexane{C}_6{H}_{12}\left(l\right)is-3923.7kJmo{l}^{-1}.$

Concept of Enthalpy

Enthalpy change is the amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as the difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in a chemical reaction.

It is represented by $\Delta H$.

Step3: Calculate the value of ΔHf°of C6H12

The enthalpy change is calculated as:

$\Delta H^{°}=n\Delta H_f^o\left(Products\right)-n\Delta H_f^o\left(Reactants\right)$ ...... (1)

Where, $\Delta H_f^{\circ }$represent the standard enthalpy value.

$$\begin{gathered} \Delta H_f^{o}of{O}_{2}is0. \\ \Delta H_f^{o}of{H}_{2}Ois-285.83kJmo{l}^{-1}. \\ \Delta H_f^{o}ofC{O}_{2}is-393.51kJmo{l}^{-1}. \\ \Delta H^{°}=-3923.7kJmolJ{m}^{-1}. \end{gathered}$$

The given expression is $C_6{H}_{1}2\left(l\right)+9O_2\left(g\right)\to 6C{O}_2\left(g\right)+6H_{2}O\left(l\right)$.

Substitute these values in equation (1).

$-3923.7kJmo{l}^{-1}=6\Delta H_f^0\left[C{O}_2\right(g\left)\right]+6\Delta H_f^0\left[H_{2}O\right(l\left)\right]-\left\{\Delta H_f^a\right[C_6{H}_{12}\left(l\right)]+9\times \Delta H_f^0[O_2\left(g\right)\left]\right\}$$$\begin{gathered} -3923.7kJmo{l}^{-1}=6\times (-393.51kJmo{l}^{-1})+6\times (-285.3kJmo{l}^{-1})-\left\{\Delta H_f^a\right[C_6{H}_{12}\left(l\right)]+9\times 0\} \\ =3923.7kJmo{l}^{-1}-2361.06kJmo{l}^{-1}-1714.98kJmo{l}^{-1}-\left\{\Delta H_f^a\right[C_6{H}_{12}\left(l\right)\left] \\ \right\{\Delta H_f^a\left[C_6{H}_{12}\right(l\left)\right]=3923.7kJmo{l}^{-1}-2361.06kJmo{l}^{-1}-1714.98kJmo{l}^{-1} \end{gathered}$$

$Therefor4e,=-152.3kJmo{l}^{-1}.$