Question

The standard enthalpy change of combustion [to $C{O}_2\left(g\right)and{H}_{2}O\left(l\right)$at 25⁰ Cof the organic liquid cyclohexane, C₆H₁₂(l), is -3731.7kJmol^-1. Determine the $\Delta H_f^0$of C₆H₁₂(l).

Short answer

The standard enthalpy value of the reactant$\left(\Delta H_f^{°}\right)of{C}_6{H}_{12}is-58.86kJ/mol$

Step-by-step solution

Given data

$\Delta H^{\circ }$(Standard enthalpy change) value of Cyclohexane) C₆H₁₂(l) is -3731.7kJmol^-1.

$C_6{H}_{1}0\left(l\right)+17/2O_2\left(g\right)\to 6C{O}_2\left(g\right)+5H_{2}O\left(l\right).$

Concept of Enthalpy

Enthalpy change is 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$\Delta H$.

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, $\Delta H_f^o$represent the standard enthalpy value.

localid="1663737494734" $\Delta H_f^{o}of{O}_{2}is0.$

$\Delta H_f^{o}of{H}_{2}Ois-285.83kJmo{l}^{-1}.$

$\Delta H_f^{o}ofC{O}_{2}is-393.51kJmo{l}^{-1}$

$\Delta H^{°}=-3731.7kJmo{l}^{-1}$

.

The given expression is localid="1663738633260" $C_6{H}_{1}0\left(l\right)+17/2O_2\left(g\right)\to 6C{O}_2\left(g\right)+5H_{2}O\left(l\right).$

Substitute these values in equation (1).

$\Delta H^{\circ }=6\Delta H_f^0\left[C{O}_2\right(g\left)\right]+5\Delta H_f^0\left[H_{2}O\right(l\left)\right]-\left\{\Delta H_f^a\right[C_6{H}_{10}\left(l\right)]+\frac{17}{2}\times \Delta H_f^0[O_2\left(g\right)\left]\right\}$

$$\begin{gathered} -3731.7kJmo{l}^{-1}=6\times \left(-393.51kJmo{l}^{-1}\right)+5\times \left(-285.83kJmo{l}^{-1}\right)-\left\{\Delta H_f^a\right[C_6{H}_{10}\left(l\right)]+17/2\times 0]\left\} \\ \Delta H_f^a\right[C_6{H}_{10}\left(l\right)]=-2361.06kJmo{l}^{-1}-1429.5kJmo{l}^{-1}+3731.7kJmo{l^{-1}}^{} \end{gathered}$$

Therefore,$=-58.86\frac{k]}{mol}$.