Question

Change in enthalpy for reaction : $2H_2{O}_2\left(\ell \right)$ $⟶$ $2H_{2}O\left(\ell \right)$ + $O_2$(g). If heat of formation of $H_2{O}_2\left(l\right)$ and $H_{2}O\left(l\right)$ are -188 and -286 kJ/mol respectively:

• A. - 196 kJ/mol
• B. + 196 kJ/mol
• C. + 948 kJ/mol
• D. - 948 kJ/mol

Answer 1

The correct option is A - 196 kJ/mol

$2H_{2}O\left(l\right)⟶2H_{2}O\left(l\right)+O_2\left(g\right)$

$△H_f\left[H_2{O}_2\right(l\left)\right]=-188kJ/mol$

For $2$ moles $H_2{O}_2=-2\times 188kJ/mol$

$△H_f\left[H_{2}O\right(l\left)\right]=-286kJ/mol$

For $2$ moles $H_{2}O=-2\times 286kJ/mol$

$△H_r=(2\times -286)-(2\times -188)$

$=-196kJ/mol$.