Question

The standard enthalpy of formation of $C{F}_{2}ClC{F}_{2}Cl\left(g\right)$ $\left({\Delta }_f{H}_{300}^{\circ }\right)$ at $1\text{bar}$ and $300\text{K}$ from its constituent elements in their standard states is $-900{\text{kJ mol}}^{-1}$. (Given, $R=8.3{\text{J K}}^{-1}{\text{mol}}^{-1}$)
The standard internal energy of formation $\left({\Delta }_f^{\circ }{U}_{300}\right)$ at the same pressure and temperature is:

• A. $-905.02{\text{kJ mol}}^{-1}$
• B. $-895.02{\text{kJ mol}}^{-1}$
• C. $895.02{\text{kJ mol}}^{-1}$
• D. $905.02{\text{kJ mol}}^{-1}$

Answer 1

The correct option is B $-895.02{\text{kJ mol}}^{-1}$

Here, $\Delta H=-900{\text{kJ mol}}^{-1}T=300\text{K}R=8.3{\text{J K}}^{-1}{\text{mol}}^{-1}\Delta U=?$

Formation of $C{F}_{2}ClC{F}_{2}Cl\left(g\right)$:
$2C\left(s\right)+2F_2\left(g\right)+C{l}_2\left(g\right)\to C{F}_{2}ClC{F}_{2}Cl\left(g\right)$
$\Delta n_g=1-3=-2\text{We know},\Delta U=\Delta H-\Delta n_{g}RT=\Delta H+2RT=-900{\text{kJ mol}}^{-1}+2\times 8.3\times 300=-900{\text{kJ mol}}^{-1}+4.98{\text{kJ mol}}^{-1}=-895.02{\text{kJ mol}}^{-1}$