Question

At $298\, K$ what will be the change in standard internal energy change for the given reaction $OF _{2}( g )+ H _{2} O ( g ) \longrightarrow O _{2}( g )+ 2 HF ( g ) \Delta H =-310\, kJ$

• A. - 312.5 kJ
• B. - 125.03 kJ
• C. - 310 kJ
• D. - 156 kJ

Answer 1

Answer: (A)

Relation between internal energy change $\left(\Delta E ^{\circ}\right)$

and enthalpy change $\left(\Delta H ^{\circ}\right)$ of a reaction is given by

$\Delta H ^{\circ}=\Delta E ^{\circ}+\Delta n _{ g } RT$ ...(i)

$\Delta n_{g}=$ moles of product $-$ moles of reactant For the reaction given,

$OF _{2}( g )+ H _{2} O ( g ) \longrightarrow O _{2}( g )+2 HF ( g )$

$\Delta n _{ g }=(1+2)-( l + l )$

$\Delta n _{ g }= l \{\Delta H =-310\, kJ ,=-310000 J \}$

Putting the values in Eq (i)

$-310000\, J =\Delta E + l \left(8.3141\, JK ^{-1}\, mol ^{-1}\right)(298\, K )$

$\Delta E =-312,477.572\, J \approx-312.5\, kJ$