Question

What will be the standard enthalpy change for the reaction $O{F}_2\left(g\right)+H_{2}O\left(g\right)\to O_2\left(g\right)+2HF\left(g\right)$ at 298K if enthalpies
of formation of $O{F}_2\left(g\right),H_{2}O\left(g\right)$ and $HF\left(g\right)$ are +20, -250 and -270kJ $mo{l}^{-1}$ respectively.

• A. + 310 kJ
• B. -310 kJ
• C. + 770 kJ
• D. - 770 kJ

Answer 1

The correct option is B -310 kJ

$O{F}_2\left(g\right)+H_{2}O\left(g\right)\to O_2\left(g\right)+2HF\left(g\right)\Delta H^{\circ }=\sum \Delta H_f^{\circ }\left(products\right)-\sum \Delta H_f^{\circ }\left(reactants\right)=\left[\Delta H_f^{\circ }{O}_2\right(g)+2\Delta H_f^{\circ }HF(g\left)\right]-\left[\Delta H_f^{\circ }O{F}_2\right(g)+\Delta H_f^{\circ }{H}_{2}O(g\left)\right]\text{Substituting the values from the available data}:\Delta H^{\circ }=[0+2(-270kJ\left)\right]-[20kJ-250kJ]=-540+230=-310kJ$